Math 75 Linear Algebra Class Notes

Transcription

1 Mth 75 Liner Algebr Clss Notes Prof. Erich Holtmnn For use with Elementry Liner Algebr, 7 th ed., Lrson Revised -Nov-5 p. i

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3 Contents Chpter : Systems of Liner Equtions. Introduction to Systems of Equtions.. Gussin Elimintion nd Guss-Jordn Elimintion. 9. Applictions of Systems of Liner Equtions. 5 Chpter : Mtrices. 9. Opertions with Mtrices. 9. Properties of Mtri Opertions.. The Inverse of Mtri. 7. Elementry Mtrices..5 Applictions of Mtri Opertions. 7 Chpter : Determinnts.. The Determinnt of Mtri.. Determinnts nd Elementry Opertions. 5. Properties of Determinnts. 5. Applictions of Determinnts. 55 Chpter : Vector Spces Comple Numbers (Optionl) Conjugtes nd Division of Comple Numbers (Optionl). 67. Vectors in R n. 7. Vector Spces. 77. Subspces of Vector Spces. 8. Spnning Sets nd Liner Independence Bsis nd Dimension Rnk of Mtri nd Systems of Liner Equtions. 5.7 Coordintes nd Chnge of Bsis. 7 Chpter 5: Inner Product Spces. 5. Length nd Dot Product in R n. 5. Inner Product Spces Orthogonl Bses: Grm-Schmidt Process Mthemticl Models nd Lest Squres Anlysis (Optionl) Applictions of Inner Product Spces (Optionl) Polr Form nd De Moivre's Theorem. (Optionl) Comple Vector Spces nd Inner Products. 6 p. i

4 Chpter 6: Liner Trnsformtions Introduction to Liner Trnsformtions The Kernel nd Rnge of Liner Trnsformtion Mtrices for Liner Trnsformtions Trnsition Mtrices nd Similrity Applictions of Liner Trnsformtions. 9 Chpter 7: Eigenvlues nd Eigenvectors. 7. Eigenvlues nd Eigenvectors. 7. Digonliztion Symmetric Mtrices nd Orthogonl Digonliztion Applictions of Eigenvlues nd Eigenvectors. 8.5 Unitry nd Hermitin Spces. p. ii

5 Chpter : Systems of Liner Equtions. Introduction to Systems of Equtions. Chpter : Systems of Liner Equtions. Introduction to Systems of Equtions. Objective: Recognize nd solve mn systems of liner equtions by hnd using Gussin elimintion nd bck-substitution n n = b is liner eqution in stndrd form in n vribles i. The first nonzero coefficient i is the leding coefficient. The constnt term is b. Compre to the fmilir forms of liner eqution s in two vribles y = m + b nd =. Emple: Liner nd Nonliner Equtions (sin ) = e sin + = Liner Nonliner An mn system of liner equtions is set of m liner equtions in n unknowns. Emple: Systems of Two Equtions in Two Vribles Solve nd grph ech system.. y = + y = 5 b. y = + y = c. y = + y = 5 For system of liner equtions, ectly one of the following is true. ) The system hs ectly one solution (consistent, nonsingulr system). ) The system hs infinitely mny solutions (consistent, singulr system). Use free prmeter or free vrible (or severl free prmeters) to represent the solution set. ) The system hs no solution (inconsistent, singulr system). p.

6 Chpter : Systems of Liner Equtions. Introduction to Systems of Equtions. To solve mn systems of liner equtions (when m nd n re lrge) we use procedure clled Gussin elimintion to find n equivlent system of equtions in row-echelon form. Then we use bck-substitution to solve for ech vrible. Row-echelon form mens tht the leding coefficients of (clled pivots ) nd the zero terms below them form stir-step pttern. You could wlk downstirs from the top left. You might hve to move more thn one column to the right to rech the net step, but you never hve to step down more thn one row t time. Row-echelon form Row-echelon form Not row-echelon form The gol of Gussin elimintion is to find n equivlent system tht is in row-echelon form. The three opertions you cn use during Gussin elimintion re ) Swp the order of two equtions. ) Multiply n eqution on both sides by non-zero constnt. ) Add multiple of one eqution to nother eqution. In Gussin elimintion, you strt with Eqution (the first eqution of your mn system). ) Find the leding coefficient in the current eqution. (Sometimes you need to swp equtions in this step.) ) Eliminte the coefficients of the corresponding vrible in ll of the equtions below the current eqution. ) Move down to the net eqution nd go bck to Step. Repet until you run out of equtions or you run out of vribles. Solve using bck-substitution: solve the lst eqution for the leding vrible, then substitute into the preceding (i.e. second-to-lst) eqution nd solve for its leding vrible, then substitute into the preceding eqution nd solve for its leding vrible, etc. Vribles tht re not leding vribles re free prmeters, nd we often set them equl to t, s,. p.

7 Chpter : Systems of Liner Equtions. Introduction to Systems of Equtions. Emples: Gussin Elimintion nd Bck-Substitution on Systems of Liner Equtions. p.

8 Chpter : Systems of Liner Equtions. Introduction to Systems of Equtions. p.

9 Chpter : Systems of Liner Equtions. Introduction to Systems of Equtions. p. 5

10 Chpter : Systems of Liner Equtions. Introduction to Systems of Equtions. p. 6

11 Chpter : Systems of Liner Equtions. Introduction to Systems of Equtions. Emple: Chemistry Appliction Write nd solve system of liner equtions for the chemicl rection ( )CH + ( )O ( )CO + ( )H O Solution: write seprte eqution for ech element, showing the blnce of tht element. C: + = + so + + = H: + = + so + + = O: + = + so + = p. 7

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13 Chpter : Systems of Liner Equtions. Gussin Elimintion nd Guss-Jordn Elimintion. p. 9. Gussin Elimintion nd Guss-Jordn Elimintion. Objective: Use mtrices nd Guss-Jordn elimintion to solve mn systems of liner equtions by hnd nd with softwre. Objective: Use mtrices nd Gussin elimintion with bck-substitution to solve mn systems of liner equtions by hnd nd with softwre. Use row-echelon form or reduced row-echelon form to determine the number of solutions of homogeneous system of liner equtions, nd (if pplicble) the number of free prmeters. A mtri is rectngulr rry of numbers, clled mtri entries, rrnged in horizontl rows nd verticl columns. Mtrices re denoted by cpitl letters; mtri entries re denoted by lowercse letters with two indices. In given mtri entry ij, the first inde i is the row, nd the second inde j is the column. The entries,,, compose the min digonl. If m = n then A is clled squre mtri. A liner system m n mn m m m n n n n b b b cn represented either by coefficient mtri A nd column vector b mn m m m n n n A nd b m b b b b or by n ugmented mtri M, which I will sometimes write s [A b] m mn m m m n n n b b b b M (The book, Mthemtic, nd the clcultor do not disply the dotted verticl line.) To crete M in Mthemtic from A nd b, type m=join[,b,] To crete M on the TI-89 from A nd b, type Mtriugment( A B M mn m m m n n n A

14 Chpter : Systems of Liner Equtions. Gussin Elimintion nd Guss-Jordn Elimintion. p. In similr mnner to tht used for n mn system of liner equtions, we cn use Gussin elimintion on the coefficient side A of the ugmented mtri [A b] to find n equivlent ugmented mtri [U c] in row-echelon form. Then we use bck-substitution to solve for ech vrible. U is clled n upper tringulr mtri becuse ll non-zero entries re on or bove the min digonl. row-echelon form row-echelon form not row-echelon form Emple: Use Gussin elimintion nd bck-substitution to solve. The three elementry row opertions you cn use during Gussin elimintion re ) Swp the two rows. ) Multiply row by non-zero constnt. ) Add multiple of one row to nother row. 8 6 z y z y z y 8 6 () ) ( so () so z y y z y z Steps:

15 Chpter : Systems of Liner Equtions. Gussin Elimintion nd Guss-Jordn Elimintion. p. Insted of using bck-substitution, we cn tke the row-echelon form [U c] nd eliminte the coefficients bove the pivots by dding multiples of the pivot rows. The result [R d] is clled reduced row-echelon form. reduced row-echelon form reduced row-echelon form Emple: Use Guss-Jordn elimintion to solve. 8 6 z y z y z y 8 6 z y

16 Chpter : Systems of Liner Equtions. Gussin Elimintion nd Guss-Jordn Elimintion. Emple Using softwre to find the Reduced Row-Echelon Form (Eercise. #9) Mthemtic Go to the Plettes Menu nd open the Bsic Mth Assistnt. Under Bsic Commnds, open the Mtri Tb. Type = nd click Use the Add Row nd Add Column buttons to epnd the mtri, so you cn enter the ugmented mtri (If you mke the mtri to lrge, use to remove rows nd columns.) Press (or on the number pd) when you re finished entering the ugmented mtri. Another wy to enter the ugmented mtri is to type (or downlod) ={{,-,,,6,6},{,-,,,,},{,,-,-,-,-},{,-,,5,5,},{,-,,,,}} MtriForm[] You cn downlod this mtri from Click on Electronic Dt Sets nd open 878_8.zip/DtSets/Mthemtic/9.nb (Ch., Section, Problem 9). Notice tht A nd re different vribles! User-defined vribles should lwys begin with lower-cse letter, becuse Mthemtic s built-in fuctions nd commnds begin with cpitl letters. (For emple, N nd C re lredy defined by Mthemtic.) On the Bsic Mth Assistnt plette, click on MtriForm RowReduce nd type so you hve MtriForm[RowReduce[]] Converting bck to system of liner equtions, we hve = = 5 = 5 = = p.

17 Chpter : Systems of Liner Equtions. Gussin Elimintion nd Guss-Jordn Elimintion. TI-89: Type Dt/Mtri Editor New... Type: Mtri Folder: min Vrible: A (A is bove. If you type = insted of, use. If lredy eists, then use Open... on the previous screen insted of New...) Use nd the rrow keys to type in the coefficient mtri (If you need to insert or delete row or column, use ) When you re finished, press. Another wy to enter the mtri is to type (from the home screen),,,,6,6,,,,,,,,,,,,,5,5,,,,,, A After the mtri is entered, type A Mtri rref( A Converting bck to system of liner equtions, we hve = = = = 5 5 = p.

18 Chpter : Systems of Liner Equtions. Gussin Elimintion nd Guss-Jordn Elimintion. Mthemtic cn help you perform row opertions. If your ugmented mtri is in, then [[]] is row of the mtri. To view in the usul formt, type MtriForm[] To swp rows nd of, type [[{,}]]=[[{,}]] To multiply row of by 7, type [[]]=7*[[]] To dd 7 times row to row, type [[]]=[[]]+7*[[]] Mthemtic performs opertions in the order tht you type them in, not s they pper on the screen. If you go bck nd edit your work, you cn use Evlute Notebook under the Evlution menu to reclculte the notebook in the order on the screen. The TI-89 cn lso help you perform row opertions. If your ugmented mtri is in, then To swp rows nd of nd store the result in, type Mtri Row opsrowswp( A,,,) A To multiply row of by 7 nd store the result in, type Mtri Row opsmrow(7, A,) A To dd 7 times row of to row nd store the result in, type Mtri Row opsmrowadd(7, A,,) A About nottion: nd re mtrices, but is determinnt (Ch. ). Theorem.. Every homogeneous (constnts on right-hnd side re ll zeroes) system of liner equtions is consistent. The number of free prmeters in the solution set is the number of vribles minus the number of pivots (leding coefficients). If there re zero free prmeters, then there is ectly one solution. Emples: Solutions: p.

19 Chpter : Systems of Liner Equtions. Applictions of Systems of Liner Equtions. p. 5. Applictions of Systems of Liner Equtions. Objective: Set up nd solve system of equtions to fit polynomil function to set of dt points. Objective: Set up nd solve system of equtions to represent network. Polynomil Curve Fitting. Given m dt points (t, y ), (t, y ),, (t m, y m ). We wnt to find polynomil of degree m tht psses through these points. p(t) = c + c t + c t + + c m t m Notice tht y j = c + c (t j )+ c (t j ) + + c m (t j ) m This produces n mm liner system: ) ( ) ( ) ( m m m m m t t t t t t c m c c = y m y y tht you cn solve using Guss-Jordn elimintion. Emple.# y t

20 Chpter : Systems of Liner Equtions. Applictions of Systems of Liner Equtions. Emple.#9 t y Network Anlysis: write system of liner equtions using Kirchoff s Lws. ) Flow into ech node (lso clled verte) equls flow out. ) In n electricl network, the sum of the products IR (I = current nd R = resistnce) round ny closed pth of edges (lines) is equl to the totl voltge in the loop from the btteries. A resistor is represented by the symbol k = Resistnce is mesured in ohms (). A bttery is represented by the symbol from the short line ( ) to the long line (+), then the voltge is positive. Current is mesure in mps (A). ma =. A. I If the current flows through the bttery Flow into node is positive. Flow out of node is negtive. To write system of equtions, ) Pick direction (t rndom) for ech current I. ) For ech node, write n eqution for the current input nd output. ) For ech loop, write the V = IR eqution. p. 6

21 Chpter : Systems of Liner Equtions. Applictions of Systems of Liner Equtions. Emple:..#. Solve for the currents. Emple:..9. The figure shows trffic flow in vehicles/hour through network of streets. ) Solve for,,, nd. b) Find the trffic flow when =. c) Find the trffic flow when =. p. 7

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23 Chpter : Mtrices.. Opertions with Mtrices. Chpter : Mtrices.. Opertions with Mtrices. Objective: Determine whether or not two mtrices re equl. Objective: Add nd subtrct mtrices nd multiply mtri by sclr. Objective: Multiply two mtrices. Objective: Write solutions to system of liner equtions in column vector nottion. Objective: Prtition mtri nd perform block multipliction. Three wys to represent the sme mtri re A = [ ij ] = m m n n mn. Two mtrices A = [ ij ] nd B = [b ij ] re equl if nd only if they hve the sme dimensions or order (mn) nd ij = b ij for ll i m nd j n. Comment on logic: An emple of P if Q (lso written P Q) is It is cloudy if it is rining. An emple of Q only if P (i.e. if Q then P, lso written Q P) is It is rining only if it is cloudy. R iff S is shorthnd for R if nd only if S (lso written R S). R iff S mens tht R is equivlent to S. To dd (or subtrct) two mtrices A = [ ij ] nd B = [b ij ] tht hve the sme dimensions, dd (or subtrct) corresponding mtri entries. A + B = [ ij + b ij ] A B = [ ij b ij ] The sum (nd difference) of two mtrices with different dimensions is undefined. To multiply mtri by sclr (number), multiply ech entry by tht sclr. Emples: ca = [c ij ] Let A = 5, B =, nd D =. Find A + B, A B, A + D, nd A. p. 9

24 Chpter : Mtrices.. Opertions with Mtrices. In Mthemtic, use +b, b, nd * or to dd mtrices, subtrct mtrices, nd multiply 5 sclrs c by mtrices. Remember, you cn enter D = from the Plettes Menu::Bsic Mth Assistnt::Bsic Commnds::Mtri Tb. On the TI-89, use b, b, nd c or to dd mtrices A nd B, subtrct mtrices, 5 nd multiply sclrs c by mtrices. Remember, you cn enter D = from the home screen s [,5;-,] D or from Dt/Mtri Editor. (Be sure to use Type: Mtri) Mtri multipliction is defined in much more comple mnner. For system, we cn write = b. We wnt definition of mtri multipliction tht llows us to write n mn system b m m m n n n mn n b n b m s A = b, n where A is the coefficient mtri n A, m m mn b nd nd b re column mtrices (or column vectors): nd b b. n b m Observe tht ech row of A ws multiplied by the column to give the corresponding row of b. If A = [ ij ] is n mn mtri nd B = [b ij ] is n np mtri, then the product AB is n mp mtri A = [c ij ] where [c ij ] = b ik kj n k = i b j + i b j + i b j + in b nj If the column dimension of A does not mtch the row dimension of B, then the product is undefined p.

25 Chpter : Mtrices.. Opertions with Mtrices. Emples 5 5 In Mthemtic, use.b to multiply mtrices.(do not use *b) On the TI-89, use b multiply mtrices. Emple: Write solutions to system of liner equtions in column vector nottion. Solve = = t Block multipliction on prtitioned mtrices works whenever the dimensions re OK. Emples A A A with A A A A nd A A B B B with B B 5 6 B B 5 6 B B Then AB = A A B B A A B B A A B B A A B B AB B A B = = B AB AB p.

26 Chpter : Mtrices.. Opertions with Mtrices. p. A= mn m m n n n = n mn m m n n n n = n mn n n m m In block mtri nottion, we write [ n ] = mn m m n n where the i re m column mtrices. The A = [ n ] n = n n Similrly, if B is n lm mtri (so BA is defined), then BA = B[ n ] = [ B B B n ], i.e. the columns of BA re B i. Notice tht B is lm, i is m, B i is l, nd BA is ln. On the other hnd, we could prtition A into m row mtrices r i A = mn m m n n = r m r r. Then A = mn m m n n n = r m r r = r r r m. Notice tht r i is n, is n, r i is (i.e. number), nd A is m. If E = lm l l m m e e e e e e e e e = e l e e then EA = e l e e A = A A A e l e e. Notice tht e i is m, A is mn, e i A is n, nd EA is ln If e = [ e e e m ], then ea = [ e e e m ] r m r r = e r + e r + + e m r m Notice tht e is m, A is mn, e i is number, r i is n, nd ea is n.

27 Chpter : Mtrices.. Properties of Mtri Opertions.. Properties of Mtri Opertions. Objective: Know nd use the properties of mtri opertions (mtri ddition nd subtrction, sclr multipliction, nd mtri multipliction), nd of the zero nd identity mtrices. Objective: Know which properties of fields do not hold for mtrices (commuttivity of mtri multipliction nd eistence of multiplictive inverses). Objective: Find the trnspose of mtri nd know properties of the trnspose. The rel numbers R, together with the opertions of ddition nd multipliction, is n emple of mthemticl field. (The comple numbers C with ddition nd multipliction is nother emple.) Fields nd their opertions hve ll of the usul properties. ) Closure under ddition: if nd b R, then + b R. ) Addition is ssocitive: ( + b) + c = + (b + c) ) Addition is commuttive: + b = b + ) Additive identity (zero): R contins, which hs the property tht + = for ll. 5) Additive inverses (opposites): every R. hs n opposite, such tht + ( ) =. We define subtrction b s + ( b). 6) Closure under multipliction: if nd b R, then b R. 7) Multipliction is ssocitive: (b)c = (bc) 8) Multipliction is commuttive: b = b 9) Multipliction distributes over ddition: (b + c) = b +c ) Multiplictive identity (one): R contins, which hs the property tht = for ll. ) Multiplictive inverses (reciprocls): every R. hs n inverse, such tht =. We define division b s b. Mtri ddition (nd subtrction) hs ll of the usul properties: closure, ssocitivity, commuttivity, zero mtrices, nd opposites. A zero mtri hs zero in ll entries, but becuse the mtri cn hve ny dimensions mn, we hve mny zero mtrices mn. The opposite of mtri [ ij ] is [ ij ] = [ ij ]. Multipliction of sclr by mtri hs ll of the usul properties: closure, ssocitivity, commuttivity, multiplictive identity (the sclr ), nd distribution. For distribution, we hve both c(a + B) = ca + cb nd (c + d)a = ca + da. Emples: ( ) = 7 9 = 7 9, = 6 5 = p.

28 Chpter : Mtrices.. Properties of Mtri Opertions. Multipliction of mtrices is closed, is ssocitive, nd distributes over mtri ddition. The multiplictive identity mtrices re squre mtrices with ones on the min digonl nd zeros everyplce else: I nn =. If A is mn, then I mm A mn = A mn nd A mn I nn = A mn We define eponents for squre mtrices A nd positive integers k: A k = define A = I. Multipliction of mtrices is not commuttive in generl. AA A. Also, we k times Mny mtrices do not hve multiplictive inverses. Division of mtrices is undefined. Emples: = 6 5 = = = = = Theorem.5: For system of liner equtions, ectly one of the following is true. ) The system hs no solution (inconsistent, singulr system). ) The system hs ectly one solution (consistent, nonsingulr system). ) The system hs infinitely mny solutions (consistent, singulr system). Use free prmeter or free vrible (or severl free prmeters) to represent the solution set. Proof using mtri opertions: Given system of liner equtions A = b. Ectly one of the following is true: the system hs no solution, the system hs ectly one solution, or the system hs t lest two solutions (cll them nd ). p.

29 Chpter : Mtrices.. Properties of Mtri Opertions. If the system hs two solutions, then A = b nd A = b so A( ) = A A = b b =. Let h =, so h is nonzero solution to the homogenous eqution A =. Then for ny sclr t, + t h is solution of A = b becuse A( + t h ) = A + ta h = b+ t = b. Thus, in the lst cse, the system hs infinitely mny solutions with prmeter t. The trnspose A T of mtri A is formed by writing its columns s rows. For emple, n m n m T if A n then A m. m m m mn n n n mn Equivlently, the trnspose of A is formed by writing rows s columns. The ij entry of A T is ji. 9 Emple: Find the trnspose of A = Mthemtic: to tke the trnspose of mtri, either type Trnspose[] (lso vilble on the Bsic Mth Assistnt plette) or type followed by tr (four seprte keystrokes). After the first three keystrokes, you will see tr. After the fourth keystroke, tr will chnge to T. Of course, t the end, type TI-89: A MtriT Properties of the trnspose: ) (A T ) T = A ) (A + B) T = A T +B T ) (ca) T =ca T ) (AB) T = B T A T * p. 5

30 Chpter : Mtrices.. Properties of Mtri Opertions. Emple of Property #: Consider A = nd B = The entry of (AB) T is the entry of AB = which is b + b + b If we look t A T nd B T, we much reverse the order of multipliction to obtin row column. B T A T = The entry B T A T is b + b + b. You cn see tht (AB) T = B T A T A mtri M is symmetric iff M T = M. 6 5 Emple: M = 6 is symmetric. Prove tht AA T is symmetric for ny mtri A. 5 5 Proof: p. 6

31 Chpter : Mtrices.. The Inverse of Mtri.. The Inverse of Mtri. Objective: Find the inverse of mtri (if it eists) by Guss-Jordn elimintion. Objective: Use properties of inverse mtrices. Objective: Use n inverse mtri to solve system of liner equtions. A squre nn mtri is invertible (or nonsingulr) when there eists n nn mtri A such tht AA = I nn nd A A = I nn where I nn is the identity mtri. A is clled the (multiplictive) inverse of A. A mtri tht does not hve n inverse is clled singulr (or noninvertible). Nonsqure mtrices do not hve inverses. Theorem.7: If A is n invertible mtri, then the inverse is unique. Proof: let B nd C be inverses of A. Then BA = I n AC = I. So B = BI = B(AC) = (BA)C = IC =C. Therefore, B = C nd the inverse of A is unique. Emple: Find the inverse of A = 5. 5 Solution: We need to solve the system AX = I, or 5 5 which gives four equtions =, To solve these four equtions, we tke the reduced row echelon forms nd, so A 5 = X = Since the row opertions performed to find the reduced row echelon form depend only on the 5 coefficient prt of the ugmented mtri, we could solve ll four equtions simultneously by using doubly ugmented mtri 5 5 [ A I ] = = [ I A ] To crete the doubly ugmented mtri in Mthemtic from A, type m=join[, IdentityMtri[],] To crete the doubly ugmented mtri on the TI-89 from A, type p. 7

32 Chpter : Mtrices.. The Inverse of Mtri. Mtri ugment( A, Mtri identity( )) M We used IdentityMtri[] nd identity() becuse we wnted mtri. To find the inverse of n nn mtri A by Guss-Jordn elimintion, find the reduced row echelon form of th nn ugmented mtri [ A I ]. If the nn block on the left cn be reduced to I, then the nn block on the right is A * [ A I ] [ I A ] If the nn block on the left cnnnot be reduced to I, then A is not invertible. Emple: Invert A = using Guss-Jordn elimintion. Emple: Invert A = using Guss-Jordn elimintion. Emple: Invert M = c b d using Guss-Jordn elimintion. You cn lso use softwre. First cler the vribles using Cler[,b,c,d] or Cler -z, then type Inverse[m] or M^ (lso on Bsic Mth Assistnt plette) p. 8

33 Chpter : Mtrices.. The Inverse of Mtri. Theorem.8 Properties of Inverse Mtrices If A is n invertible mtri, k is positive integer, nd c is nonzero sclr, then A, A k, ca, nd A T re invertible, nd ) (A ) = A ) (A k ) = (A ) k ) (ca) = c A ) (A T ) = (A ) T * Proof: ) ) Proof by Induction (see the Appendi) When k =, If (A k ) = (A ) k, ) By mthemticl induction, we conclude tht (A k ) = (A ) for ll positive integers k. ) p. 9

34 Chpter : Mtrices.. The Inverse of Mtri. Theorem.9 Inverse of Product * If A nd B re invertible nn mtrices, then (AB) = B A. Proof: Theorem. Cncelltion Properties Let C be n invertible mtri. * ) If AC = BC, then A = B. (Right cncelltion property) ) If CA = CB, then A = B. (Left cncelltion property) Theorem. Systems of Equtions with Unique Solutions. * If A is n invertible mtri, then the system A = b hs unique solution = A b. Review: Wht is wrong with the following proof tht if AB = I nd CA = I then B = C? AB CA A A B = C Wht is wrong with the following proof tht if AB = I nd CA = I nd A is invertible then B = C? AB = CA A AB = CAA IB = CI B = C p.

35 Chpter : Mtrices.. Elementry Mtrices. p.. Elementry Mtrices. Objective: Fctor mtri into product of elementry mtrices. Objective: Find the PA = LDU fctoriztion of mtri. An elementry mtri is squre mtri of dimensions nn (or of order n) is mtri tht cn be obtined from the nn identity mtri by single elementry row opertion. The three elementry row opertions, with emples of corresponding elementry mtrices, re ) Swpping two rows, e.g. R R E = ) Multiplying single row by nonzero constnt, e.g. R R E = ) Adding multiple of one row to nother row, e.g. R + R R E = Theorem. Representing Elementry Row Opertions If we premultiply (multiply on the left) mtri A by n elementry mtri, we obtin the sme result s if we hd pplied the corresponding elementry row opertion to A. Emples: = r r r r = R R = R R

36 Chpter : Mtrices.. Elementry Mtrices. p = R + R R Gussin elimintion cn be represented by product of elementry mtrices. For emple, A = R R E = R + R R E = R R E = 6 / 5 so 5 = E (E (E A)) = 6 / Two nn mtrices A nd B re row-equivlent when there eist finite number of elementry mtrices such tht B = E n E n E E A.

37 Chpter : Mtrices.. Elementry Mtrices. p. A squre mtri L is lower tringulr if ll entries bove the min digonl re zero, i.e. l ij = whenever i < j. A squre mtri U is upper tringulr if ll entries below the min digonl re zero, i.e. u ij = whenever i > j. A squre mtri D is digonl if ll entries not on the min digonl re zero, i.e. d ij = whenever i j. L = U = D = Theorem.+ Elementry Mtrices re Invertible If E is n elementry mtri, then E eists nd is n elementry mtri. Moreover, if E is lower tringulr, then E is lso lower tringulr. And if E is digonl, then E is lso digonl. Emples: E = R R E = R R E = R + R R E = R + R R E = 6 / 6 R R E = 6 6R R You cn check these using mtri multipliction. E.g. = nd =. Theorem. Invertible Mtrices re Row-Equivlent to the Identity Mtri A squre mtri A is invertible if nd only if it is row equivlent to the identity mtri: A = E n E E I = E n E E if nd only if A is product of elementry mtrices. Proof of If A is invertible, then A is product of elementry mtrices : Since A is invertible, A = b hs unique solution (nmely, = A b). But this mens tht we cn use row opertions to reduce [ A b ] to [ I c ] (where c = A b, of course). If the

38 Chpter : Mtrices.. Elementry Mtrices. p. corresponding elementry mtrices re E, E,, E n, then I = E n E n E E A so A = E E E n E n, which is product of elementry mtrices. Proof of If A is product of elementry mtrices, then A is invertible : If A = E E E n E n, then A eists nd A = E n E n E E becuse every elementry mtri is invertible.- Theorem.5 Equivlent Conditions for Invertibility If A is n n n mtri, then the following sttements re equivlent. ) A is invertible. ) A = b hs unique solution for every n column mtri b (nmely, = A b). ) A = hs only the trivil solution. ) A is row-equivlent to I nn. 5) A cn be written s product of elementry mtrices. Returning to our emple of Gussin elimintion using elementry mtrices, we found the reduced echelon form of the ugmented mtri = 6 / Look just t the coefficient mtri insted of the ugmented mtri. We hve U form row-echelon = mrows 6 / mrowadds swps row A 9 6 The row-echelon form is upper tringulr. In generl, we my hve more thn one mrow (multiply row by nonzero constnt) elementry mtri, more thn one mrowadd (dd multiple of one row to nother row) elementry mtri, nd more thn one row swp mtri. The product of nd rbitrry number of row swp mtrices is clled permuttion mtri P. U= mrows F F F n mrowadds E E E m PA

39 Chpter : Mtrices.. Elementry Mtrices. mrowadds mrows E E E m F F F n U= PA lower tringulr digonl Lemm The product of digonl mtrices is digonl. The product of lower tringulr mtrices is lower tringulr. LDU = PA E E Em L F F Fn / 6 D row-echelon form U = row swps P 6 9 A Theorem LU-Fctoriztion Every squre mtri cn be fctored s PA = LDU, where P is permuttion mtri, L is lower tringulr with ll ones on the min digonl, D is digonl, nd U is upper tringulr with ll ones on the min digonl. A vrition on this is PA = LU, where this L equls the LD from bove, nd does not necessrily hve ones on the digonl. The PA = LU fctoriztion is the usul method used by computers for solving systems of liner equtions, finding inverse mtrices, nd clculting determinnts (Chpter ). It is lso useful in proofs. p. 5

40 Chpter : Mtrices.. Elementry Mtrices. p. 6 * Emple: Find the PA = LDU fctoriztion of A = 6 Solution: A = 6 R R 6 R R R R R R R R = U U = F E E P A 6 E E F U = P A 6 L D U = P A 6

41 Chpter : Mtrices..5 Applictions of Mtri Opertions..5 Applictions of Mtri Opertions. Objective: Write nd use stochstic (Mrkov) mtri. Objective: Use mtri multipliction to encode nd decode messges. Objective: Use mtri lgebr to nlyze nd economic system (Leontief input-output model). Consider sitution in which members of popultion occupy finite number of sttes {S, S,, S n }. For emple, multintionl compny hs $ trillion in ssets (the popultion). Some of the money is in the Americs, some in Asi, nd the rest is in Europe (the three sttes). In Mrkov process, t ech discrete step in time, members of the popultion my move from one stte to nother, subject to the following rules: ) The totl number of individuls stys the sme. ) The numbers in ech stte never become negtive. ) The new stte depends only on the current stte (history is disregrded). The behvior of Mrkov process is described by mtri of trnsition probbilities (or stochstic mtri or Mrkov mtri). p ij is the probbility tht member of the popultion will chnge from the j th stte to the i th stte. The rules bove become ) Ech column of the trnsition mtri dds up to one. ) Every probbility entry is p ij. from S S Sn p p p n p p pn P pn pn pnn S S to S n Emple: Stochstic Mtri. A chemistry course is tught in two sections. Every week, of the students in Section A nd of the students in Section B drop, nd 6 of ech section trnsfer to the other section. Write the trnsition mtri. At the beginning of the semester, ech section hs students. Find the number of students in ech section nd the number of students who hve dropped fter one week nd fter two weeks. Solution: 7 A 6 B * (A) P = (B) 6 6 =. 6 (d) 8 (A) 8 79 (A) P = 96 (B). P 96 = 66 (B) 8 (d) 8 (d) 6 drop p. 7

42 Chpter : Mtrices..5 Applictions of Mtri Opertions. Mtri multipliction cn be used to encode nd decode messges. The encoded messges re clled cryptogrms. To begin, ssign number to ech letter of the lphbet (nd ssign to spce) _ A B C D E F G H I J K L M N O P Q R S T U V W X Y Z Use these numbers to convert messge in to row mtri, including spces but ignoring punctution. Then prtition the row mtri into uncoded row mtrices. Emple: M A K E _ I T _ S O [ ] [5 9] [ 9] [5 ] Emple: Use the invertible mtri A = Solution: 7 6 to encode the messge MAKE IT SO. *-- [ ] 7 6 = [ 56 6] [ 9] 7 6 = [ 99 7] [5 9] 7 6 = [5 9 ] [5 ] 7 6 = [5 6 5] The sequence of encoded mtrices is [ 56 6] [5 9 ] [ 99 7] [5 6 5] Removing the brckets yields the cryptogrm In order to decrypt messge, we need to know the encryption mtri A. p. 8

43 Chpter : Mtrices..5 Applictions of Mtri Opertions. Emple: Use the invertible mtri A = 7 6 to decode Solution: A = * [ 8 6 ] = [ 5 ] [9 97 ] 6 7 = [9 ] 7 [ 7 5] = [ ] [6 6 6] 6 7 = [6 ] [ 5 ] [ ] [9 ] [6 ] B E A M _ U S _ U P In economics, n input-output model (developed by Leontief) consists of n different industries I n, ech of which needs inputs (e.g. steel, food, lbor, ) nd hs nd output. To produce unit (e.g. $ million) of output, n industry my use the outputs of other industries nd of itself. For emple, production of steel my use steel, food, nd lbor. User (Output) I I I n d d d d d d D d n d n d n n nn I I I n Supplier (Input) Let d ij be the mount of output the industry j needs from industry i to produce one unit of output per yer. (We ssume the d ij re constnt, i.e. fied prices.) The mtri of these coefficients is clled the input-output mtri or consumption mtri D. A column represents ll of the inputs to given industry. For this model to work, d ij nd the sum of the entries in ech column must be less thn or equl to. (Otherwise, it costs more thn one unit to produce unit in tht industry.) Let i be the totl output mtri of industry i, nd X = [ i ]. If the economic system is closed (selfsustining: totl output = intermedite demnd, i.e. wht is needed to produce it), then X = DX. If the system is open with eternl demnd mtri E (e.g. eports) then X = DX + E. To find wht output mtri is needed to produce given eternl demnd mtri, we solve p. 9

44 Chpter : Mtrices..5 Applictions of Mtri Opertions. X = DX + E X DX = E (I D)X = E X = (I D) E Emple: Input-Output Economic Model Production of one unit of steel requires. units of steel, no food, nd.5 units of lbor. Production of one unit of food requires no steel,. units of food, nd.7 units of lbor. Production of one unit of lbor requires. units of steel,.8 units of food, nd. units of *- - lbor. Find the output mtri when the eternl demnds re units of steel, units of food, nd no lbor. Solution: D = E =.. X = We need 88 units of steel, 76 units of food, nd 86 units of lbor. p.

45 Chpter : Determinnts.. The Determinnt of Mtri. Chpter : Determinnts.. The Determinnt of Mtri. Objective: Find the determinnt of mtri. Objective: Find the minors nd cofctors of mtri. Objective: Use epnsion by cofctors to find the determinnt of mtri. Objective: Find the determinnt of tringulr mtri. Every squre mtri cn be ssocited with sclr clled its determinnt. Historiclly, determinnts were recognized s pttern of nn systems of liner equtions. The system b b hs the solution b b nd b b. The determinnt of mtri A = is det(a) = A =. The symbols men determinnt, not bsolute vlue. The determinnt of mtri A = is det(a) = A = =. Geometriclly, the signed re of prllelogrm with vertices t (, ), (, y ), (, y ), nd ( +, y + y ) is y A = y (, y ) (The re is positive if the ngle from (, y ) to (, y ) is counterclockwise; otherwise, the re is negtive.) (, y ) Proof: The re A of the prllelogrm is A = re of lrge rectngle res of four tringles res of two smll rectngles = ( + )(y + y ) y y y y y = y + y + y + y y y y y = y y = y p.

46 Chpter : Determinnts.. The Determinnt of Mtri. To define the determinnt of squre mtri A of order (dimensions) higher thn, we define minors nd cofctors. The minor M ij of the entry ij is the determinnt of the mtri obtined by deleting row i nd column j of A. The cofctor C ij of the entry ij is C ij = ( ) i+j M ij. Notice tht ( ) i+j is checkerbord pttern: ( ) i+j = Emples: Finding Cofctors. Let A = Find C. Solution: C = Find C. Solution: C = + = + = The determinnt of n nn mtri A (n ) is the sum of the entries in the first row of A multiplied by their respective cofctors. n det(a) = C j j = C + C + + n C n j Emple: = = ( 7 5) +(5 ) + ( + ) = ( 7) +() +(5) = 6 Theorem. Epnsion by Cofctors Let A be squre mtri of order n. Then the determinnt of A is given by n epnsion in ny row i det(a) = C ij ij n j = i C i + i C i + + in C in nd lso by n epnsion in ny column j det(a) = C ij ij n i = j C j + j C j + + nj C nj p.

47 Chpter : Determinnts.. The Determinnt of Mtri. When epnding, you don t need to find the cofctors of zero entries, becuse ij C ij = ()C ij =. The definition of the determinnt is inductive, becuse it uses the determinnt of mtri of order n to define the determinnt of mtri of order n. Emple: Epnding by Cofctors to Find Determinnt = = + ( ) = ( +7 ) +( ) = (() + 7( 6)) + ( )() = (6 ) = 6 To find determinnt using Mthemtic, type Det[] (lso on Bsic Mth Assistnt, More drop-down menu) To find determinnt on the TI-89, type Mtri det( A To find determinnt of mtri, you cn lso use the following shortcut. Copy Columns nd into Columns nd 5. To clculte the determinnt, dd nd subtrct the indicted products. subtrct Emple: 5 = dd 5 6 so = = p.

48 optionl Chpter : Determinnts.. The Determinnt of Mtri. Theorem. Determinnt of Tringulr Mtri The determinnt of tringulr mtri A of order n is the product of the entries on the min digonl. det(a) = nn Proof by Induction for upper tringulr mtrices: When k =, A = [ ] so A = Assume tht the theorem holds for ll upper tringulr mtrices of order k. Let A be n upper tringulr mtri of order k +. Then epnding in the lst row, k, k k k, k k A = = k+,k+ kk k, k kk k, k = k+,k+ ( kk ) = kk k+,k+ The proof for lower tringulr mtrices is similr. Optionl ppliction to multivrible clculus: Remember integrtion by substitution: f ( u) du = du f ( u( )) d d cos( ) For emple, cot( ) d = d sin( ) du cos( ) du Let u = sin(), = cos() so d d = sin( ) d u d u u = du u = ln u + C = ln sin() + C In multivrible clculus, V f u, v, w) dudvdw ( = V u v f ( u(, y, z), v(, y, z), w(, y, z)) w u y v y w y u z v ddydz z w z The determinnt is clled the Jcobin. p.

49 Chpter : Determinnts.. Determinnts nd Elementry Opertions.. Determinnts nd Elementry Opertions. Objective: Use elementry row opertions to evlute determinnt. Objective: Use elementry column opertions to evlute determinnt. Recognize conditions tht yield zero determinnts. In prctice, we rrely evlute determinnts using epnsion by cofctors. The properties of determinnts under elementry opertions provide much quicker wy to evlute determinnts. Theorem.9 det(a T ) = det(a). [Proof is in Section.] Theorem. Elementry Row (Column) Opertions nd Determinnts. Let A nd B be nn squre mtrices. ) When B is obtined from A by swpping two rows (two columns) of A, det(b) = det(a). b) When B is obtined from A by dding multiple of one row of A to nother row of A (or one column of A to nother column of A), det(b) = det(a). c) When B is obtined from A by multiplying of row (column) of A by nonzero constnt c, det(b) = c det(a). Theorem. Conditions tht Yield Zero Determinnt. If A is n nn squre mtri nd ny one of the following conditions is true, then det(a) = ) An entire row (or n entire column) consists of zeros. b) Two rows (or two columns) re equl. c) One row is multiple of nother row (or one column is multiple of nother column). Proof by Induction of. (for rows): When k =, A = nd B = so det(b) = = ( ) = det(a) Assume tht the theorem holds for ll mtrices of order k. Let A be mtri of order k + nd B be mtri obtined by swpping two rows of A. To find det(a) nd det(b), epnd in ny row other thn the swpped rows. The respective cofctors re opposites, becuse they come from kk mtrices tht hve two rows swpped. Thus, det(b) = det(a). Proof of. (for rows): Suppose tht row i of A is ll zeroes. Epnd by cofctors in row i. det(a) = C ij ij n j n = C ij = j p. 5

50 Chpter : Determinnts.. Determinnts nd Elementry Opertions. Proof of.b: Let B be the mtri obtined from A by swpping the two identicl rows (columns) of A, so det(b) = det(a). But B = A, so det(a) = det(a) so det(a) =. Proof of.b (for rows): Suppose B is obtined from A by dding c times row k to row i. Epnd by cofctors in row i. Note tht the cofctors of C ij re the sme for mtrices A nd B, becuse the mtrices re the sme everywhere ecept row i. det(b) = b C ij ij n j becuse C kj ij n j See Theorem.b. n = ( ckj ij ) Cij = C kj ij j n c j n + C ij ij j = c + det(a) = det(a). is the determinnt of mtri with two identicl rows (row k nd row i). Another wy of writing this is k k n i kn n = c ( c ) ( c kn nn in ) k k n n kn kn nn + k i n n kn in nn = c + det(a) = det(a). Proof of.c (for rows): Suppose B is obtined from A by multiplying row i by nonzero sclr c. Epnd by cofctors in row i. det(b) = b C ij ij n j n = c C ij ij j n = c j ij C ij = c det(a) Proof of.c: Suppose B is mtri with two equl rows (or two equl columns), nd A is obtined from B by multiplying one of those rows (or columns) by nonzero sclr c. Using.c on tht row (or column), det(a) = c det(b). Using.b, det(b) =. Thus, det(a) =. Geometriclly, the signed re of prllelogrm with edges from (, ) to (, y ) nd from (, ) to (, y ) hs the sme properties y s when you perform n elementry row opertion. Also, y the signed re of prllelepiped with edges from (, ) to (, y, z ), from (, ) to (, y, z ), nd from (, ) to (, y, z ) hs the y z sme properties s y z when you perform n elementry row opertion. y z p. 6

51 Chpter : Determinnts.. Determinnts nd Elementry Opertions. p. 7 Row Swpping (Theorem.) If we swp two rows, e.g. y y y y or z y z y z y z y z y z y then the sign of the re/volume chnges becuse we chnge from right-hnd orienttion to left-hnd orienttion. Adding multiple of one row to nother row (Theorem.b) If we dd multiple of one row to nother row, e.g. y y.5.5 y y y, then the A = bh is unchnged.

52 Chpter : Determinnts.. Determinnts nd Elementry Opertions. Adding multiple of one row to nother row (Theorem.b) If we multiply row by nonzero constnt c, e.g. (cb)h = ca is lso multiplies by the constnt c. y y c cy y, then the A = bh Emple: Finding Determinnt Using Elementry Row Opertions 5 = 5 R + R R = [( ) ( )] = 5 R R R = (6) = 6 = p. 8

53 Chpter : Determinnts.. Determinnts nd Elementry Opertions. Emple: Finding Determinnt Using Elementry Column Opertions = ( ) = ( ) = 7 9 C C C C C C C C C C C = ( )( ) = ( )( )( ) = p. 9

54

55 Chpter : Determinnts.. Properties of Determinnts.. Properties of Determinnts. Objective: Find the determinnt of mtri product nd of sclr multiple of mtri. Find the determinnt of n inverse mtri nd recognize equivlent conditions for nonsingulr mtri. Find the determinnt of the trnspose of mtri. Theorem.5 Determinnt of Mtri Product If A nd B re squre mtrices of the sme order, then det(ab) = det(a) det(b). Proof: To begin, let E be n elementry mtri. By Thm., EB is the mtri obtined from pplying the corresponding row opertion to B. By Thm.., det( B) echnging two rows det(eb) = det( B) if the row opertion is dding multiple of one row to nother c det( B) multiplying row by nonzero constnt c Also by Thm., echnging two rows det(e) = det(ei) = if the row opertion is dding multiple of one row to nother c multiplying row by nonzero constnt c Thus, det(eb) = det(e) det(b). This cn be generlized by induction to conclude tht E k E E B = E k E E B where the E i re elementry mtrices. If A is nonsingulr, then by Thm.., it cn be written s the product A = E k E E so AB = A B. If A is singulr, then A is row-equivlent to mtri with n entire row of zeroes (for emple, the reduced row echelon form). From Thm., we know A =. Moreover, becuse A is singulr, it follows tht AB must be singulr. (Proof by contrdiction: if AB were nonsingulr, then A[B(AB) - ] = I would show tht A is not singulr, becuse A = B(AB) -.) Therefore, AB = = A B. Comment on Proof by Contrdiction: P implies Q is equivlent to not Q implies not P. Theorem.6 Determinnt of Sclr Multiple of Mtri If A is squre mtri of order n nd c is sclr, then det(ca) = c n det(a). Proof: Apply Property (c) of Thm.. to ech of the n rows of A to obtin n fctors of c. Theorem.7 Determinnt of n Invertible Mtri A squre mtri A is invertible (nonsingulr) if nd only if det(a). Proof: On the one hnd, if A is invertible, then AA = I, so. A A = I =. Therefore, A. On the other hnd, ssume det(a). Then use Guss-Jordn elimintion to find the reduced row-echelon form R. Since R is in reduced row-echelon form, it is either the identity mtri or p. 5

56 Chpter : Determinnts.. Properties of Determinnts. it must hve t lest one row of ll zeroes. The second cse is not possible: if R hd row of ll zeroes, then det(r) =, but then det(a) = (which contrdicts the ssumption). Therefore, A is row-equivlent to R = I, so A is invertible. Theorem.8 Determinnt of n Inverse Mtri If A is n invertible mtri, then det (A ) = det( A) Proof: AA = I, so. A A = I = nd A, so A = A. Equivlent Conditions for Nonsingulr nn Mtri (Summry) ) A is invertible. ) A = b hs unique solution for every n column mtri b. ) A = hs only the trivil solution for the n column mtri. ) A is row-equivlent to I. 5) A cn be written s product of elementry mtrices. 6) det(a). Theorem.9 Determinnt of the Trnspose of Mtri If A is squre mtri, then det(a T ) = det(a). Proof: Let A be squre mtri of order n. From Section., we know tht A cn be fctored s PA = LDU, where P is permuttion mtri, L is lower tringulr with ll ones on the min digonl, D is digonl, nd U is upper tringulr with ll ones on the min digonl. L is obtined from I by dding multiple of the rows contining the digonl ones to the rows below the digonl, so L = I = by Thm..b. Likewise, U is obtined from I by dding multiple of the rows contining the digonl ones to the rows bove the digonl, so U = I = by Thm..b. By Thm.., D = d d d nn i.e. the product of its digonl elements. P is product of elementry row-swp mtrices, ech of which hs determinnt. So P is the product of some number of s. P = if the number of row swps is even; P = if the number of row swps is odd. p. 5

57 Chpter : Determinnts.. Properties of Determinnts. p. 5 Let e =, e =,, e n = be n mtrices. Then P = T i T i T i n e e e where i, i,, i n is some permuttion of,,, n. Now P T = i n i i e e e, so PP T = n n n n n n i T i i T i i T i i T i i T i i T i i T i i T i i T i e e e e e e e e e e e e e e e e e e = = I, nd by Thm..5, det(p) det(p T ) = det(pp T ) = det(i) =. Then either det(p) = so det(p T ) =, or det(p) = so det(p T ) =. In both cses, det(p) = det(p T ). So we hve PA = LDU which gives us P A = L D U = () D () = D, so A = P D. Tking the trnspose, we hve A T P T = U T D T L T which gives us A T P T = U T D T L T. Now P T = P ; D T = D becuse D T = D since D is digonl; L T = becuse L T is upper tringulr with ll ones on the min digonl; nd U T = becuse U T is lower tringulr with ll ones on the min digonl. Thus, A T P T = U T D T L T becomes A T P T = () D (), so A T = T P D = P D = A.

58

59 Chpter : Determinnts.. Applictions of Determinnts.. Applictions of Determinnts. Objective: Find the djoint of mtri nd use it to find the inverse of mtri. Objective: Use Crmer s Rule to solve system of n liner equtions in n unknowns. Objective: Use determinnts to find re, volume, nd the equtions of lines nd plnes. Using the djoint of mtri to clculte the inverse is time-consuming nd inefficient. In prctice, Guss-Jordn elimintion is used for mtrices nd lrger. However, djoint is vocbulry you my be epected to know in future clsses. Recll from Section. tht the cofctor C ij of mtri A is ( ) i+j times the determinnt of the mtri obtined by deleting row i nd column j of A. The mtri of cofctors of A is C C Cn C C C n C C C n n nn. The djoint of A is the trnspose of mtri of cofctors: dj(a) = C C C n C C C n C C C n n nn Theorem. The Inverse of Mtri Given by Its Adjoint If A is n invertible mtri, then A = det( A) dj(a). optionl Proof: Consider A[dj(A)] = i n i n n n ki nn C C C n C C C n C C C j j jn C C C n n nn The ij entry of this product is i C j + i C j + + in C jn. If i = j, this is det(a) (epnded by cofctors in row i). If i j, this is the determinnt of the mtri B, which is the sme s A ecept tht row j hs been replced with row i. p. 55

60 Chpter : Determinnts.. Applictions of Determinnts. n n n n C C C j Cn A = i i ki, B[dj(A)] = i i ki C C C j Cn j j jn i i in C n Cn C jn Cnn n n nn n n nn The j column of the cofctor mtri is unchnged, becuse it does not depend on the j row of A or B. Since two rows of B re the sme, the cofctor epnsion for i j is zero. det( A) Thus, A[dj(A)] = det( A) = det(a)i. det( A) det(a) becuse A is invertible, so we cn write A[ det( A) dj(a)] = I, so A = det( A) dj(a). For mtrices nd lrger, Guss-Jordn elimintion is much more efficient thn the djoint method for finding the inverse of mtri. However, for mtri A = c b d b d, dj(a) = c so A = d bc c b d. Crmer s Rule to solve n liner eqution in n vribles is time-consuming nd inefficient. In prctice, Gussin elimintion is used to solve liner systems. However, Crmer s Rule is vocbulry you my be epected to know in future clsses. Theorem. Crmer s Rule If n nn system A = b hs coefficient mtri with nonzero determinnt A, then det( A ) det( A, ) det( A ), det( A) det( A) n n det( A) where A i is the mtri A but with column i replce by b. p. 56

61 Chpter : Determinnts.. Applictions of Determinnts. C C Cn b Proof: = A b = det( A) dj(a)b = C C C n b det( A) n Cn Cn Cnn b n so i = det( A ) (b C i + b C i + + b n C ni ). The sum in prentheses is the cofctor epnsion of det( A ) det(a i ), so i = i det( A) Emple: Solve y y 5 using Crmer s Rule. Solution: = 5 = = 5; y = 5 = 5 5 = Are, Volume nd Equtions of Lines nd Plnes: We lredy know tht the signed re of prllelogrm y is given by determinnt (Section.). A = y The signed re of tringle with vertices (, y ), (, y ), nd (, y ) is A = y y y Proof: The re of tringle is hlf of the re of prllelogrm. So the re of the tringle we wnt is y pos y + y neg y + y = y neg For tringle the signed re is A = y y y y y y (, y ) (, y ) (, y ) If the vertices (, y ), (, y ), nd (, y ) re ordered clockwise then the re is positive; otherwise, it is negtive. (The homework sks for the bsolute vlue of the re.) p. 57

62 Chpter : Determinnts.. Applictions of Determinnts. p. 58 The re of the tringle is zero if nd only if the three points re colliner. (, y ), (, y ), nd (, y ) re colliner if nd only if y y y = The eqution of line through distinct points (, y ) nd (, y ) is y y y = Similrly to the two-dimensionl cse of prllelogrm, the signed volume of prllelepiped is given by determinnt (Section.). V = z y z y z y Let s find the volume of the tetrhedron (pyrmid with four tringulr fces) with vertices t (, y, z ), (, y, z ), (, y, z ), nd (, y, z ). The volume of the tetrhedron with vertices t (,, ), (, y, z ), (, y, z ), nd (, y, z ) is 6 times the volume of the prllelepiped, i.e. 6 z y z y z y. For the tringle bove ( y y y ), we hd sides nd the res of tringles to dd/subtrct. Now we hve fces nd the volumes of tetrhedrons to dd/subtrct. The signed volume of the tetrhedron with vertices t (,, ), (, y, z ), (, y, z ), nd (, y, z ) is (imgine for the moment tht (, y, z ) in front of the tringle (, y, z ), (, y, z ), (, y, z ) V = 6 z y z y z y + 6 z y z y z y + 6 z y z y z y + 6 z y z y z y For tetrhedron, the signed volume is V = 6 z y z y z y z y

63 Chpter : Determinnts.. Applictions of Determinnts. If when you wrp the fingers of your right from (, y, z ) to (, y, z ) to (, y, z ), your thumb points towrd (, y, z ), then the signed volume is positive. If when you wrp the fingers of your left from (, y, z ) to (, y, z ) to (, y, z ), your thumb points towrd (, y, z ), then the signed volume is negtive. (The homework sks for the bsolute vlue of the volume.) Four points (, y, z ), (, y, z ), (, y, z ), nd (, y, z ) re coplnr if nd only if y z y y y z z z = becuse tht is when the tetrhedron hs zero volume. The eqution of plne through distinct points (, y, z ), (, y, z ), nd (, y, z ) is y z y y y z z z = p. 59

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65 Chpter : Vector Spces. 8. Comple Numbers (Optionl). Chpter : Vector Spces. 8. Comple Numbers (Optionl). Objective: Use the Qudrtic Formul to find ll zeroes of qudrtic polynomil with rel coefficients. Objective: Add nd subtrct two comple numbers nd multiply comple number by rel sclr. Objective: Grphiclly represent comple numbers in the comple plne s directed line segments. Objective: Multiply two comple numbers. Objective: Multiply nd find determinnts of mtrices with comple entries. Objective: Perform Gussin nd Guss-Jordn elimintion on mtrices with comple entries. So fr, the sclrs we hve been using hve been rel numbers R. However, ny mthemticl field cn be used s the sclrs. Properties of Field F Let, b, nd c be ny elements of F. Then F is field if it hs two opertions, ddition nd multipliction, nd it contins two distinct elements nd, such tht the following properties re true. ) + b is n element of F. Closure under ddition ) + b = b + Commuttive property of ddition ) ( + b) + c = + (b + c) Associtive property of ddition ) + = Additive identity property 5) There eists such tht + ( ) = Additive inverse property 6) b is n element of F. Closure under multipliction 7) b = b Commuttive property of multipliction 8) (b)c = (bc) Associtive property of multipliction 9) = Multiplictive identity property ) Ecept for =, there eists n such tht ( ) = Multiplictive inverse property ) (b + c) = b + c Distributive property Three fmilir emples of fields re the rtionl numbers Q =, the rel numbers R, nd the comple numbers C. A fmilir set tht is not field is the integers Z = {,,,,,, }. Why not? We re going to need to use the field of comple numbers s our sclrs, becuse we will need to solve polynomil equtions. All polynomils cn be solved using comple numbers (comple numbers re n lgebriclly closed field ), but the sme is not true of rel numbers. p. 6

66 Chpter : Vector Spces. 8. Comple Numbers (Optionl). The imginry unit is i, so i =. def (Electricl engineers often write j becuse they use i for electric current.) Emple: Solve = Aside: the comple roots of polynomil with rel coefficients re comple conjugtes (Section 8.) of ech other ( + bi nd bi). Notice tht we hve ssumed definition of multipliction by rel number: (6 + 8i) = ( 6) + ( 8)i Emple: Solving the polynomil eqution = using softwre. Solution: Mthemtic: inputting Cler[];Solve[^+^+5+75==,] yields output You cn lso find Solve in the Plettes Menu::Bsic Mth Assistnt::y = menu. TI-89: ComplecSolve(^+^+5+75=,) A comple number is number of the form + bi, where nd b re rel numbers. is the rel prt nd bi is the imginry prt of + bi. The form + bi is the stndrd form of comple number, for emple + i, + i, nd + i. Geometriclly, comple number + bi is represented in the comple plne by directed line segment from the origin to (, b), where nd b re Crtesin coordintes. In other words, the horizontl is is the rel is nd the verticl is is the imginry is. Opertions in the Set of Comple Numbers C Addition: ( + bi) + (c + di) def ( + c) + (b + d)i Multipliction by rel number: c( + bi) def c + cbi Negtive: ( + bi) def + bi. Notice tht ( + bi) = ( )( + bi). Subtrction: ( + bi) + (c + di) def ( + bi) + (c + di) = ( c) + (b d)i p. 6

67 Chpter : Vector Spces. 8. Comple Numbers (Optionl). Emples Let z = + i nd w = + i. Illustrte the following grphiclly. z, w, z + w,.5z, w, w z p. 6

68 Chpter : Vector Spces. 8. Comple Numbers (Optionl). Multipliction of comple numbers is defined using the distributive property nd using i =. ( + bi)(c + di) def (c bd) + (d + bc)i becuse ( + bi)(c + di) = c + di + bci + bdi = c + di + bci + bdi = c + di + bci bd Wrning: When you multiply squre roots of negtive numbers, convert into stndrd form + bi (not using the squre root of negtive number) before you multiply. For emple, = i i =, not = ( )( ) = =. Appliction (Electricl Engineering): V(t)= I(t)Z, I(t)= current = I cos(t) + i I sin(t) V(t) = voltge, Z = impednce = R + + i L, ic R = resistnce, C = cpcitnce, L = inductnce Emple: Use softwre to check tht i is zero of the polynomil Solution: Mthemtic: to input the imginry unit i, denoted by i in Mthemtic, type ii (four seprte keystrokes). After the first three keystrokes, you will see ii. After the fourth keystroke, ii will chnge to i. Type =/5+ii/5 5^-6+5 TI-89: (/5+ /5) 5^-6+5 Comple mtrices Emple: Find the inverse of A = i i 5 by hnd. Check your nswer by multiplying. Solution: A 5 i = det( A). i det(a) = 5() ( + i)( i) = 5 ( + i i + 6) = A 5 i = i p. 6

69 Chpter : Vector Spces. 8. Comple Numbers (Optionl). p. 65 Check: AA = 5 i i 5 i i = i i i i i i i i = = Emple: By hnd, find the determinnt of Emple: Perform Guss-Jordn elimintion using row opertions to solve i z i y i i w i z i y i iw z iy i w ) 7 (6 ) 6 ( ) (6 ) 5 ( ) ( ) (

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71 9 Chpter 8 Comple Vector Spces 8. Comple Numbers Use the imginry unit i to write comple numbers. Grphiclly represent comple numbers in the comple plne s points nd s vectors. Add nd subtrct two comple numbers, nd multiply comple number by rel sclr. Multiply two comple numbers, nd use the Qudrtic Formul to find ll zeros of qudrtic polynomil. Perform opertions with comple mtrices, nd find the determinnt of comple mtri. REMARK When working with products involving squre roots of negtive numbers, be sure to convert to multiple of i before multiplying. For instnce, consider the following opertions. i i i Correct Incorrect COMPLEX NUMBERS So fr in the tet, the sclr quntities used hve been rel numbers. In this chpter, you will epnd the set of sclrs to include comple numbers. In lgebr it is often necessry to solve qudrtic equtions such s. The generl qudrtic eqution is b c, nd its solutions re given by the Qudrtic Formul b b c where the quntity under the rdicl, b c, is clled the discriminnt. If b c, then the solutions re ordinry rel numbers. But wht cn you conclude bout the solutions of qudrtic eqution whose discriminnt is negtive? For instnce, the eqution hs discriminnt of b c 6, but there is no rel number whose squre is 6. To overcome this deficiency, mthemticins invented the imginry unit i, defined s i where i. In terms of this imginry unit, 6 i. With this single ddition of the imginry unit i to the rel number system, the system of comple numbers cn be developed. Definition of Comple Number If nd b re rel numbers, then the number bi is comple number, where is the rel prt nd bi is the imginry prt of the number. The form bi is the stndrd form of comple number. Some emples of comple numbers written in stndrd form re i, i, nd 6i 6i. The set of rel numbers is subset of the set of comple numbers. To see this, note tht every rel number cn be written s comple number using b. Tht is, for every rel number, i. A comple number is uniquely determined by its rel nd imginry prts. So, two comple numbers re equl if nd only if their rel nd imginry prts re equl. Tht is, if bi nd c di re two comple numbers written in stndrd form, then bi c di if nd only if c nd b d.

72 Imginry is b (, b) or + bi Rel is THE COMPLEX PLANE 8. Comple Numbers 9 Becuse comple number is uniquely determined by its rel nd imginry prts, it is nturl to ssocite the number bi with the ordered pir, b. With this ssocition, comple numbers cn be represented grphiclly s points in coordinte plne clled the comple plne. This plne is n dpttion of the rectngulr coordinte plne. Specificlly, the horizontl is is the rel is nd the verticl is is the imginry is. The point tht corresponds to the comple number bi is, b, s shown in Figure 8.. The Comple Plne Figure 8. Plotting Numbers in the Comple Plne Plot ech number in the comple plne.. i b. i c. i d. 5 SOLUTION Figure 8. shows the numbers plotted in the comple plne.. Imginry b. is + i or (, ) Rel is Imginry is i or (, ) Rel is c. Imginry d. is i or (, ) Rel is Imginry is 5 or (5, ) Rel is 5 Figure 8. Another wy to represent the comple number bi is s vector whose horizontl component is nd whose verticl component is b. (See Figure 8..) (Note tht the use of the letter i to represent the imginry unit is unrelted to the use of i to represent unit vector.) Imginry is Horizontl component Rel is Verticl component i Vector Representtion of Comple Number Figure 8.

73 9 Chpter 8 Comple Vector Spces ADDITION, SUBTRACTION, AND SCALAR MULTIPLICATION OF COMPLEX NUMBERS Becuse comple number consists of rel prt dded to multiple of i, the opertions of ddition nd multipliction re defined in mnner consistent with the rules for operting with rel numbers. For instnce, to dd (or subtrct) two comple numbers, dd (or subtrct) the rel nd imginry prts seprtely. REMARK Note in prt () of Emple tht the sum of two comple numbers cn be rel number. Definition of Addition nd Subtrction of Comple Numbers The sum nd difference of bi nd c di re defined s follows. bi c di c b di Sum bi c di c b di Difference Adding nd Subtrcting Comple Numbers. b. i i i 5 i i i i Using the vector representtion of comple numbers, you cn dd or subtrct two comple numbers geometriclly using the prllelogrm rule for vector ddition, s shown in Figure 8.. Imginry is z + w = 5 z = + i 5 6 Rel is Imginry is w = + i Rel is w = i Addition of Comple Numbers Figure 8. Mny of the properties of ddition of rel numbers re vlid for comple numbers s well. For instnce, ddition of comple numbers is both ssocitive nd commuttive. Moreover, to find the sum of three or more comple numbers, etend the definition of ddition in the nturl wy. For emple, i i i i i. z = i z w = i Subtrction of Comple Numbers

74 8. Comple Numbers 95 Another property of rel numbers tht is vlid for comple numbers is the distributive property of sclr multipliction over ddition. To multiply comple number by rel sclr, use the definition below. Definition of Sclr Multipliction If c is rel number nd bi is comple number, then the sclr multiple of c nd bi is defined s c bi c cbi. Sclr Multipliction with Comple Numbers. b. 7i 8 i 6 i i 8 7i i i i i 6 i i 6i Geometriclly, multipliction of comple number by rel sclr corresponds to the multipliction of vector by sclr, s shown in Figure 8.5. Imginry is Imginry is z = + i z = 6 + i 5 6 Rel is z = i z = + i Rel is Multipliction of Comple Number by Rel Number Figure 8.5 With ddition nd sclr multipliction, the set of comple numbers forms vector spce of dimension (where the sclrs re the rel numbers). You re sked to verify this in Eercise 55. LINEAR ALGEBRA APPLIED Comple numbers hve some useful pplictions in electronics. The stte of circuit element is described by two quntities: the voltge V cross it nd the current I flowing through it. To simplify computtions, the circuit element s stte cn be described by single comple number z V li, of which the voltge nd current re simply the rel nd imginry prts. A similr nottion cn be used to epress the circuit element s cpcitnce nd inductnce. When certin elements of circuit re chnging with time, electricl engineers often hve to solve differentil equtions. These cn often be simpler to solve using comple numbers becuse the equtions re less complicted. Adrio Communictions Ltd/shutterstock.com

75 96 Chpter 8 Comple Vector Spces MULTIPLICATION OF COMPLEX NUMBERS The opertions of ddition, subtrction, nd sclr multipliction of comple numbers hve ect counterprts with the corresponding vector opertions. By contrst, there is no direct vector counterprt for the multipliction of two comple numbers. Definition of Multipliction of Comple Numbers The product of the comple numbers bi nd c di is defined s bic di c bd d bci. TECHNOLOGY Mny grphing utilities nd softwre progrms cn clculte with comple numbers. For emple, on some grphing utilities, you cn epress comple number bi s n ordered pir, b. Try verifying the result of Emple (b) by multiplying, nd,. You should obtin the ordered pir,. Rther thn try to memorize this definition of the product of two comple numbers, simply pply the distributive property, s follows. bic di c di bic di c di bci bdi c di bci bd c bd di bci c bd d bci. i 6i b. i i 8 6i i i i Multiplying Comple Numbers 8 6i i 8 6i i Distributive property Distributive property Use i. Commuttive property Distributive property Comple Zeros of Polynomil REMARK A well-known result from lgebr sttes tht the comple zeros of polynomil with rel coefficients must occur in conjugte pirs. (See Review Eercise 8.) Use the Qudrtic Formul to find the zeros of the polynomil p 6 nd verify tht p for ech zero. SOLUTION Using the Qudrtic Formul, b b c 6 6 Substitute ech vlue of into the polynomil p to verify tht p. p i i 6 i i i 6 i 9 6i 6i 8 i p i i 6 i i i 6 i 9 6i 6i 8 i In Emple 5, the two comple numbers i nd i re comple conjugtes of ech other (together they form conjugte pir). More will be sid bout comple conjugtes in Section i i.

76 COMPLEX MATRICES 8. Comple Numbers 97 Now tht you re ble to dd, subtrct, nd multiply comple numbers, you cn pply these opertions to mtrices whose entries re comple numbers. Such mtri is clled comple. Definition of Comple Mtri A mtri whose entries re comple numbers is clled comple mtri. All of the ordinry opertions with mtrices lso work with comple mtrices, s demonstrted in the net two emples. Opertions with Comple Mtrices Let A nd B be the comple mtrices nd B i A i i i i nd determine ech of the following.. A b. ib c. A B d. BA SOLUTION. A i i b. c. d. ib i i i A B i i BA i i i i 6 7 i i i i i i i i i 9i i 6 9i i i i i i i i i i i i i 8i i 5 i Finding the Determinnt of Comple Mtri TECHNOLOGY Mny grphing utilities nd softwre progrms cn perform mtri opertions on comple mtrices. Try verifying the clcultion of the determinnt of the mtri in Emple 7. You should obtin the sme nswer, 8, 6. Find the determinnt of the mtri A i SOLUTION 5 i. deta i 5 i i5 i i 6i 6 8 6i

77 98 Chpter 8 Comple Vector Spces 8. Eercises Simplifying n Epression In Eercises 6, determine the vlue of the epression i 5. i 6. i 7 Equlity of Comple Numbers In Eercises 7, determine such tht the comple numbers in ech pir re equl. 7. i, 6 i 8. 8 i, i 9. 6 i, 5 6i. i, i Plotting Comple Numbers In Eercises 6, plot the number in the comple plne.. z 6 i. z i. z 5 5i. z 7 5. z 5i 6. z 5i Adding nd Subtrcting Comple Numbers In Eercises 7, find the sum or difference of the comple numbers. Use vectors to illustrte your nswer. 7. 6i i 8. i i 9. 5 i 5 i. i i. 6 i. 7i i. i i. i i Sclr Multipliction In Eercises 5 nd 6, use vectors to illustrte the opertions geometriclly. Be sure to grph the originl vector. 5. u nd u, where u i 6. u nd u, where u i Multiplying Comple Numbers In Eercises 7, find the product i i i7 i. i i i i. bi. bi bi. i. i i i Finding Zeros In Eercises 5, determine ll the zeros of the polynomil function. 5. p 5 6. p 7. p p 5 9. p 6. p 9 Finding Zeros In Eercises, use the given zero to find ll zeros of the polynomil function.. p Zero:. p 5 Zero:. p 5 75 Zero: 5i. p 9 9 Zero: i Opertions with Comple Mtrices In Eercises 5 5, perform the indicted mtri opertion using the comple mtrices A nd B. nd B i i A i i i i 5. A B 6. B A 7. A 8. B 9. ia 5. ib 5. deta B 5. detb 5. 5AB 5. BA 55. Proof Prove tht the set of comple numbers, with the opertions of ddition nd sclr multipliction (with rel sclrs), is vector spce of dimension. 56. Consider the functions p 6 nd q 6. () Without grphing either function, determine whether the grphs of p nd q hve -intercepts. Eplin your resoning. (b) For which of the given functions is i zero? Without using the Qudrtic Formul, find the other zero of this function nd verify your nswer. 57. () Evlute i n for n,,,, nd 5. (b) Clculte i. (c) Find generl formul for i n for ny positive integer n. 58. Let A i i. () Clculte A n for n,,,, nd 5. (b) Clculte A. (c) Find generl formul for A n for ny positive integer n. True or Flse? In Eercises 59 nd 6, determine whether ech sttement is true or flse. If sttement is true, give reson or cite n pproprite sttement from the tet. If sttement is flse, provide n emple tht shows the sttement is not true in ll cses or cite n pproprite sttement from the tet Proof Prove tht if the product of two comple numbers is zero, then t lest one of the numbers must be zero.

78 Answer Key Section Imginry. Imginry is is z = 5 + 5i 5 Rel 6 is z = 6 i Rel is 5 5. Imginry is 5 Rel 5 is 7. 5 i 9. i Imginry is i. i Imginry is z = + 5i u = + 6i u + v = 5 + i 6 8 v = i Rel is Imginry is 6 Imginry is u v = i u = 5 + i v = 5 i 8 Rel is 5. 5 i i 5i 5 i 55. Proof 57. () i i (b) i i i i i i 5 i (c) n k i n k i n, where k is n integer. n k i n k 59. Flse. See the Remrk, pge Proof 5. u = 6 v = i Imginry is u v = 6 + i Rel is u + v = + i v = + i u = + i Rel is u = + i Rel 6 is u = 6 i u = i 7. i b bi. i 5. ± i 7., 9. ±, ±., ± i i., ±5i 5. i i i i i i 6i i 6

79 Chpter : Vector Spces. 8. Conjugtes nd Division of Comple Numbers (Optionl). 8. Conjugtes nd Division of Comple Numbers (Optionl). Objective: Find the conjugte of comple number. Objective: Find the modulus of comple number. Objective: Divide comple numbers. Objective: Perform Gussin on nd find the inverses of mtrices with comple entries. The conjugte of the comple number z = + bi is denoted by z or z* nd is given by z* = bi Theorem 8. Properties of Comple Conjugtes For comple numbers z = + bi, ) zz* = zz = + b ) zz* = zz ) zz* = zz = if nd only if z = ) (z*)* = (z ) = z The modulus of the comple number z = + bi is denoted by z nd is given by z = b Theorem 8. The modulus of comple number. z = zz * = zz Emple: Find z* nd z if z = i. Solution: z* = + i z = ( ) = 7 The quotient of two comple numbers z = + bi nd w = c + di, w, is z z = w w w * zw * ( bi)( c di) = = w * w c d c bd ( bc d) i c d = 7i Emple: Find. i 7i ( 7i)( i) Solution: = i ( i)( i) = 8 6i 8i ( ) = i 5 5 p. 67

80 Chpter : Vector Spces. 8. Conjugtes nd Division of Comple Numbers (Optionl). Theorem 8. Properties of Comple Conjugtes For comple numbers z nd w (w ), ) (z + w)* = z* + w* i.e. z + w = z + w ) (z w)* = z* w* i.e. z w = z w ) (zw)* = z*w* i.e. zw = z w ) (z/w)* = z*/w* i.e. z/w = z /w Emple: Find the inverse of A = Solution 5 5i i i i by hnd. p. 68

81 Chpter : Vector Spces. 8. Conjugtes nd Division of Comple Numbers (Optionl). Emple: Perform Gussin elimintion using row opertions to solve ( i) ( 8 i) y i (5 5i) Solution: (9 i) z 5 i 7iy ( i) z 67 57i iy (5 6i) z 57 i p. 69

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83 8. Conjugtes nd Division of Comple Numbers Conjugtes nd Division of Comple Numbers Find the conjugte of comple number. Find the modulus of comple number. Divide comple numbers, nd find the inverse of comple mtri. COMPLEX CONJUGATES In Section 8., it ws mentioned tht the comple zeros of polynomil with rel coefficients occur in conjugte pirs. For instnce, in Emple 5 you sw tht the zeros of p 6 re i nd i. In this section, you will emine some dditionl properties of comple conjugtes. You will begin with the definition of the conjugte of comple number. Definition of the Conjugte of Comple Number The conjugte of the comple number z bi is denoted by z nd is given by z bi. REMARK In prt (d) of Emple, note tht 5 is its own comple conjugte. In generl, it cn be shown tht number is its own comple conjugte if nd only if the number is rel. (See Eercise 9.) z = + i Imginry is Comple Number. z i b. z 5i c. z i d. z 5 Finding the Conjugte of Comple Number Conjugte z i z 5i z i z 5 Geometriclly, two points in the comple plne re conjugtes if nd only if they re reflections in the rel (horizontl) is, s shown in Figure 8.6. Comple conjugtes hve mny useful properties. Some of these re shown in Theorem 8.. z = i Rel is THEOREM 8. Properties of Comple Conjugtes For comple number z bi, the following properties re true.. zz b. zz. zz if nd only if z.. z z Imginry is z = 5i Conjugte of Comple Number Figure 8.6 z = + 5i Rel is PROOF To prove the first property, let z bi. Then z bi nd zz bi bi bi bi b i b. The second nd third properties follow directly from the first. Finlly, the fourth property follows from the definition of the comple conjugte. Tht is, z bi bi bi z. Finding the Product of Comple Conjugtes When z i, you hve zz i i 5.

84 Chpter 8 Comple Vector Spces REMARK The modulus of comple number is lso clled the bsolute vlue of the number. In fct, when z is rel number, z. THE MODULUS OF A COMPLEX NUMBER Becuse comple number cn be represented by vector in the comple plne, it mkes sense to tlk bout the length of comple number. This length is clled the modulus of the comple number. Definition of the Modulus of Comple Number The modulus of the comple number z bi is denoted by nd is given by z b. z Finding the Modulus of Comple Number For z i nd w 6 i, determine the vlue of ech modulus.. b. c. z w zw SOLUTION. z b. w 6 7 c. Becuse zw i6 i 5 6i, you hve zw zw z w. Note tht in Emple, In Eercise, you re sked to prove tht this multiplictive property of the modulus lwys holds. Theorem 8. sttes tht the modulus of comple number is relted to its conjugte. THEOREM 8. The Modulus of Comple Number For comple number z, z zz. PROOF Let z bi, then z bi nd zz bi bi b z. LINEAR ALGEBRA APPLIED Frctls pper in lmost every prt of the universe. They hve been used to study wide vriety of pplictions such s bcteri cultures, the humn lungs, the economy, nd glies. The most fmous frctl is clled the Mndelbrot Set, nmed fter the Polish-born mthemticin Benoit Mndelbrot (9 ). The Mndelbrot Set is bsed on the following sequence of comple numbers. z n z n c, z c The behvior of this sequence depends on the vlue of the comple number c. For some vlues of c, the modulus of ech term z n in the sequence is less thn some fied number N, nd the sequence is bounded. This mens tht c is in the Mndelbrot Set, nd its point is colored blck. For other vlues of c, the moduli of the terms of the sequence become infinitely lrge, nd the sequence is unbounded. This mens tht c is not in the Mndelbrot Set, nd its point is ssigned color bsed on how quickly the sequence diverges. Andrew Prk/Shutterstock.com

85 8. Conjugtes nd Division of Comple Numbers DIVISION OF COMPLEX NUMBERS One of the most importnt uses of the conjugte of comple number is in performing division in the comple number system. To define division of comple numbers, consider z bi nd w c di nd ssume tht c nd d re not both. For the quotient z yi w to mke sense, it hs to be true tht z w yi c di yi c dy d cyi. But, becuse z bi, you cn form the liner system below. c dy d cy b Solving this system of liner equtions for nd y yields c bd ww nd y bc d ww. Now, becuse zw bic di c bd bc di, the following definition is obtined. REMARK If c d, then c d, nd w. In other words, s is the cse with rel numbers, division of comple numbers by zero is not defined. Definition of Division of Comple Numbers The quotient of the comple numbers z bi nd w c di is defined s z bi w c di c bd bc d c d c d i w zw provided c d. In prctice, the quotient of two comple numbers cn be found by multiplying the numertor nd the denomintor by the conjugte of the denomintor, s follows. bi bi c di c di c dic di bic di c dic di c bd bc di c d c bd bc d c d c d i. b. i i i i i i i i i i Division of Comple Numbers i i i i i i

86 Chpter 8 Comple Vector Spces Now tht you cn divide comple numbers, you cn find the (multiplictive) inverse of comple mtri, s demonstrted in Emple 5. Finding the Inverse of Comple Mtri TECHNOLOGY If your grphing utility or softwre progrm cn perform opertions with comple mtrices, then you cn verify the result of Emple 5. If you hve mtri A stored on grphing utility, evlute A. Find the inverse of the mtri A i i nd verify your solution by showing tht AA I. SOLUTION Using the formul for the inverse of mtri from Section., A Furthermore, becuse A A it follows tht i6 i 5 i i 6i i 5 6i 5i i A 6 i i i To verify your solution, multiply A nd A s follows. AA i i 5 i 6 i 6 i i 5 i i i 6 i i i 5 i i. i i 7 i 7 i. 5 6 i 5 i i i i 7 i 7 i The lst theorem in this section summrizes some useful properties of comple conjugtes. THEOREM 8. Properties of Comple Conjugtes For the comple numbers z nd w, the following properties re true.. z w z w. z w z w. zw z w. zw zw PROOF To prove the first property, let z bi nd w c di. Then z w c b di c b di bi c di z w. The proof of the second property is similr. The proofs of the other two properties re left to you.

87 8. Eercises 8. Eercises Finding the Conjugte In Eercises 6, find the comple conjugte z nd geometriclly represent both z nd z.. z 6 i. z 5i. z 8i. z i 5. z 6. z Finding the Modulus In Eercises 7, find the indicted modulus, where z i, w i, nd v 5i. 7. z 8. z 9. zw. wz. v. zv. Verify tht wz wz zw, where z i nd w i.. Verify tht zv zv zv, where z i nd v i. Dividing Comple Numbers In Eercises 5, perform the indicted opertions. 5. i 6. i 6 i 7. i 5 i 8. i i 9. i i i. i i5 i Opertions with Comple Rtionl Epressions In Eercises, perform the opertion nd write the result in stndrd form. i.. i 5 i i i i i i.. i i i i Finding Zeros In Eercises 5 8, use the given zero to find ll zeros of the polynomil function. 5. p 8 8 Zero: i 6. p Zero: i 7. p 5 Zero: i 8. p Zero: i Powers of Comple Numbers In Eercises 9 nd, find ech power of the comple number z. () z (b) z (c) z (d) z 9. z i. z i Finding the Inverse of Comple Mtri In Eercises 6, determine whether the comple mtri A hs n inverse. If A is invertible, find its inverse nd verify tht AA I. 6 i i i. A. A i i i.. A i A i i i A A i i i i Singulr Mtrices In Eercises 7 nd 8, determine ll vlues of the comple number z for which A is singulr. (Hint: Set deta nd solve for z. ) i i 7. A 5 z 8. A i i z i i 9. Proof Prove tht z z if nd only if z is rel.. Consider the quotient () Without performing ny clcultions, describe how to find the quotient. (b) Eplin why the process described in prt () results in comple number of the form bi. (c) Find the quotient. i 6 i.. Proof Prove tht for ny two comple numbers z nd w, ech of the sttements below is true. zw z w () (b) If w, then zw z w.. Grphicl Interprettion Describe the set of points in the comple plne tht stisfies ech of the sttements below. () z (b) z i 5 (c) z i (d) z 5. () Evlute i n for n,,,, nd 5. (b) Clculte i nd i. (c) Find generl formul for i n for ny positive integer n.. () Verify tht i i i. (b) Find the two squre roots of i. (c) Find ll zeros of the polynomil.

88 Answer Key Section i. 5. Imginry is Imginry is z = 6 + i 6 z = 6 i Rel is 8i 8 Imginry is 8 8 z = 8i z = 8i 8 z nd z = 5 Rel is wz i w z 5 zw i 5. i i i i i 5., ± i 7.,, ± i 9. () i (b) i (c) 5 5i (d). A i i i 6. Not invertible 5. 5 A 7. i i i i 9. Proof. () nd (b) Proofs. () i i i, i, i i, i, i 5 i (b) i, i i (c) i n i 5 5i n k n k, where k is n integer n k n k

89 Chpter : Vector Spces.. Vectors in Rn.. Vectors in R n. Objective: Represent vector in the plne s directed line segment. Objective: Perform bsic vector opertions in R nd represent them grphiclly. Objective: Perform bsic vector opertions in R n. Prove bsic properties bout vectors nd their opertions in R n. In physics nd engineering, vector is n object with mgnitude nd direction nd represented grphiclly by directed line segment. In mthemtics we hve much more generl definition of vector. Geometriclly, vector in the plne is represented by directed line segment with its initil point t the origin nd its terminl (finl) point t (, ). The sme ordered pir used to represent the terminl point is used to represent the vector. Tht is, = (, ). The coordintes nd re clled the components of the vector. Two vectors u = (u, u ) nd v = (v, v ) re equl iff u = v nd u = v. Vector opertions in R Vector Addition: u + v = (u, u ) + (v, v ) def (u + v, u + v ). Sclr Multipliction: cu = c(u, u ) def (cu, cu ). Negtive: u def ( u, u ). Notice tht u = ( )u. Subtrction: u v def u + ( v) = (u, u ) + ( v, v ) = (u v, u v ). The zero vector in R is = (, ). p. 7

90 Chpter : Vector Spces.. Vectors in Rn. Emples Let u = (, ) nd v = (, ). Illustrte the following grphiclly. u, v, u + v,.5u, v, v u p. 7

91 Chpter : Vector Spces.. Vectors in Rn. Theorem. Properties of Vector Addition nd Sclr Multipliction in the Plne (R ) Let u, v, nd w be vectors in R, nd let c nd d be sclrs. ) u + v is vector in R. Closure under ddition ) u + v = v + u Commuttive property of ddition ) (u + v) + w = u + (v + w) Associtive property of ddition ) u + = u Eistence of dditive identity 5) u + ( u) = Eistence of dditive inverses 6) cv is vector in R. Closure under sclr multipliction 7) c(u + v) = cu + cv Distributive property over vector ddition 8) (c + d)u = cu + du Distributive property over sclr ddition 9) c(du) = (cd)u Associtive property ) u = u Multiplictive identity property Proof of (): Associtive property of ddition (u + v) + w = [(u, u ) + (v, v )] + (w, w ) = = = Assoc. prop. of ddition of rel numbers = = = u + (v + w) p. 7

92 Chpter : Vector Spces.. Vectors in Rn. Proof of (8): Distributive property of sclr multipliction over rel number ddition (c + d)u = (c + d)(u, u ) = = Distributive property of rel numbers = = = cu + du To dd (, ) + (, ) in Mthemtic, type {,}+{,-} You cn lso ssign vrible by typing u={,} You cn perform sclr multipliction by u or *u To dd (, ) + (, ) on the TI-89, you cn type, +,- or + - You cn lso ssign vrible by typing, U You cn perform sclr multipliction by u or u Vector opertions in R n We cn generlize from the -dimensionl plne R to n n-spce R n of ordered n-tuples. For emple, R = R = set of ll rel numbers; R = -spce = set of ll ordered pirs of rel numbers; R = -spce = set of ll ordered triples of rel numbers.\ An n-tuple (,,, n ) cn be viewed s point in R n with the i s its coordintes, or s vector with the i s its components. The stndrd vector opertions in R n re Vector Addition: u + v = (u, u,, u n ) + (v, v,, v n ) def (u + v, u + v,, u n + v n ). Sclr Multipliction: cu = c(u, u,, u n ) def (cu, cu,, cu n ). Negtive: u def ( u, u,, u n ). Notice tht u = ( )u. Subtrction: u v def u + ( v) = (u, u,, u n ) + ( v, v,, v n ) = (u v, u v,, u n v n ). The zero vector in R n is = (,,, ). p. 7

93 Chpter : Vector Spces.. Vectors in Rn. Theorem. Properties of Vector Addition nd Sclr Multipliction in the Plne (R n ) Let u, v, nd w be vectors in R n, nd let c nd d be sclrs. ) u + v is vector in R n. Closure under ddition ) u + v = v + u Commuttive property of ddition ) (u + v) + w = u + (v + w) Associtive property of ddition ) u + = u Eistence of dditive identity 5) u + ( u) = Eistence of dditive inverses 6) cv is vector in R n. Closure under sclr multipliction 7) c(u + v) = cu + cv Distributive property over vector ddition 8) (c + d)u = cu + du Distributive property over sclr ddition 9) c(du) = (cd)u Associtive property ) u = u Multiplictive identity property The vector is clled the dditive identity in R n nd v is the dditive inverse of v. Theorem. Properties of Vector Addition nd Sclr Multipliction in R n Let v be vector in R n nd let c be sclr. Then ) The dditive identity is unique. Tht is, if v + u = v, then u =. ) The dditive inverse of v is unique. Tht is, if v + u =, then u = v. ) v = ) c = 5) If cv =, then c = or v =. 6) ( v) = v Proof of (): Uniqueness of the dditive identity v + u = v Given (v + u) + ( v) = v + ( v) Add v to both sides = v + ( v) = v + ( v) = u = p. 75

94 Chpter : Vector Spces.. Vectors in Rn. Proof of (): Uniqueness of the dditive inverse v + u = Given ( v) + (v + u) = ( v) + = ( v) + = ( v) + = ( v) + u = v p. 76

95 Chpter : Vector Spces.. Vector Spces.. Vector Spces. Objective: Define vector spce nd recognize some importnt emples of vector spces. Objective: Show tht given set is not vector spce. (Optionl) Theorem. listed ten properties of vector ddition nd sclr multipliction in R n. However, there re mny other sets (C n, sets of mtrices, polynomils, functions) besides R n tht cn be given suitble definitions of vector ddition nd sclr multipliction so tht they too stisfy the sme ten properties. Hence, one brnch of mthemtics, liner lgebr, cn study ll of these. Definition of Vector Spce Let V be set on which two opertions (vector ddition nd sclr multipliction) re defined. If the ioms listed below re stisfied for every u, v, nd w in V nd every sclr c nd d in given field F (usully, F = R or F = C), then V is clled vector spce over F. * ) u + v is in V. Closure under ddition ) u + v = v + u Commuttive property ) (u + v) + w = u + (v + w) Associtive property ) V hs zero vector such tht Eistence of dditive identity for every u in V, u + = u 5) For every u in V, there is vector Eistence of dditive inverses (opposites) denoted by u such tht u + ( u) = 6) cv is vector in V. Closure under sclr multipliction 7) c(u + v) = cu + cv Distributive property over vector ddition 8) (c + d)u = cu + du Distributive property over sclr ddition 9) c(du) = (cd)u Associtive property ) u = u Sclr identity Notice tht vector spce ctully consists of four entities: set V of vectors, field F of sclrs, nd two defined opertions (vector ddition nd sclr multipliction). Be sure ll four entities re clerly understood. (For emple, I could keep the set V of vectors, the field F of sclrs, the sme definition of sclr multipliction, but chnge the definition of how to dd vectors nd end up with different vector spce, or end up with something tht is no longer vector spce.) Emples of Vector Spces. (Unless otherwise stted, ssume the field is R.) R with the stndrd opertions * u + v = (u, u ) + (v, v ) def (u + v, u + v ). cu = c(u, u ) def (cu, cu ). = (, ) u = ( u, u ) R n with the stndrd opertions. Note tht this includes R, which is just R with the usul ddition nd multipliction. C n over the field C with the stndrd opertions p. 77

96 Chpter : Vector Spces.. Vector Spces. More Emples of Vector Spces. (Unless otherwise stted, ssume the field is R.) The vector spce M, of ll rel mtrices with the stndrd opertions A + B = nd ca = c b b b b b b + = b b b b b b c c c = c c c The vector spce M m,n of ll mn rel mtrices with the stndrd opertions. The vector spce P of ll polynomils of degree or less with the usul opertions. Let p() = + + nd q() = b + b + b. Define the usul opertions (p + q)() def p() + q() nd (cp)() def c[p()]. We cn verify closure under ddition: (p + q)() = p() + q() = b + b + b = ( + b ) + ( + b ) + ( + b ) which is polynomil of degree or less (less if + b = ). Notice tht we hve used the commuttive nd distributive properties of rel numbers. The other ioms cn be verified in similr mnner. Note tht () = + +. The vector spce P n of ll polynomils of degree n or less with the usul opertions. The vector spce P of ll polynomils with the usul opertions. The vector spce C(,) of continuous relvlued functions on the domin (,) For emple, 8 +,, sin(), nd e re vectors in this spce. Addition nd sclr multipliction re defined in the usul wy. (f + g)() def f() + g() nd (cf )() def c[f ()] f, g, nd f + g re vectors in C(,), just s u, v, nd u + v nd re vectors in R n. f () cn be thought of s component of f, just s u i is component of u. u hs n components: u, u,, u n. f hs n infinite number of components:, f ( ),, f ( ),, f (),, f ( ),. The dditive identity (zero function) is f () = (the -is), nd given f (), the dditive inverse of f is [ f ]() = [ f ()]. p. 78

97 Chpter : Vector Spces.. Vector Spces. Another Emple of Vector Spces. The vector spce C[, b] of continuous rel-vlued functions on the domin [, b] over the field R. The most importnt reson for defining n bstrct vector spce using the ten ioms bove is tht we cn mke generl sttements bout ll vector spces. I.e. the sme proof cn be used for R n nd for C[, b]. Theorem. Properties of Vector Addition nd Sclr Multipliction Let v be vector in V n nd let c be sclr. Then ) v = ) c = ) If cv =, then c = or v =. ) v = v Proof of (): c = c = Given c = c( + ) Additive identity c = c + ( c) c + ( c) = = c Additive inverse Proof of (): If cv =, then c = or c. If c, then cv = c Given c cv = c Multiply both sides by Multiplictive inverse in R p. 79

98 optionl Chpter : Vector Spces.. Vector Spces. c(c )v = c Commuttive property of multipliction v = c Multiplictive inverse in R v c Sclr identity v = Theorem.() just proved Thus, either c = or v =. Emples tht re not Vector Spces Z (ordered pir of integers) over the field R. Z is not closed under sclr multipliction, for emple (, ) = (, ) Z. Aside on nottion: Z mens is n element of (is member) of the set of integers. Z mens is not n element of the set of integers. Although Z stisfied Aioms 5 nd of vector spce, it is not vector spce becuse not ll ioms re stisfied..() V = The set of second-degree polynomils is not vector spce becuse it is not closed under ddition. For emple, let p() = nd q() = + +. Then p() + q() = + is first degree polynomil. Let V = R with the stndrd vector ddition but nonstndrd sclr multipliction defined by c(u, u ) = (cu, ). Show tht V is not vector spce. It turns out tht the only iom tht is not stisfied in this cse is () Sclr identity. For emple, (, ) = (, ) (, ). p. 8

99 optionl Chpter : Vector Spces.. Vector Spces. Another Emple tht is not Vector Spces Rottions in three dimensions represented s rrows using the right-hnd rule. The direction of the rrow represents the direction of the rottion, vi the right-hnd rule, while the length of the rrow represents the mgnitude of the direction in degrees. Sclr multipliction is the stndrd opertion (stretching the rrow, or reversing the direction if the sclr is negtive). Vector ddition (e.g. ) is the first rottion followed by the second. This is not vector spce becuse vector ddition is not commuttive.. (In Chpter 6, we will see tht rottions cn be represented not s vectors, but s mtrices.) p. 8

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101 Chpter : Vector Spces.. Subspces of Vector Spces.. Subspces of Vector Spces. Objective: Determine whether subset W of vector spce V is subspce of V. Objective: Determine subspces of R n. Mny vector spces re subspces of lrger spces..() A V nonempty = subset W of vector spce V is clled subspce of V when is vector spce under the opertions of vector ddition nd sclr multipliction defined in V. Theorem.5 Test for Subspce If W is nonempty subset of V, then W is subspce of V if nd only if the following.() conditions V = hold. Proof: ) W is not empty. ) If u nd v re in W, then u + v is in W. Closure under ddition ) If u is in W nd c is sclr, then cu is in W. Closure under sclr multipliction If W is subspce of V, then W is vector spce stisfying the closure ioms, so u + v is in W nd cu is in W. On the other hnd, ssume W is closed under vector ddition nd sclr multipliction. By ssumption, two ioms re stisfied. ) u + v is in W. Closure under ddition 6) cv is vector in W. Closure under sclr multipliction Then if u, v, nd w re in W then they re lso in V, so the following ioms re utomticlly stisfied. ) u + v = v + u Commuttive property ) (u + v) + w = u + (v + w) Associtive property 7) c(u + v) = cu + cv Distributive property over vector ddition 8) (c + d)u = cu + du Distributive property over sclr ddition 9) c(du) = (cd)u Associtive property ) u = u Sclr identity Becuse W is closed under sclr multipliction, we know tht for ny v in W, v nd ( )v re lso in W. From Thm.., we know tht v = nd ( )v = v so the remining ioms re lso stisfied. ) W contins the zero vector. Additive identity 5) For every v in W, W contins v. Additive inverse (opposite) p. 8

102 Chpter : Vector Spces.. Subspces of Vector Spces. Emples of Subspces nd Sets tht re not Subspces. Show tht the set W = {(v,, v ): v nd v re rel numbers} is subspce of R with the stndrd opertions. Grphiclly, W is the -z plne in R. W is nonempty, becuse it contins (,, ). W is closed under ddition becuse if u, v W, then u + v = (u,, u ) + (v,, v ) = (u + v,, u + v ) W. W is closed under sclr multipliction becuse if c is sclr nd v W, then cv = c(v,, v ) = (cv,, cv ) W. Is Z (ordered pir of integers) with the stndrd opertions subspce of R? Z is closed under ddition. But s we sw in., Z is not closed under sclr multipliction, for emple (, ) = (, ) Z. So Z is not subspce of R. Show tht the set W = {(v, v ): v = or v = } with the stndrd opertions is not subspce of R. * W is closed under sclr multipliction, but W is not closed under ddition. For emple, (, ) + (, ) = (, ) W. Is tht the set W = {(v, v ): v nd v } with the stndrd opertions subspce of R? W is closed under ddition, but W is not closed under sclr multipliction when c < nd v. For emple, ( )(, ) = (, ) W. In generl, subspce is stright (line) or flt (plne or higher-dimensionl object), is infinite in ll directions, hs no holes, nd contins the origin. Let W be the set of ll symmetric mtrices. Show tht W is subspce of M, with the stndrd opertions. A is symmetric mens tht A = A T. W is nonempty, becuse it contins. W is closed under ddition becuse if A, B W, then (A + B) T = A T + B T = A + B, so A + B is symmetric. W is closed under sclr multipliction becuse if c is sclr nd A W, then (ca) T = c(a T ) ca, so ca is symmetric. p. 8

103 Chpter : Vector Spces.. Subspces of Vector Spces. More Emples of Subspces nd Sets tht re not Subspces. Let W be the set of ll singulr mtrices of order. Show tht W is not subspce of M, with the stndrd opertions. nd re singulr mtrices (det = ), but + = is not singulr, so W is not closed under ddition nd is therefore not subspce. (Notice tht b W = { : d bc = }, nd d bc = is not liner eqution.) c d Let W be the unit circle in R, i. e. W = {(v, v ): v + v = }, with the stndrd opertions. Is W subspce of R No. W is not closed under ddition: (, ) + (, ) = (, ) W. Another reson is tht W is not closed under sclr multipliction: (, ) = (, ) W. Is W = {(, )} with the stndrd opertions subspce of R Yes, the only ddition to check is + = W. Sclr multipliction is lso esy to check: c = W. Is W = {} nd W = V re clled the trivil subspces of c. Which of these two subsets is subspce of R with the stndrd opertions? ) U = {(v, v, v ) R : v + v + v = } This is the eqution of plne through the points (6,, ), (,, ), nd (,, ). U is not closed under ddition, becuse (6,, ) + (,, ) = (6,, ), but (6) + () + () =. Moreover, U is not closed under sclr multipliction, becuse (6,, ) = (,, ), but () + () + () =. Either one of these resons is enough to show tht U is not subspce. b) W= {(v, v, v ) R : v v + v = } This is the eqution of plne prllel to U, but W contins. If v nd w W, then v + w = (v, v, v ) + (w, w, w ) = (v + w, v + w, v + w ), nd (v + w ) (v + w ) + (v + w ) = v + w v w + v + w = (v v + v ) + (w w + w ) = + =, so v + w W. Also, if c is sclr nd v W, then cv = (cv, cv, cv ) nd (cv ) (cv ) + (cv ) = c(v v + v ) = c =, so cv W. Therefore, W is subspce. p. 8

104 Chpter : Vector Spces.. Subspces of Vector Spces. More Emple of Subspces. P[, ] is the set of ll rel-vlued polynomil funtions on the domin [, ]. C [, ] is the set of ll rel-vlued, continuously differentible functions on the domin [, ]. C [, ], lso written s C[, ], is the set of ll rel-vlued, continuous functions on the domin [, ]. Let W be the set of ll rel-vlued, integrble functions on the intervl [, ]. Let V be the set of ll rel-vlued functions on the intervl [, ]. If we tke the usul definitions of vector ddition nd sclr multipliction, nd we use R s the field of sclrs, then V is vector spce, nd the other four sets re subspces of V. In fct, P[, ] is subspce of C [, ], which is is subspce of C [, ], which is is subspce of W, which is is subspce of V. Theorem.6 The Intersection of Two Subspces is Subspce. If V nd W re both subspces of vector spce U, then the interction of V nd W (denoted V W) is lso subspce of U. (Note: V W is the set of ll vectors tht re in both V W.) Proof: Becuse V nd W re both subspces of U, we know tht they both.() V = contin, so V W contins nd is not empty. To show tht V W is closed under ddition, let u nd u be two vectors in V W. Becuse u nd u re both in V, nd being subspce V is closed, u + u is lso in V. Likewise, u + u is lso in W. Since u + u is in both V nd W, u + u is in V W, so V W is closed under vector ddition. A similr rgument shows tht V W is closed under sclr multipliction, so V W is subspce of U. V W V W U p. 85

105 Chpter : Vector Spces.. Spnning Sets nd Liner Independence.. Spnning Sets nd Liner Independence. Objective: Write vector s liner combintion of other vectors in vector spce V. Objective: Determine whether spnning set S of vectors in vector spce is spnning set of V. Objective: Determine whether set of vectors in vector spce V is linerly independent. Objective: Prove results bout spnning sets nd liner independence. A vector v in vector spce V is clled liner combintion of the vectors u, u,, u k in V if nd only if v cn be written in the form v = c u + c u + + c k u k where c, c,, c k re sclrs. Emple: Write the vector v = (,, ) s liner combintion of the vectors in the set S = {(,, ), (5,, )} (if possible). Solution: We need to solve v = c u + c u for the sclrs c i. Substituting in the given vectors, we hve (,, ) = c (,, ) + c (5,, ) (,, ) = (c, c, c ) + (5c, c, c ) (,, ) = (c + 5c, c + c, c + c ) c 5c 6 which gives the system c c or c c c c Solve this system by finding the reduced row-echelon form (using softwre) 6 or c so c = nd c =. c Answer: (,, ) = (,, ) (5,, ). Emple: Write the vectors v = (, 5, 8) nd w = (,, ) s liner combintions of the vectors in the set S = {(,, 7), (,, 5), (,, )} (if possible). Solution: We need to solve p. 86

106 Chpter : Vector Spces.. Spnning Sets nd Liner Independence.. Emple: Write the vector v = 57 8 s liner combintions of the vectors in the set S = {, 5, 6, } (if possible). 5 Solution: p. 87

107 Chpter : Vector Spces.. Spnning Sets nd Liner Independence. Let S = {v, v,, v k } be subset of vector spce V. Then S is clled spnning set of V if nd only if every vector in V cn be written s liner combintion of vector in S. In such cses, we sy tht S spns V. Emples of spnning sets: The set {(,, ), (,, ), (,, )} spns R becuse ny vector v = (v, v, v ) in R cn be written s v = v (,, ) + v (,, ) + v (,, ). The set {,, } spns P becuse ny vector (polynomil) p() = + b + c in P cn be written s p() = c() + b() + ( ). Emple: Determine whether the set S = {(5, 7, 6), (,, ), (,, )} spns R. R consists of ll the vectors of the form (v, v, v ), where v, v, nd v re rel numbers. S spns R if we cn lwys solve for the sclrs c, c, nd c in the eqution (v, v, v ) = c (5, 7, 6) + c (,, ) + c (,, ) (v, v, v ) = (5c, 7c, 6c ) + (c, c, c ) + (c, c, c ) (v, v, v ) = (5c + c + c, 7c + c c, 6c c + c ) This vector eqution is equivlent to the system 5 or the mtri eqution 7 6 c c c v v v. v v v 5c c 7c c 6c c c c c p. 88

108 Chpter : Vector Spces.. Spnning Sets nd Liner Independence. 5 This eqution cn lwys be solved for c, c, nd c, becuse 7, so the mtri is 6 invertible. (To be precise, the determinnt is.) Therefore, S spns R. Emple: Determine whether the set S = {(5, 7, 6), (,, ), (,, )} spns R. only this number hs chnged. Solution: Geometriclly, the vectors in S ll lie in the sme plne ( -dimensionl object), while the vectors in S do not. We need -dimensionl spce to contin S. S S. p. 89

109 Chpter : Vector Spces.. Spnning Sets nd Liner Independence. If S = {v, v,, v k } is set of vectors in vector spce V, then the spn of S is the set of ll liner combintions of the vectors in S. spn(s) = {c v + c v + + c k v k : c, c, c k, re sclrs} We sometimes write spn{v, v,, v k } insted of spn(s). Another nottion for spn(s) which we will void becuse it will be confusing lter is v, v,, v k. Theorem.7 Spn(S) is Subspce of V If S = {v, v,, v k } is set of vectors in vector * spce V, then spn(s) is subspce of V. Moreover, spn(s) is the smllest subspce of V tht contins (S), in the sense tht every subspce tht contins S must lso contin spn(s). Proof First, we wnt to show tht spn(s) is subspce of V. So we need to show tht Let c be sclr nd let u nd w be ny vectors in spn(s). Then Net, we wnt to show tht every subspce tht contins S must lso contin spn(s). This is Lb Problem..55. p. 9

110 Chpter : Vector Spces.. Spnning Sets nd Liner Independence. Sometimes, one vector cn be written in terms of other vectors. For emple, in S = {(5, 7, 6), ( 5,7,6) (,,) (,, ), (,, )} from bove, (,, ) =. We could sy tht (,, ) is dependent upon (5, 7, 6) nd (,, ). But we could just s esily solve for (5, 7, 6); there is no good reson to tret (,, ) s specil. A more equitble eqution is (5, 7, 6) (,, ) + (,, ) = (,, ). In fct, there re n infinite number of solutions to c (5, 7, 6) + c (,, ) + c (,, ) =. c = t, c = t, c = t. (This includes c = c = c =.) On the other hnd, for S = {(5, 7, 6), (,, ), (,, )}, the only solution to is c c c 5 c (5, 7, 6) + c (,, ) + c (,, ) = (,, ) or This is clled the trivil solution. c c c If S = {v, v,, v k } is set of vectors in vector spce V, then S is clled linerly independent if nd only if the eqution c v + c v + + c k v k = hs only the trivil solution c = c = = c k =. S is clled linerly dependent if nd only if there re lso nontrivil solutions. Emples #-: Let w = (7,, 5), w = (,, 6), w = (9,, 7), w = (, 5, ), nd w 5 = ( 6, 5, 8). Emple #: Is {w, w } linerly independent? Does {w, w } spn R? Solution: To decide liner independence, we wnt to solve c w + c w =. To decide spnning R, we wnt to solve c w + c w = (v, v, v ) for rbitrry v, v, nd v. Both equtions look like c (7,, 5) + c (,, 6) = (v, v, v ). (For liner independence, the right-hnd side is v = v = v =.) p. 9

111 Chpter : Vector Spces.. Spnning Sets nd Liner Independence. So we hve (7c, c, 5c ) + (c, c, 6c ) = (v, v, v ) (7c + c, c + c, 5c + 6c ) = (v, v, v ) 7 5 c 6 c v v v is not squre, so we cnnot tke the determinnt s we did before. Insted, find the reduced row-echelon form: 7 5 v v 6 v (Don t worry bout reproducing this result; I ll eplin soon why Mthemtic nd the TI-89 give different nswer.) Writing this s system of equtions, we hve When the right-hnd side is v = v = v =, we hve only the trivil solution c = c =, becuse ech vrible c i corresponds to pivot. Therefore, {w, w } is linerly independent. On the other hnd, there re mny choices of v, v, nd v 7 5 for which the lst eqution = v 8 v + 8 v cn be solved. This hppens whenever the (reduced) rowechelon form hs row of ll zeroes. Therefore,{w, w } does not spn R. Notice tht to nswer these questions we didn t need to py ttention to the coefficients on the right-hnd side of the line in the ugmented mtri. All tht we needed to know ws the coefficient mtri on the left-hnd side of the line. 7 v (Mthemtic nd the TI-89 both give v becuse they ssume 5 6 v 7 5 you cn divide Row by v 8 v + 8 v, nd they do not know tht we intend the lst p. 9

112 Chpter : Vector Spces.. Spnning Sets nd Liner Independence. column to be on the right-hnd side of the eqution. However, the right-hnd column is not importnt to deciding liner independence nd spnning.) Emple #: Is {w, w, w } linerly independent? Does {w, w, w } spn R? Solution: To decide liner independence, we wnt to solve c w + c w + c w =. To decide spnning R, we wnt to solve c w + c w + c w = v for rbitrry v. c v Using block multipliction nottion, we write [ w w w ] c = v c v 7 9 c v so c = v c v 7 9 is singulr (not invertible) becuse its determinnt (found using softwre) is zero. This tells us tht sometimes we cnnot solve c w + c w + c w = v, so {w, w, w } does not spn R. This lso tells us tht c w + c w + c w = hs n infinite number of solutions, so {w, w, w } is linerly dependent. 7 9 For nother perspective, the reduced row-echelon form of is To investigte liner independence, we set the right-hnd side equl to zero. The third column in the mtri, which doesn t hve pivot, gives us free prmeter. (Fewer pivots thn vribles.) c c c c t c c c hs the solutions c t where t is free prmeter. c c c c t Since we hve non-trivil solutions, {w, w, w } is linerly dependent. To investigte spnning, we set the right-hnd side equl to rbitrry numbers. The third row of ll zeroes in the mtri gives us c + c + c = #, which does not hve solution when the right-hnd side is not zero. (Fewer pivots thn equtions.) Therefore, {w, w, w } does not spn R. Emple #: Is {w, w, w } linerly independent? Does {w, w, w } spn R? Solution: To decide liner independence, we wnt to solve c w + c w + c w =. To decide spnning R, we wnt to solve c w + c w + c w = v for rbitrry v. p. 9

113 Chpter : Vector Spces.. Spnning Sets nd Liner Independence. p. 9 Using block multipliction nottion, we write [ w w w ] c c c = v v v so c c c = v v v is invertible (non-singulr) becuse its determinnt (found using softwre) is 59. This tells us tht we cn solve c w + c w + c w = v, so {w, w, w } spns R. This lso tells us tht c w + c w + c w = hs unique of solution, so {w, w, w } is linerly independent. Emple #: Is {w, w, w, w 5 } linerly independent? Does {w, w, w, w 5 } spn R? Solution: To decide liner independence, we wnt to solve c w + c w + c w + c 5 w 5 =. To decide spnning R, we wnt to solve c w + c w + c w + c 5 w 5 = v for rbitrry v. Using block multipliction nottion, we write [ w w w w 5 ] 5 c c c c = v v v so c c c c = v v v. We cnnot tke the determinnt of non-squre mtri. The reduced row-echelon form of is To investigte liner independence, we see tht the fourth column in the mtri (which doesn t hve pivot) gives us free prmeter, so {w, w, w, w 5 } is linerly dependent. To investigte spnning, we see tht there is no row of ll zeroes (every row hs pivot), so we will never hve n inconsistent eqution = #. Therefore, {w, w, w, w 5 } spns R.

114 Chpter : Vector Spces.. Spnning Sets nd Liner Independence. Summry: {w,, w k } in R n is linerly independent if nd * only if the reduced row-echelon form of the mtri [w w k ] hs s mny pivots s columns (vribles). {w,, w k } spns R n if nd only if the reduced row-echelon form of the mtri [w w k ] hs s mny pivots s rows (equtions), i.e. no rows of ll zeroes. Theorem.8 A Property of Linerly Dependent Sets A set S = {v, v,, v k }, k, is linerly dependent if nd only if t lest one of the vectors vj cn be written s liner combintion of the other vectors in S. Proof To prove only if (), ssume set S is linerly dependent nd k. Then Becuse one of the coefficients must be nonzero, we cn ssume without loss of generlity (WOLOG) tht c. Then * Conversely (which mens chnging to ), suppose v is liner combintion of the other vectors in S. Then v = = Thus, {v, v,, v k } is linerly dependent becuse Emple: Given tht {w, w, w, w 5 } is linerly dependent, nd tht c5 t 6 c 59t c w + c w + c w + c 5 w 5 = hs the solutions, write w 56 s liner c 59t 8 c t 59 combintion of the other vectors in the set. p. 95

115 Chpter : Vector Spces.. Spnning Sets nd Liner Independence. Solution: tw tw + 59 tw + tw 5 = so 8w + 6w + 59w 5 = 56w so w = 56 w + 56 w + 56 w 5 Theorem.9 Corollry Two vectors v nd v in vector spce V re linerly dependent if nd only if one is sclr multiple of the other. * Proof: From the proof of Theorem.8, Theorem Any set S = {, v, v,, v k }, contining the zero vector is linerly dependent. * Proof: + v + v + + v k =. Since not ll coefficients re zero, S is linerly dependent. p. 96

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117 Chpter : Vector Spces..5 Bsis nd Dimension..5 Bsis nd Dimension. Objective: Recognize bses in the vector spces R n, P n, nd M m,n. Objective: Find the dimension of vector spce. Objective: Prove results bout spnning sets nd liner independence. A set of vectors S = {v, v,, v k } in vector spce V is clled bsis for V when ) S spns V, nd ) S is linerly independent. S spns V sys tht the set S is not too smll to be bsis; S is linerly independent sys tht the set S is not too big to be bsis. (The plurl of bsis is bses. ) In the digrm on the left, you cn see tht S = {u, u } is too smll to spn R. For emple, v is outside of spn(s ). In the figure on the right, you cn see tht S = {u, u, u } is lrge enough to spn R. For emple, you cn see the liner combintion tht gives v. Finlly, In the figure on the right, you cn see tht S = {u, u, u, v} is lrge to be linerly independent, becuse v is liner combintion of u, u, nd u. p. 97

118 Chpter : Vector Spces..5 Bsis nd Dimension. Emple. The stndrd bsis for R : Show tht {(,, ), (,, ), (, )} is bsis for R. * Solution: To decide liner independence, we wnt to solve c (,, ) + c (,, ) + c (, ) =. To decide spnning R, we wnt to solve c (,, ) + c (,, ) + c (, ) = v for rbitrry v. These equtions re c c = c nd c v c = v c v The first eqution hs only the trivil solution, nd the second eqution lwys hs solution, so {(,, ), (,, ), (, )}is linerly independent nd it spns R. Therefore, {(,, ), (,, ), (, )} is bsis for R. The stndrd bsis in R n is e e e n,,,,,,,,, Emple: Show tht {(7,, 5), (, 9, ), (, 7, 7)} is bsis for R. Solution: To decide liner independence, we wnt to solve. To decide spnning R, we wnt to solve * As mtri equtions, we obtin The mtri is invertible (its determinnt is 7). Becuse of this, the first eqution hs only the trivil solution, nd the second eqution lwys hs solution, so {(7,, 5), (, 9, ), (, 7, 7)} is linerly independent nd it spns R. Therefore, {(7,, 5), (, 9, ), (, 7, 7)} is bsis for R. p. 98

119 Chpter : Vector Spces..5 Bsis nd Dimension. The stndrd bsis for P n is {,,,, n }. The stndrd bsis for M, is {,,,,, }. Theorem.9 Uniqueness of Bsis Representtion If S = {v, v,, v k } is bsis for vector spce V, then every vector in V, cn be written in one nd only one wy s liner combintion of vectors in S. Proof: Let u be n rbitrry vector in V, u cn be written in t lest one wy s liner combintion of vectors in S becuse Now suppose tht u cn be written s liner combintion of vectors in S in two wys: u = b v + b v + + b k v k nd u = c v + c v + + c k v k Subtrcting these two equtions gives * Since S is linerly independent, Thus, u cn be written s liner combintion of vectors in S in only one wy. Emple: Using the bsis {(7,, 5), (, 9, ), (, 7, 7)} for R from the previous emple, find the unique representtion of u = (u, u, u ) using this bsis. In other words, find the unique solution for the constnts c, c, c in the eqution u = c (7,, 5), + c (, 9, ), + c (, 7, 7)} p. 99

120 Chpter : Vector Spces..5 Bsis nd Dimension. p. Solution: As mtri eqution, we obtin Lemm.9½ Dimension of Spnning Sets nd Linerly Independent Sets If S = {v, v,, v n } spns vector spce V nd S = {u, u,, u m } is set of m linerly independent vectors in V, then m n. Proof by contrdiction: Suppose m > n. To show tht S is linerly dependent ( contrdiction), we need to find sclrs k, k,, k m (not ll zero) such tht k u + k u + + k m u m = m m m k m k u u = Becuse S spns V, ech u i is liner combintion of vectors in S : n n n m m m C v v u u i.e. n mn m m n n C C C C v v u v v u Substituting into the first eqution gives n n n m m k m C k v v = Now consider ] [ n m m m n T k k C This is n equtions for m unknowns (k i ), with n < m, so we hve (infinitely mny) nontrivil (non-zero) solutions (k, k,, k m ). Tking the trnspose, we hve non-zero solution to

121 Chpter : Vector Spces..5 Bsis nd Dimension. =k k n Cm n n m Therefore, we hve non-zero solution to v u = k k m C m mn = k k m m v n n u m m This sys tht S = {u, u,, u m } is linerly dependent, which contrdicts the premise, thus completing the proof by contrdiction. Theorem. Bses nd Liner Dependence * If S = {v, v,, v n } is bsis for vector spce V, then every set contining more thn n vectors in linerly dependent. Proof: S = {v, v,, v n } spns V so by Lemm.9½, ny set of m linerly independent vectors in V hs m n. Therefore, ny set of m vectors in V where m > n must be linerly dependent. Note: the lst step is true becuse P Q is logiclly equivlent to its contrpositive (not Q) (not P). Theorem. Bses nd Liner Dependence If vector spce V hs one bsis with n vectors, then every bsis for V hs n vectors. Proof: Let S = {u, u,, u n } be one bsis for Vnd let S = {v, v,, v m } be nother bsis for V. Becuse S is bsis, nd S is linerly independent, Thm.. tells us tht m < n. Similrly, becuse S is bsis, nd S is linerly independent, Thm.. tells us tht n < m. Therefore, m = n. * If vector spce V hs bsis consisting of n vectors, * then the number n is clled the dimension of V, denoted by dim(v) = n. We define the dimension of the trivil vector spce {} to be n =. The dimension of vector spce is well-defined (unmbiguous) becuse of Thm... p.

122 Chpter : Vector Spces..5 Bsis nd Dimension. Emples: dim(r n ) = n dim(p n ) = n + dim(m m,n ) = mn Emple: Find the dimension of the subspce of R given by W= {(, b, + b): nd b re rel numbers}.. Solution:We cn write (, b, + b) = (,, ) + b(,, ), so W is spnned by S = {(,, ), (,, )}. Moreover, S is linerly independent, becuse the reduced row echelon form of is (every column hs pivot). The only solution to is = b =. So S is bsis, nd dim(w) =. = b Emple: Find the dimension of the subspce W of P spnned by S = { 5 + +, 5 +, + +, } Solution: S spns W, but it might not be linerly independent. To decide liner independence, we solve p.

123 Chpter : Vector Spces..5 Bsis nd Dimension. Theorem. Bsis Tests in n n-dimensionl Spce Let V be vector spce of dimension n. ) If S = {v, v,, v n }is linerly independent set of vectors in V, then S is bsis for V. ) If S = {v, v,, v n }spns V, then S is bsis for V. Proof by Contrdiction of Prt() Assume tht S is not bsis for V. Since S is linerly independent, it must not spn V. Choose vector u in V tht is not in spn(s) Then the set {v, v,, v n, u} is lso linerly independent. To see this, note tht c v + c v + + c n v n = u hs no solution. c v + c v + + c n v n = c n+ u hs solution only when c n+ =, nd in tht cse, c = c = = c n = becuse {v, v,, v n } is linerly independent. {v, v,, v n, u} being linerly independent contrdicts Thm.. we hve n + linerly independent vectors in n n-dimensionl vector spce. So S must be bsis for V * Proof by Contrdiction of Prt() Assume tht S is not bsis for V. Since S spns V, it must not be linerly independent. So c v + c v + + c n v n = hs solution where not ll of the c i re zero. Without loss of generlity, we cn ssume tht c n. Then the set {v, v,, v n }lso spns V. To see this, first observe tht v n = Then c c v v c n c n cn v n. c n p.

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125 Chpter : Vector Spces..6 Rnk of Mtri nd Systems of Liner Equtions..6 Rnk of Mtri nd Systems of Liner Equtions. Objective: Find bsis for the row spce, bsis for the column spce, nd the rnk of mtri. Find the nullspce of mtri. Find the solution of consistent system A = b in the form p + h. Objective: Prove results bout subspces ssocited with mtrices. Given n mn mtri A, in this section we will spek of its row vectors, which re in R n A = m n,, n,, m n mn nd its column vectors, which re in R m m, m, n mn A = m m n n mn m m n n mn Given n mn mtri A. The row spce of A is the subspce of R n spnned by the row vectors of A. The column spce of A is the subspce of R m spnned by the column vectors of A. Theorem. Row-Equivlent Mtrices Hve the Sme Row Spces.( If n mn mtri A is row-equivlent to n mn mtri U, then the row spce of A is equl to the row spce of U. Proof Becuse the rows of U cn be obtined from the rows of A by elementry row opertions (sclr multipliction nd ddition), it follows tht the row vectors of U re liner combintions of the row vectors of A. The row vectors of U lie in the row spce of A, nd the subspce.( spnned by the row vectors of U is contined in the row spce of A: spn(u) spn(a). Since the rows of A cn lso be obtined from the rows of U by elementry row opertions, we lso hve tht subspce spnned by the row vectors of A is contined in the row spce of U: spn(a) spn(u). Therefore, the two row spces re equl: spn(u) = spn(a). p. 5

126 Chpter : Vector Spces..6 Rnk of Mtri nd Systems of Liner Equtions. p. 6.( Theorem. Bsis for the Row Spce of Mtri If mtri A is row-equivlent to mtri U in row-echelon form, the then the nonzero row vectors of U form bsis for the row spce of A. Emple: Show tht the nonzero rows of 6 9 ( mtri in row-echelon form) re linerly independent. Solution: We wnt to solve = 6 9 c c c c The first column of the right-hnd side sum is c, so c =. The second column of the righthnd side sum is 9c + c = + c, so c =. The fourth column of the right-hnd side sum is + + c (becuse c = c = ), so c =. Finlly, the fifth column of the right-hnd side sum is c (becuse c = c = c = ), so c =. Therefore the nonzero rows re linerly independent. Alternte Solution: The trnspose of = 6 9 c c c c is = c 9 + c 6 + c + c = 6 9 c c c c. Solving this by forwrd substitution (first row down to the lst row) yields c c c c =.

127 Chpter : Vector Spces..6 Rnk of Mtri nd Systems of Liner Equtions. p. 7 Emple: Find bsis for the row spce of Solution: The row-echelon form is , so bsis for the row spce is {[, 8, 9, ], [,, 8, 6], [,,, ]}. You could lso tke the reduced row-echelon form 8 55 to obtin the bsis {[,, 55, ], [,, 8, ], [,,, ]}. Emple: Find bsis for the subspce of R spnned by {[, 6, ], [, 5, 7], [ 7, 8, 7]}. Solution: Lemm: The pivot columns of mtri U in row-echelon form (or reduced row-echelon form) re linerly independent.

128 Chpter : Vector Spces..6 Rnk of Mtri nd Systems of Liner Equtions. p. 8.( Emple: U =. If we look only t the pivot columns nd solve c c c c =, we see by bck substitution tht the unique solution is c c c c =, so the pivot columns of U re linerly independent. Lemm: The pivot columns of mtri U in row-echelon form (or reduced row-echelon form) spn the column spce of U. Emple: U =. The column spce of U is the subset of R 5 consisting of vectors whose fifth component is zero (becuse we re tking liner combintions of vectors whose fifth component is zero). If we look only t the pivot columns, we cn lwys solve c c c c = v v v v, by bck substitution so the pivot columns of U spn the column spce of U. Theorem Bsis for the Column Spce of Mtri Given n mn mtri A nd its row-echelon (or reduced row-echelon) form U. Then the columns of A tht correspond to the pivot columns of U form bsis for the column spce of A. Proof: by combining the two preceding lemms, we know tht the pivot columns of U form bsis for the column spce of U. Let i be the column vectors of A nd u i be the column vectors of U, so A = [ n ] nd U = [u u n ]. We cn fctor PA = LU, so P i = Lu i for ech column vector, u i = L P i, nd i = P Lu i. (essentil ide)

129 Chpter : Vector Spces..6 Rnk of Mtri nd Systems of Liner Equtions. To show tht the pivot columns of A re linerly independent, consider the eqution = ci, where the terms in the sum include only the pivot columns. Then i ipivots L P = L P ci, so = i ci L ipivots ipivots P c u = i i ipivots Since the pivot columns of U re linerly independent, ll of the c i (i pivots) re zero, so the pivot columns of A re lso linerly independent. To show tht the pivot columns of A spn the column spce of A, consider ny vector v in the n c i i column spce of A. Then v cn be written s v = L Pv = L P n n c i i = i i c L i P n i =c i i u i so Since L Pv is liner combintion of the u i, it cn be written s liner combintion of the pivot columns (which re bsis for the column spce of U): L Pv = diu so v = P L(L Pv) = P L i diu = i dip Lu = i dii ipivots i ipivots Therefore, the pivot columns of A spn the column spce of A. i ipivots ipivots Tken together, the two prts of this proof show tht the pivot columns of A form bsis for the column spce of A. Emple: Find bsis for the column spce of 8 9 A = ( p. 9

130 Chpter : Vector Spces..6 Rnk of Mtri nd Systems of Liner Equtions. Theorem.5 Row Spce nd Column Spce hve the Sme Dimension..( If A is n mn mtri, then the row spce nd column spce of A hve the sme dimension. Proof: The dimension of the row spce nd the dimension of the column spce re both equl to the number of pivots in the row-echelon form (or in the reduced row echelon form) of A..( The rnk r = rnk(a) of mtri is the dimension of its row spce, the dimension of its column spce, nd the number of pivots in the row-echelon form (or in the reduced row echelon form). Theorem.6 Solutions to Homogeneous System A = If A is n mn mtri, then the set of ll solutions to the homogeneous system of liner equtions A = is subspce of R n. Proof: Becuse A is mn, is n, so the set of ll solutions to the homogeneous system of liner equtions A = is subset of R n. We will cll this subset N. N is not empty becuse it contins : A =. Now we must show.( If A is n mn mtri, the subspce of R n consisting of ll solutions to the homogeneous system A = is clled the nullspce of A nd is denoted by N(A). We sometimes cll N(A) the solution spce of A =. The dimension of N(A) is clled the nullity of A, nd is sometimes written s (the Greek letter nu). p.

131 Chpter : Vector Spces..6 Rnk of Mtri nd Systems of Liner Equtions. Theorem.7+ Solutions of Homogeneous System A = If A is n mn mtri of rnk r nd nullity, then the dimension of the solution spce of A =, i.e. N(A), is = n r. Tht is, n = rnk(a) + nullity(a), or the number of vribles = number of pivot vribles + number of free vribles. Moreover, the solution set to the homogeneous system A = is the sme s the solution set to the reduced row-echelon form R =. This solution hs the form.( h = t + t + + t where the t i re free prmeters, one for ech free (non-pivot) vrible, nd {,,, } is bsis for N(A). Proof A = nd R = hve the sme solution set, i.e. N(A) = N(R), becuse A nd R re rowequivlent. A = iff E n E E A = E n E E iff R = Ech column of R is multiplied by component of. R hs r pivots, corresponding to r pivot vribles in. The remining n r vribles re free vribles, which we cn replce with free prmeters t, t,, t n r. Now we cn solve r equtions (on for ech row of R tht hs pivot, nd is therefore not ll zeroes) for r pivot vribles by bck-substitution. The set of these solutions h is N(A). If we collect like terms in t i, nd fctor out the t i, we hve h = t + t + + t so N(A) is spnned by {,,, }. Moreover, {,,, } is linerly independent, becuse the solution to = t + t + + t = h is = free(i) = t i nd for ech free vrible nd = pivot(j) = (some liner combintion of the t i ) for ech pivot vrible. Since ll of the t i re, {,,, } is linerly independent, so {,,, } is bsis for N(A), nd N(A) hs dimension = nullity(a) = n r. Tht is, the number of vribles = number of pivot vrible. + number of free vribles. Emple: Find bsis for nullspce of A = p.

132 Chpter : Vector Spces..6 Rnk of Mtri nd Systems of Liner Equtions. p. Solution: The reduced row-echelon form is [ R ] =. Now we need to solve the system 5 = or the equivlent form The two free (non-pivot) vribles re nd 5. We cn choose prmeters = s nd 5 = t, nd solve for the pivot vribles,, nd using bck substitution. t t s t s = t, = s t, nd = s + t Tht mens tht N(A) is the set of ll vectors of the form 5 = t t s t s t s = s + t So {, } is bsis for N(A). pivots

133 Chpter : Vector Spces..6 Rnk of Mtri nd Systems of Liner Equtions. Emple: Find bsis for nullspce of A =. 8 Solution: Now tht we hve solved A =, let us consider A = b, where b. We hve lredy seen (Ch..) tht A = b hs n infinite number of solutions if there re more vribles thn equtions (n > m). And since you cn t hve more pivots thn rows, m r, so n > r, or >. Theorem.8 Solutions of Nonhomogeneous System A = b If p is prticulr solution to the nonhomogeneous eqution A = b, then every solution of this system cn be written in the form = p + h, where S* h is solution of the corresponding homogeneous system A =, i.e. h is in N(A). The generl solution of A = b, prmeterized by the free prmeters t, t,, t is = p + t + t + + t Proof Let be ny solution of A = b. Then ( p ) is solution of the homogeneous system A =, becuse.( A( p ) = A A p = b b = If we let p = p, then = p + h. (For ese of clcultion, we usully find p by solving A = b when ll of the free vribles re set to zero.) From Thm..7+ we know tht h = t + t + + t. p.

134 Chpter : Vector Spces..6 Rnk of Mtri nd Systems of Liner Equtions. p..( Emple: Find the generl solution (the set of ll solutions) of the system of liner system Solution: rref 6 5 To find p, set the free vribles =, 5 =. () 6() 5() () So =, =, =. p = To find h, set = s, 5 = t. A = 6 5 t t t s So = t, = 6t, = s + 5t. h = t t t s t s 6 5 = s + t 6 5 = + s + t 6 5

135 Chpter : Vector Spces..6 Rnk of Mtri nd Systems of Liner Equtions. p. 5 Emple: Find the generl solution (the set of ll solutions) of the system of liner system z y z y z y z y Solution: Emple: Find the generl solution (the set of ll solutions) of the system of liner system z y w z w z y z y Solution:

136 Chpter : Vector Spces..6 Rnk of Mtri nd Systems of Liner Equtions. Theorem.9 Consistency of System of Liner Equtions The system A = b is consistent if nd only if b is in the column spce of A. Proof Suppose A is mn, so is n nd b is m. n Then A = n m m mn n m S* m n n = n mn (see Section.) So A = b is if nd only if b is liner combintion of the columns of A, i.e. b is in the column spce of A. Summry of Equivlent Conditions for Squre Mtrices If A is n nn mtri, then the following conditions re equivlent. ) A is invertible. ) A = b hs unique solution for ny n mtri b. ) A = hs only the trivil solution. ) A is row-equivlent to I n. 5) det(a) 6) rnk(a) = n 7) The n row vectors of A re linerly independent. 8) The n column vectors of A re linerly independent. p. 6

137 Chpter : Vector Spces..7 Coordintes nd Chnge of Bsis..7 Coordintes nd Chnge of Bsis. Objective: Represent coordintes in generl n-dimensionl spces. Objective: Find coordinte mtri reltive to bsis in R n. Objective: Find the trnsition mtri from the bsis B to the bsis B in R n. Let B = {v, v,, v n } be n ordered bsis for vector spce V nd let be vector in V such tht = c v + c v + + c n v n The sclrs c, c,, c n re clled the coordintes of the vector reltive to (or with respect to) the bsis B. The coordinte vector (or mtri) of reltive B is the column mtri in R n whose components re the coordintes of. S* c [] B = c c n Emple: The coordinte mtri of = (,, ) reltive to the stndrd bsis S = {e, e, e } = {(,, ), (,, ), (, )} in R is simply [] S = becuse = e + e + e Emple: Find the coordinte mtri of p() = reltive to the stndrd ordered bsis S = {,, } in P Solution: p() = 6() 5() + ( ) so [p] S = Emple: Find the coordinte mtri of A = S = {,,, } in M, 7 reltive to the stndrd ordered bsis Solution: A = so [A] S = 8 7 p. 7

138 Chpter : Vector Spces..7 Coordintes nd Chnge of Bsis. Emple : Given tht the coordinte mtri of in R reltive to the nonstndrd ordered bsis B = {v, v } = {(, ), (, )} is [] B = Find the coordinte mtri of reltive to the stndrd bsis B = {u, u } = {(, ), (, )} = S. Solution: [] B = mens tht = v + v = (, ) + (, ) = (5, ). 5 Now (5, ) = 5(, ) + (, ) = 5u + u so [] B=S =. Another method is to tke the coordinte mtrices reltive to B of = v + v = v v to obtin []B=S = [ v ] B' [ v ] B' = 5 =. This procedure is clled chnge of bsis from B to B. Given [] B, we found [] B using n eqution of the form [] B = P [] B where P = [ v ] B' [ v ] B'. The mtri P is clled the trnsition mtri from B to B. Emple : Given tht the coordinte mtri of in R reltive to the stndrd ordered bsis B = S = {e, e, e } is [] B =. Find the coordinte mtri of reltive to the nonstndrd ordered bsis B = {u, u, u } = {(,, ), (,, ), (,, 5)}. p. 8

139 Chpter : Vector Spces..7 Coordintes nd Chnge of Bsis. p. 9 S* Solution: Write s liner combintion of the vectors in B: = c u + c u + c u = u u u c c c = [] B = B B B ] [ ] [ ] [ u u u c c c Since B = S, = 5 c c c so c c c = 5 = 8 5. Thus we hve [] B=S = c c c = 8 5. Moreover, we found tht [] B = B B B ] [ ] [ ] [ u u u [] B Compring to [] B = P [] B fter Emple bove, we must hve [] B = P[] B where P = B B B ] [ ] [ ] [ u u u. The mtri P is clled the trnsition mtri from B to B. Given nonstndrd bses B = {v, v,, v n }, B = {u, u,, u n }, nd the stndrd bsis S = {e, e,, e n } of R n. Define mtrices B BS = S n S S ] [ ] [ ] [ v v v nd (B) BS = S n S S ] [ ] [ ] [ u u u. (B nd B re squre, invertible mtrices.) B is the trnsition mtri from B to S; B is the trnsition mtri from B to S. To get P, the trnsition mtri from B to B, first pply B BS, then pply (B) SB. P = (B) B or (P ) BB = (B) SB B BS The inverse is P = B B or P BB = (B ) SB (B) BS

140 Chpter : Vector Spces..7 Coordintes nd Chnge of Bsis. Emple: Given B = {v, v, v } = {(,, ), (,, ), (,, )} nd [] B =. Find [] S Solution: Emple: Given B = {u, u, u } = {(8,, ), (7,, ), (,, 6)} nd =(, 9, ). Find [] B Solution: Emple: Given B = {v, v, v } = {(,, ), (,, ), (,, )}, B = {u, u, u } = {(,, ), (,, ), (,, )}, nd [] B =. Find the trnsition mtri P from B to B the trnsition mtri P from B to B, nd [] B. Solution: p.

141 Chpter 5: Inner Product Spces. 5. Length nd Dot Product in Rn. Chpter 5: Inner Product Spces. 5. Length nd Dot Product in R n. Objective: Find the length of vector nd find unit vector. Objective: Find the distnce between two vectors. Objective: Find dot product nd the ngle between two vectors, determine orthogonlity, nd verify the Cuchy-Schwrz Inequlity, the tringle inequlity, nd the Pythgoren Theorem. Objective: Use mtri product to represent dot product. From the Pythgoren Theorem, we know tht the length, or norm, of vector v in R is v = v = v v. Also from the Pythgoren Theorem, in R we hve v v v = v v v. In R n we define the length or norm of v to be v = S* v Notice tht v v = if nd only if v =. v vn A unit vector is vector whose length is one. In R nd R lternte nottion is sometimes used for the stndrd unit vectors i = (, ), j = (, ) in R i = (,, ), j = (,, ), k = (,, ) in R Two nonzero vectors u nd v re prllel when one of them is sclr multiple of the other: u = cv. If c >, then u nd v hve the sme direction. If c <, then u nd v hve opposite directions. p.

142 Chpter 5: Inner Product Spces. 5. Length nd Dot Product in Rn. Theorem 5. Length of sclr multiple Let v be vector in R n nd let c be sclr. Then cv = c v, where c is the bsolute vlue of c. Proof: cv = ( cv ) ( cv ) ( cvn ) = c v v n v = c v Theorem 5. Unit Vector in the Direction of v If v is nonzero vector in R n, then the vector u = v v hs length one nd hs the sme direction s v. u is clled the unit vector in the direction of v. Proof: Becuse v, we know tht v > so direction s v. Moreover, u = v v = >. Then u = v v so u hs the sme v v =, so u is unit vector. v The process of finding the unit vector in the sme direction s v is clled normlizing v. The distnce between two vectors u nd v is d(u, v) = u v. In R n, d(u, v) = ( u v ) ( u v ) ( u n vn ) Notice tht d(u, v) d(u, v) = if nd only if u = v. d(u, v) = d(v, u) Emple: Let u = (,, ), v = (,, ). Find the length of u, normlize v, nd find the distnce between u nd v. Solution by hnd: u = ( ) = 9 or v (,, ) = v ( ) =,, d(u, v) = ( ) ( ) ( ( )) = p.

143 Chpter 5: Inner Product Spces. 5. Length nd Dot Product in Rn. ( ) 6 = 6 = 6.78 Solution using Mthemtic: u={-,,} Out = {-,,} Norm[u] Out = v={,,-} Out = {,,-} Normlize[v] Out = Norm[u-v] Out = Solution using TI-89: [-,,] U [,,-] V Mtri Norms norm( U ) Mtri Vector ops unitv( V ) Mtri Norms norm( U - V ) To find the ngle between to vectors in R n, we strt with the Lw of Cosines in R. u v = u + v u v cos( ) (u v ) + (u v ) = u + u + v + v u v cos( ) v u v u u u v + v + u u v + v = u + u + v + v u v cos( ) u v u v = u v cos( ) u v + u v = u v cos( ) This eqution leds us to the following definitions. The dot product between two vectors u = (u, u,, u n ) nd v = (v, v,, v n ) in R n u v = u v + u v + + u n v n S* The ngle between two nonzero vectors u nd v in R n is = u v rccos,. u v p.

144 Chpter 5: Inner Product Spces. 5. Length nd Dot Product in Rn. Two vectors u nd v in R n re orthogonl (perpendiculr) when u v =. (Notice tht the vector is orthogonl to every vector.) We often represent vectors u = (u, u,, u n ) nd v = (v, v,, v n ) in R n by their coordinte mtrices reltive to the stndrd bsis {e, e,, e n } u v u = u nd v = v u n v n We then write u T v for u v becuse u T v = [u v + u v + + u n v n ]. Wrning: In our formuls, we will be sloppy nd pretend tht u T v is equl to the sclr u v. However, Mthemtic nd the TI-89 know tht u T v is mtri, nd will not let you use it s sclr. Emple: Using u = (,, ), v = (,, ) from the previous emple, find u v nd the ngle between u nd v. period, not sterisk Solution by hnd: u v = ( )() + ()() + ()( ) = u = v = = = ( ) = 9 or ( ) = rccos.8 or. 9 Solution using Mthemtic: u.v Out = ArcCos[u.v/Norm[u]/Norm[v]] Out = Out =.899 Solution using TI-89: Mtri Vector ops dotp( U, V ) Mtri Vector ops dotp(u, v) Mtri Norms norm(u) Mtri Norms norm(v)) Mtri Vector ops dotp(u, v) p.

145 Chpter 5: Inner Product Spces. 5. Length nd Dot Product in Rn. Mtri Norms norm(u) Mtri Norms norm(v)) To obtin deciml pproimtion insted of n ect formul, use N[...] in Mthemtic or insted of on the TI-89. You cn chnge modes between degrees nd rdins on the TI-89 using Angle DEGREE Theorem 5. Properties of the Dot Product If u, v, nd w re vectors in R n nd c is sclr, then ) u v = v u ) u (v + w) = u v + u w ) c(u v) = (cu) v = u (cv) ) v v = v 5) v v nd v v = if nd only if v =. Proof of 5..: optionl u (v + w) = R n combined with the usul opertions of vector ddition, sclr multipliction, vector length, nd the dot product is clled Eucliden n-spce. optionl (Optionl side: non-eucliden spces include elliptic spce, which is curved, cnnot be smoothly mpped R n nd hs different definition of distnce; hyperbolic spce, which is curved nd hs different definition of distnce; nd Minkowski spce, used in Einstein s specil reltivity, which replces the non-negtive dot product with n intervl tht cn be negtive.) p. 5

146 Chpter 5: Inner Product Spces. 5. Length nd Dot Product in Rn. Theorem 5. The Cuchy-Schwrz Inequlity If u nd v re vectors in R n, then u v u v, where u v is the bsolute vlue of u v. Proof If u =, then equlity holds, becuse u v = v = =, nd u v = v = v =. If u, then let t be ny rel number nd consider the vector tu + v. Then tu + v = S* The left-hnd side is qudrtic function f (t) = t + bt + c, where =, b =, c = The qudrtic formul is t = b± b c Becuse f (t), we hve either one or zero roots, so b c Theorem 5.5 The Tringle Inequlity If u nd v re vectors in R n, then u + v u + v. Proof u + v = S* p. 6

147 Chpter 5: Inner Product Spces. 5. Length nd Dot Product in Rn. Theorem 5.6 The Pythgoren Theorem If u nd v re vectors in R n, then u nd v re orthogonl if nd only if u + v = u + v. Proof By definition, u nd v re orthogonl if nd only if S* u + v = The conclusion follows from these two equtions. p. 7

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149 * Chpter 5: Inner Product Spces. 5. Inner Product Spces. 5. Inner Product Spces. Objective: Determine whether function defines n inner product. Objective: Find the inner product of two vectors in R n, M m,n, P n, nd C[,b]. Objective: Use the inner product to find ngles between two vectors nd determine whether two vectors re orthogonl. Objective: Find n orthogonl projection of vector onto nother vector in n inner product spce. The inner product is n etension of the concept of the dot product from R n to generl vector spce. The stndrd dot product, lso clled the Eucliden inner product, on R n is written s u v, while the generl inner product on vector spce V is written s u, v. Definition of Inner Product Let u, v, nd w be vectors in vector spce V, nd let c be ny sclr. An inner product on V is function tht ssocites rel number u, v with ech pir of vectors u nd v. nd stisfies the following ioms. ) u, v = v, u Symmetric ) u, v + w = u, v + u, w ) cu, v = cu, v ) v, v nd v, v = if nd only if v =. Nonnegtive A vector spce V with n inner product is clled n inner product spce. Emple: Consider R u v. Let u = nd v = u. Insted of the Eucliden inner product v u T v, let u, v = u T v = u v u v u v + u v. Show tht u, v is n inner product. Solution: ) v, u = v T u = v u v u v u + v u = u, v ) Let A =. We know tht for mtri multipliction, u, v + w = u T A(v + w) = (u T A)(v + w) = (u T A)v + (u T A)w = u T Av + u T Aw = u, v + u, w. ) We know tht for mtri multipliction, cu, v = c(u T v) = (cu T ) v = = cu, v. p. 9

150 Chpter 5: Inner Product Spces. 5. Inner Product Spces. ) v, v = v v v + v = v + v + v v v + v = v + v + (v v ), nd v, v = if nd only if v =, v =, nd v v = if nd only if v =. ut vt Emple: Consider R. Let u = u nd v = v. Insted of the Eucliden inner product u y v y u z v z c u T v, let u, v = u T v, where c is positive constnt (the speed of light), so u, v = c u t v t + u v + u y v y + u z v z. Show tht u, v is not n inner product. (Optionl side: u, v is clled the spcetime intervl.) Solution: ) v, v = c v t + v + v y + v z which cn be less thn zero, for emple when v =. Note: Aioms nd re stisfied becuse of the rules of mtri multipliction. Aiom is stisfied becuse the mtri is symmetric (A T = A). Theorem 5.7 Properties of Inner Products If u, v, nd w re vectors in n inner product spce V over the rel numbers nd c is sclr, then ), v = v, = ) u + v, w = u, w + v, w ) u, cv = cu, v Proof of 5.7.:, v = v, v = v, v =. By the symmetric property, v, =, v = optionl Proof of 5.7.: u + v, w = w, u + v Symmetry = w, u + w, v Aiom = u, w + v, w Symmetry Proof of 5.7.: u, cv = cv, u Symmetry = cv, u Aiom = cu, v Symmetry p.

151 Chpter 5: Inner Product Spces. 5. Inner Product Spces. Together with the inner product ioms, Thm 5.7 sys tht the inner product is liner in ech of its rguments (we will study liner opertors much more in Chpter 6.) Linerity is essentilly the sme s distributivity: Dot product: Generl inner product: Let u nd v be vectors in n inner product spce V. (u + v) (5 + 7y) = (u ) + (u y) + 5(v ) + (v y) u + v, 5 + 7y = u, + u, y + 5v, + v, y The length, or norm, of u is u = u, u. * The distnce between two vectors u nd v is d(u, v) = u v. The ngle between two nonzero vectors u nd v is given by = u, v rccos,. u v Two vectors u nd v in R n re orthogonl (perpendiculr) when u, v =. (Notice tht the vector is orthogonl to every vector.) If u = then u is clled unit vector. If v is ny nonzero vector, then u = in the direction of v. Notice tht the definition of ngle presumes tht gurnteed by the Cuchy-Schwrz Inequlity (Theorem 5.8.). Properties of Length ) u ) u = if nd only if u =. ) cu = c u v v u, v u is the unit vector v. This is These follow from the ioms in the definition of inner product nd the definition of norm. Properties of Distnce ) d(u, v) ) d(u, v) = if nd only if u = v. ) d(u, v) = d(v, u) These follow from the ioms in the definition of inner product nd the definition of distnce. p.

152 Chpter 5: Inner Product Spces. 5. Inner Product Spces. Emple: Let A = nd B = b b product A, B = b + b + b + b. Let I =, R = nd L = 5 5 b b. be two mtrices in M, nd define the inner ) Find the norm of I =. Solution: b) Find the inner product of nd ngle between R nd L. Are they orthogonl?.( Solution: c) Find the inner product of nd ngle between I nd L. Are they orthogonl? Solution: p.

153 Chpter 5: Inner Product Spces. 5. Inner Product Spces. Emple: Let p nd q be two polynomils in P nd define the inner product p, q = p( ) q( ) d ) Find the norm of p() =. (Sometimes we will just sy Find the norm of. ) Solution: b) Find the inner product of p() = nd q() =. Are they orthogonl?.( Solution: c) Find the ngle between nd. Are they orthogonl? Solution: Theorem 5.8 Let u nd v be vectors in n inner product spce V. ) Cuchy-Schwrz Inequlity: u, v u v ) Tringle inequlity: u + v u + v ) Pythgoren Theorem: u nd v re orthogonl if nd only if u + v = u + v. The proofs re the sme s the proofs of Theorems 5., 5.5, nd 5.6, ecept tht we replce u v with u, v. A result such s the tringle inequlity, which seems intuitive in R nd R, cn be generlized through liner lgebr to some less obvious sttements bout functions in C[, b]. p.

154 Chpter 5: Inner Product Spces. 5. Inner Product Spces. Emple: Consider f () = cos() nd g() = / (/ )(/ + ) in C[ /, /] with the inner product f, g = f ( ) g( ) d. Verify the tringle inequlity by direct clcultion. / Solution: The tringle inequlity is f + g f + g. f = g = f + g = f + g = f + g = Orthogonl Projections: Let u nd v be two vectors in R. If v is nonzero, then u cn be orthogonlly projected onto v. This projection is denoted by proj v u. Becuse proj v u is sclr multiple of v, we cn write proj v u = v v (Recll tht v v Then = u cos = u is the unit vector in the direction of v.) u v u v = u v v u proj v u v proj v u = u v v v u v = v v v p.

155 Chpter 5: Inner Product Spces. 5. Inner Product Spces. * Let u nd v be vectors in generl inner product spce V. The orthogonl projection of u onto v is u, v proj v u = v, v v Emple: Use the Eucliden inner product find the orthogonl projection of u = (,, ) onto v = (,, ) nd of v onto u..( Solution: proj v u = proj v u = p. 5

156 Chpter 5: Inner Product Spces. 5. Inner Product Spces. Theorem 5.9 Orthogonl Projection nd Distnce u u Let u nd v be two vectors in n inner product spce V such tht v. Then d(u, proj v u) d(u, cv) d(u, proj v u) < d(u, cv) whenever c u, v v, v proj v u v cv v Tht is to sy, the orthogonl projection proj v u is the vector prllel to v tht is closest to u. u, v u, v Proof: Let b =, c, p = cv bv, q = u bv, nd t = u cv. v, v v, v The p is orthogonl to q, becuse p, q = cv bv, u bv bv u q t = u cv p = (c b)v = cv, u cbv, v bv, u + b v, v u, v u, v u, v = cu, v c v, v u, v + v, v v, v v, v v, v = cv, u cv, u u, v v, v u, v + v, v So the Pythgoren Theorem tells us t = p + q > q becuse p becuse c b. Therefore, d(u, proj v u) = q < t = d(u, cv). =. Emple: Consider C[, ] with the inner product f, g = f ( ) g( ) d. Find the projection of f () = onto g() = sin(). Solution:.( f, g proj g f = g( ). g, g p. 6

157 Chpter 5: Inner Product Spces. 5. Orthogonl Bses: Grm-Schmidt Process. 5. Orthogonl Bses: Grm-Schmidt Process. Objective: Show tht set of vectors is orthogonl nd forms n orthonorml bsis. Objective: Represent vector reltive to n orthonorml bsis. Objective: Apply the Grm-Schmidt orthonormliztion process. Although vector spce cn hve mny different bses, some bses my be esier to work with thn others. For emple, in R, we often use the stndrd bsis S = {e, e, e } = {(,, ), (,, ), (, )}. Two properties tht mke the stndrd bsis desirble re tht ) The vectors re normlized, i.e. ech bsis vector is unit vector: e i = for i =,,. ) The vectors re mutully orthogonl: e i e j = whenever i j, i.e. e e =, e e =, e e = A set S of vectors in n inner product spce V, is clled orthogonl when every pir of vectors in S is orthogonl. If, in ddition, ech vector in the set is unit vector, then S is clled orthonorml. If S = {v S*, v,, v n } then these definitions cn be written s S is orthogonl if nd only if v i v j = whenever i j. S is orthonorml if nd only if v i v j = whenever i j, nd e i = for i =,,, n. Emple: Show tht S = {(cos, sin, )}, ( sin, cos, ), (,, )} is n orthonorml set. Solution: S is orthogonl, becuse (cos, sin, ) ( sin, cos, ) = cos sin + sin cos = (cos, sin, ) (,, ) = nd ( sin, cos, ) (,, ) = S* S is normlized, becuse (cos, sin, ) = ( sin, cos, ) = cos sin sin = cos = p. 7

158 Chpter 5: Inner Product Spces. 5. Orthogonl Bses: Grm-Schmidt Process. (,, ) = = Theorem 5. Orthogonl Sets re Linerly Independent If S = {v, v,, v n } is n orthogonl set of nonzero vectors in n inner product spce V, then S is linerly independent. Proof: suppose c v + c v + + c n v n = Then (c v + c v + + c n v n ), v i =, v i for ech of the v i We cn distribute the inner product on the left-hnd side to obtin Becuse S is orthogonl, when j i, so eqution becomes Becuse v i, Emple: Consider C[, ] with the inner product f, g = f ( ) g( ) d. Show tht the set {, cos(), sin(), cos(), sin(),, cos(n), sin(n)} is linerly independent orthogonl set. Solution: We use softwre to integrte the following epressions, nd we use these identities to evlute the results: sin(k ) cos(k ) whenever k is n integer cos( n), sin(n) = sin( n ) d = = n = n n, cos(n) = cos( n ) d = sin( n n) = = p. 8

159 Chpter 5: Inner Product Spces. 5. Orthogonl Bses: Grm-Schmidt Process. sin(m), sin(n) when m n is sin(( m n) ) sin(( m n) ) sin( m )sin( n) d = ( ) ( ) m n m n cos(m), cos(n) when m n is sin(( m n) ) sin(( m n) ) cos( m )cos( n) d = ( ) ( ) m n m n cos(m), sin(n) when m n is cos(( m n) ) cos(( m n) ) cos( m )sin( n) d = ( ) ( ) m n m n ( m n) ( m n) ( m n) ( m n) = = ( ) ( ) = = ( + ) ( + ) = = cos(n), sin(n) = cos( n )sin( n) d = cos ( n) = n = n n Corollry Orthogonl Bses S* If V is n inner product spce of dimension n, then ny orthogonl set of n nonzero vectors in V is bsis for V. Emple: Show tht S = {(,,, ), (,,, ), (,,, ), (,,, )} is n orthogonl bsis of R. Solution: S is orthogonl, becuse (,,, ) (,,, ) = + + = (,,, ) (,,, ) = + = S* (,,, ) (,,, ) = + + = (,,, ) (,,, ) = = (,,, ) (,,, ) = + = nd (,,, ) (,,, ) = + = By the Corollry to Theorem 5., S is n orthogonl bsis for R. p. 9

160 Chpter 5: Inner Product Spces. 5. Orthogonl Bses: Grm-Schmidt Process. Theorem 5. Coordintes Reltive to n Orthonorml Bsis If B = {v, v,, v n } is n orthonorml bsis for n S* inner product spce V, then the coordinte representtion of vector w reltive to B is w = w, v v + w, v v + +w, v n v n Proof: Becuse B is bsis for V, there re unique sclrs c, c,, c n such tht w = c v + c v + + c n v n Tking the inner product with v i of both sides of the eqution gives w, v i = (c v + c v + + c n v n ), v i We cn distribute the inner product on the left-hnd side to obtin Becuse S is orthonorml, when so eqution becomes so The coordintes of w reltive to n orthonorml bsis B re clled the Fourier coefficients of w reltive to B. The corresponding coordinte mtri is c w, v [w] B = c = w, v c n w, v n Note: contrst Thm 5. with [w] B = [ ] [ ] [ ] v S v S v n S [w] S from Section.7 for generl, non-orthonorml bsis B. p.

161 Chpter 5: Inner Product Spces. 5. Orthogonl Bses: Grm-Schmidt Process. Emple: Find the coordinte mtri of w = (, 6, 5) reltive to the orthonorml bsis B = {(,, ), (,, ), (,, )} of R 6 6. Solution: S* so the coordinte mtri is Given ny bsis for vector spce, we cn construct n orthonorml bsis using procedure clled the Grm-Schmidt Orthonormliztion. Theorem 5.(Alt) (Alterntive) Grm-Schmidt Orthonormliztion Process Let B = {v, v,, v n } be bsis for n inner product spce V. Let B = {u, u,, u n }, where u i is given by w = v,...u = w = v v, u u,...u = S* w = v v, u u v, u u,...u = w n = v n v n, u u v n, u u v n, u n u n,...u n = w w w w w w w w n n Then B is n orthonorml bsis for V. Moreover, spn{ u, u,, u k } = spn{ v, v,, v k } for k =,,, n. Proof: First, B is normlized, since ll of the u i re unit vectors: u i =. Second, observe tht the terms of the form v j, u i u i re projections onto u i : v j, u i u i = proj When we subtrct orthogonl to u i v u i j proj v u i j from v j, the resultnt vector is p.

162 Chpter 5: Inner Product Spces. 5. Orthogonl Bses: Grm-Schmidt Process. Now we prove by induction tht B is lso n orthonorml set. If n =, then there re no pirs of vectors in B, so there is nothing to prove. If {u, u,, u k } is n orthonorml set, then consider u k = w w k k where w k = v k v k, u u v k, u u v k, u j u j v k, u k u k. Let j =,,, k. Then u k, u j = = = (vk v k, u u v k, u u v k, u j u j v k, u k u k ), u j w n (v k, u j v k, u u, u j v k, u u, u j v k, u j u j, u j w n w n v k, u k u k, u j ) (v k, u j v k, u () v k, u () v k, u j () v k, u k ()) = (v k, u j v k, u j ) = w n So {u, u,, u k } is lso n orthonorml set. By mthemticl induction, B is orthonorml, nd since it hs n vectors in n n-dimensionl spce, B is n orthonorml bsis. Emple: Use the Grm-Schmidt orthonormliztion process to construct n orthonorml bsis from B = {(,, ), (,,), (,, )}. Solution: w = (,, ) u = w = (,,) ((,,) (,, ))(,, ) (,,) (,,) = (,, ) = (,,) ( )(,, ) S* = (,, ) u = w = ( (,,,, ) ) = (,, ) u = p.

163 Chpter 5: Inner Product Spces. 5. Orthogonl Bses: Grm-Schmidt Process. Emple: Use the Grm-Schmidt orthonormliztion process to construct n orthonorml bsis from B = {,, } in P with the inner product p(), q() = p ( ) q( ) d. Solution: Let v =, v =, nd v =. w = u = = d = u = = w = u = p.

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165 Chpter 5: Inner Product Spces. 5. Mthemticl Models nd Lest Squres Anlysis (Optionl). 5. Mthemticl Models nd Lest Squres Anlysis (Optionl). Objective: Find the orthogonl complement of subspce. Objective: Solve lest squres problem. In this section, we consider inconsistent systems of liner equtions, nd we find the best pproimtion to solution. Emple: Given tble of dt. We wnt to find the coefficients c nd c of the line y = c + c t tht best fits these points. Solution: The system of liner equtions tht we hve to solve comes from plugging the three points (t i, y i ) into the eqution c + c t i = y i t y c c c c c c or c c. Let A =, b = c, nd = c The system is inconsistent: reduced row-echelon form is. (We lso knew tht there would be no solution becuse the three points in the grph re not colliner.) But wht is the best pproimtion? Recll tht A = c c is lwys in the column spce of A. But since A = b hs no solution, b is not in the column spce of A. We wnt to find n tht gives the A tht is closest to b. Lest Squres Problem: Given n mn mtri A nd vector b in R m, find in R n, such tht A b is minimized. This gives the A tht is closest to b. It is customry to minimize A b insted of A b becuse A b involves squre root. Intuitively, we see tht A b is orthogonl to the column spce of A. p. 5

166 Chpter 5: Inner Product Spces. 5. Mthemticl Models nd Lest Squres Anlysis (Optionl). To solve the lest squres problem, it helps to use the concept of orthogonl subspces. Let S nd S be two subspces of n n-dimensionl vector spce V. S nd S re orthogonl when v, v = for ll v in S nd v in S. The orthogonl complement of S is the set S = {u V: u, v= for ll vectors v S }. S is pronounced S perp. If every vector V cn be written uniquely s sum of vector s from S nd vector s from S, = s + s, then V is the direct sum of S nd S, nd we write V = S S. Theorem 5. Properties of Orthogonl Complements Let S be subspce of n n-dimensionl vector spce V. Then ) dim(s) + dim(s ) = n ) V = S S ) (S ) = S Emples: Consider R. {} = R. (R ) = {}. Let S = spn( ), S y = spn( ), S z = spn( ), S y = spn(, ), nd S yz = spn(, ). Then S is orthogonl to S y. S y is orthogonl to S z. S y is not orthogonl to S yz. S y = S S y. (S z ) = S y, nd (S y ) = S z. Theorem 5.6 Fundmentl Subspces of Mtri Let A be n mn mtri. The nottion R(A) mens the column spce of A (R for rnge). R(A T ) is the row spce of A. N(A) is the nullspce of A. ) R(A T ) nd N(A) re orthogonl complements. Tht is: ll vectors in the row spce of A re orthogonl to ll vectors in the nullspce of A; dim(r(a T )) + dim(n(a)) = n, nd R(A T ) N(A) = R n. ) R(A) nd N(A T ) re orthogonl complements. Tht is: ll vectors in the column spce of A re orthogonl to ll vectors in the nullspce of A T ; dim(r(a)) + dim(n(a T )) = m, nd R(A) N(A T ) = R m. p. 6

167 Chpter 5: Inner Product Spces. 5. Mthemticl Models nd Lest Squres Anlysis (Optionl). p. 7 Sketch of the proof: ) N(A) ={ R n : A = } shows tht the nullspce is orthogonl to the row spce, becuse the null vector is orthogonl to ech row of A. A = is equivlent to ) of (row ) (row of ) (row of A m A A dim(r(a T )) + dim(n(a)) = n is rnk + nullity = n, which we lredy know. The bsis vectors of R(A T )) re liner independent of the bsis of N(A) becuse they re orthogonl to the bsis of N(A), so the union of bsis of R(A T )) nd bsis of N(A) is bsis of R n, so R(A T ) N(A) = R n. ) Replce A with A T in Prt (), nd use (A T ) T = A. When we replce A with A T, we must interchnge m nd n. Emple: Find the orthogonl complement of S = spn(, ). Solution: S is the column spce R(A), where A =, so S = N(A T ). To find N(A), solve A = rref so + + s t = nd + + s + t = so = t nd = t = t s t t = s + t so S = N(A T ) = spn(, ) If S is subspce of n n-dimensionl vector spce V nd v is vector in V, then v cn be written uniquely s the sum of vector from S nd vector from S v = s + s, where s S nd s S becuse V = S S. Then s S is the orthogonl projection of v onto S, written v proj S

168 Chpter 5: Inner Product Spces. 5. Mthemticl Models nd Lest Squres Anlysis (Optionl). Theorem: v proj S v is orthogonl to every vector in S. Proof: v cn be written uniquely s v = v v S proj + s, where s = v proj v S. Since proj S, v proj v is orthogonl to every vector in S. S v S Theorem 5.5 Orthogonl Projection nd Distnce Let S be subspce of n n-dimensionl vector spce V nd v be vector in V. Then for ll u S, u proj S v, v proj S v < v u. In other words, proj S v is the vector in S tht is closest to v. S Proof: Let ll u S, u proj S v so v u = (v proj S v ) + ( proj S v u) Now v proj S v is orthogonl to ( proj S v u), so by the Pythgoren Theorem, v u = v proj + proj v u v S where proj S v u > becuse u proj S v. Therefore, v u > v proj S v v u > v proj S v S To solve the lest squres problem, we need A b (R(A)) = N(A T ), so A T (A b) =, or A T A = A T b Norml equtions Solution to Lest Squres Problem p. 8

169 Chpter 5: Inner Product Spces. 5. Mthemticl Models nd Lest Squres Anlysis (Optionl). p. 9 Emple: Let s finish the emple from the beginning of this section. Given tble of dt. We wnt to find the coefficients c nd c of the line y = c + c t tht best fits these points. Solution: c c c c c c or c c. Let A =, b =, nd = c c So A T A = A T b, i.e. 6 6 c c =, so c c = 6 6 = / 5/. Therefore, the lest-squres regression line for the dt is y = t 5 Emple: Given the tble of dt. Find the coefficients c, c, nd c of the qudrtic y = c + c t + c t tht best fits these points. Solution:.5.5 c c c c c c c c c c c c so 9 c c c =.5.5 A = 9, = c c c, nd b =.5.5. A T A = A T b so = (A T A) A T b = y =.95.8t +.5t t y t y.5.5

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171 Chpter 5: Inner Product Spces. 5.5 Applictions of Inner Product Spces (Optionl). 5.5 Applictions of Inner Product Spces (Optionl). Objective: Find the n th -order Fourier pproimtion of function. Objective: Given subspce with n orthogonl or orthonorml bsis, find the projection of vector onto tht subspce. Objective: Find the cross product of two vectors in R. Recll from Thm. 5.5 tht proj S v is the unique vector in S tht is closest to v. Theorems 5. & 5.9 Projection onto Subspce & Lest Squres Approimtion If {u, u,, u n } is n orthonorml bsis for subspce S of vector spce V, nd v V, then proj S v = v, u u + v, u u + + v, u n u n If V is spce of functions, e.g. V = C[, b] (so the u i re functions), then proj f = f, u u + f, u u + + f, u n u n S is the lest-squres pproimting function of f with respect to S. Proof: we will show tht proj S v = v, u u + v, u u + + v, u n u n by showing tht v cn be written uniquely s the sum of vector from S nd vector from S v = s + s, where s S nd s S. Then by definition, proj S v = s. Let s = v, u u + v, u u + v, u n u n nd s = v v, u u v, u u v, u n u n Then s S becuse it is liner combintion of vectors in S. For ech bsis vector u i of S u i, s = u i, (v v, u u v, u u v, u i u i v, u n u n ) = u i, v v, u u i, u v, u u i, u v, u i u i, u i v, u n u i, u n = u i, v v, u i () = Since ll vectors in S re liner combintions of the {u i }, s is orthogonl to ll vectors in S, so s S. Emple: we know from 5. tht set {, cos(), sin(), cos(), sin(),, cos(n), sin(n)} is n orthogonl set using the inner product f, g = f ( ) g( ) d. Let s normlize the set. p. 5

172 Chpter 5: Inner Product Spces. 5.5 Applictions of Inner Product Spces (Optionl). = ( )() d =. For n, cos(n) = cos( n )cos( n) d = nd sin(n) = sin( So {, cos(), sin(), orthonorml set. n )sin( n) d = cos(), sin( n)cos( n) n sin( n)cos( n) n sin(),, = cos(n), = sin(n)} is n The n th -order Fourier pproimtion of function f on the intervl [, ] is the projection of f onto spn { cos(), sin(), cos(), sin(),, cos(n), sin(n)}., Fourier pproimtions re useful in modeling periodic functions such s sound wves, hert rhythms, nd electricl signls. Emple: Find the second-order Fourier pproimtion to the periodic function f () = for [, ), nd f ( + ) = f () Solution: g() = f, + f, cos() cos() + f, sin() + f, cos() cos() + f, sin() sin() sin() Using softwre to evlute the integrls, we find f, f, = ) ( d = cos() = ( )cos( ) d = f, sin() = ( )sin( ) d = f, cos() = ( )cos() d = f, sin() = ( )sin() d = So the nd order Fourier pproimtion is g () = sin() + sin() p. 5

173 Chpter 5: Inner Product Spces. 5.5 Applictions of Inner Product Spces (Optionl). See It is common nottion with Fourier series to write the coefficients s = ) f ( d k = f ( ) cos( k) d b k = f ( )sin( k) d so tht the n th -order Fourier pproimtion is g n () = + cos() + b sin() + cos() + b sin() + + n cos(n) + b n sin(n) Emple: Given tht S = {(,, ), (,, )} is n orthonorml set in R 6 6. Find the projection of v = (,, ) onto spn{(,, ), (,, )}. 6 6 Solution: proj S v = ((,, ) (,, ))(,, ) ((,, ) (,, ))(,, ) = (,, ) + (,, ) = (,, ) In R, the cross product of two vectors u = (u, u, u ) = u i + u j + u k nd v = (v, v, v ) = v i + v j + v k is i j k uv = (u v u v )i + (u v u v )j + (u v u v )k = = where the right-hnd side is determinnt contining the vectors i, j, nd k. The cross product is undefined for vectors in vector spces other thn R. u v u v u v Theorem 5.7 Algebric Properties of the Cross Product If u, v, nd w re vectors in R nd c is sclr, then p. 5

174 Chpter 5: Inner Product Spces. 5.5 Applictions of Inner Product Spces (Optionl). ) uv = vu ) u(v + w) = uv + uw ) c(uv) = (cu)v = u(cv) ) u = u = 5) uu = 6) u (vw) = (uv) w = u v w u v w u v w p. 5

175 Chpter 5: Inner Product Spces. 5.5 Applictions of Inner Product Spces (Optionl). Theorem 5.8 Geometric Properties of the Cross Product If u, v, nd w re nonzero vectors in R, then ) uv is orthogonl both u nd v ) The ngle between u nd v is given by uv = u v sin( ) ) u nd v re prllel if nd only if uv = ) The prllelogrm hving u nd v s djcent sides hs n re of uv 5) The prllelepiped hving u, v, nd w s edges hs volume of u (vw) Emple: Find the re of the prllelogrm with vertices t (5,, ), (, 6, 7), (7,, 8), nd (5,, 5) Solution: re = uv. uv = i j 8 k 7 8 = 6 + j + k 6i + j + k = ( 6) () () = Emple: Find vector orthogonl to u = (,, ) nd v = (, 7, ). Solution: uv = i j k 7 = i j k p. 55

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177 8. Polr Form nd De Moivre's Theorem. (Optionl) 8. Polr Form nd De Moivre's Theorem. (Optionl) Objective: Determine the polr form of comple number nd convert between the polr form nd the stndrd form of comple number. Objective: Determine the eponentil polr form of comple number nd convert between the eponentil polr form nd the stndrd form of comple number using Euler s formul. Objective: Multiply nd divide comple numbers in polr form nd in eponentil polr form. Objective: Use DeMoivre s Theorem to find powers of numbers in polr form nd in eponentil polr form. Multipliction, division, nd powers of comple number re tedious to clculte in stndrd form. However, when comple numbers re written in polr form, multipliction, division, nd powers become esy to clculte nd interpret. The polr form of nonzero comple number z = + bi is given by where = r cos, b = r sin, r = z = r (cos + i sin ) b = z, nd tn = b/. The number r is the modulus of z nd is the rgument of z. There re infinitely mny choices for the rgument, becuse cosine nd sine hve period of. Usully, we choose the principl rgument, which stisfies <. The principl rgument = Arg(z) of nonzero comple number z = + bi is given by tn = b/ nd < Emple: Grph nd find the polr form (using the principl rgument) of ) b) i c) i Solution: Arg(z) = tn (b/) + Arg(z) = / Imginry is Arg(z) = tn (b/) Rel is Arg(z) = Arg(z) = Arg(z) = tn (b/) Arg(z) = tn (b/) ) Arg(z) = / c) b) p. 57

178 8. Polr Form nd De Moivre's Theorem. (Optionl) ) = (cos + i sin) b) r = ( ) ( ) = 5 = 5 Arg(z) = + tn ( ).785 i 5(cos(.785) + i sin(.785)) c) r = ( ) ( ) = = Arg(z) = tn ( i = (cos( ) + i sin( )) Emple: Find the stndrd form of (cos( 6 ) + i sin( 6 )) Solution: (cos( 6 ) + i sin( 6 )) = Theorem. Euler s Formul i = 5 5i ) = cos + i sin = e i where e is Euler s number, e.788 Proof: From clculus, we know tht the Mclurin series (the Tylor series epnsion round zero) of function f is f () = f () + f () + f () + f () + f () () + f (5) () 5 +!!!!! 5! Therefore, the Mclurin series for e is e!!!! 5! 5 6! 6 7! 7... Substitute = i. Note tht i =, i = i, i =, i 5 = i, etc. So e i i!! i!! 5 6 i 5! 6! Therefore, the Mclurin series for cos nd sin re 7 i 7!... cos( )!! 6 6!... sin( )!! 5 5! 7 7!... p. 58

179 8. Polr Form nd De Moivre's Theorem. (Optionl) Compring the series for e i, cos, sin, we see tht e i = cos + i sin Emple: Grph nd find the polr form (using the principl rgument) of ) i b) i c) + i c) Solution i / ) i = e b) r = ( ) = = = tn ( /) = / i / i = e c) r = =.66 = tn (/) i =.66 e.588i d) Emple: Grph nd find the polr form (using the principl rgument) of ) i b) i c) + i Emple: Find the stndrd form of e 5i/ Solution: e 5i/ 5 5 = (cos + i sin ) = i ) = i b) Theorem 8. Product nd Quotient of Two Comple Nubmers i i r e e re r e ( ) r = r i r e r (cos isin) r (cos isin) = r r [cos( ) isin( )] i i = r r i( e ) r (cos i sin ) r = [cos( ) i sin( )], z r (cos i sin ) r Proof: The eponentil formuls follow directly from the lws of eponents. The polr forms follow from the eponentil polr forms nd Euler s formul. Emple: Sketch nd simplify. ) e i/ e i/ = (cos + i sin )(cos + i sin ) b) e e i [cos( ) i sin( )] [cos( ) i sin( i / )] p. 59

180 8. Polr Form nd De Moivre's Theorem. (Optionl) Solution: ) e i/ e i/ = ()()e i(/+/) 6 5 /6 = / + / = 6e i5/6 (cos + i sin )(cos + i sin ) / / / = 6[cos( 6 5 ) + i sin( 6 5 )] / = 9, / = 6, 6 5 = 5 i e b) i / e = i( ( / )) e = i7 / e = i( 7 / ) e = / e i 7 / = ( /) [cos( ) isin( )] [cos( ) isin( )] 7 / [cos( 7 7 ) isin( )] [cos( / ) i sin( / )] ¾ / = 7 / / Theorem 8.5 DeMoivre s Theorem P (re i ) n = r n e in [r (cos + i sin )] n = r n [cos(n ) + i sin(n )] Proof: The eponentil formul follows directly from the lws of eponents. The polr forms follow from the eponentil polr forms nd Euler s formul. p. 6

181 8. Polr Form nd De Moivre's Theorem. (Optionl) Emple: Sketch nd simplify. ) e ik/ for k =,,,,, b) e ik/ for k =,,,,,,, 5 Solution ) e i( )/ = i = e i/ e i()/ = = e e i()/ = + i = e i/ e i()/ = i = e i/ e i()/ = = e e i()/ = + i = e i/ b) e i( )/ = e i = = e i e i( )/ = i = e i/ e i()/ = = e e i()/ = i = e i/ e i()/ = = e i e i()/ = i = e i/ e i()/ = = e e i(5)/ = i = e i/ p. 6

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183 Chpter 8 Comple Vector Spces 8. Polr Form nd DeMoivre s Theorem Determine the polr form of comple number, convert between the polr form nd stndrd form of comple number, nd multiply nd divide comple numbers in polr form. Use DeMoivre s Theorem to find powers nd roots of comple numbers in polr form. (, b) REMARK The polr form of z is epressed s z cos i sin, where is ny ngle. b r Imginry is Rel is Stndrd Form: + bi Polr Form: r(cos θ + i sin θ) Figure 8.7 θ POLAR FORM OF A COMPLEX NUMBER At this point you cn dd, subtrct, multiply, nd divide comple numbers. However, there is still one bsic procedure tht is missing from the lgebr of comple numbers. To see this, consider the problem of finding the squre root of comple number such s i. When you use the four bsic opertions (ddition, subtrction, multipliction, nd division), there seems to be no reson to guess tht Tht is, i i. i i. To work effectively with powers nd roots of comple numbers, it is helpful to use polr representtion for comple numbers, s shown in Figure 8.7. Specificlly, if bi is nonzero comple number, then let be the ngle from the positive rel is to the rdil line pssing through the point, b nd let r be the modulus of bi. This leds to the following. r cos b r sin r b So, bi r cos r sin i, from which the polr form of comple number is obtined. Definition of the Polr Form of Comple Number The polr form of the nonzero comple number z bi is given by z rcos i sin where r cos, b r sin, r b, nd tn b. The number r is the modulus of z nd is the rgument of z. Becuse there re infinitely mny choices for the rgument, the polr form of comple number is not unique. Normlly, the vlues of tht lie between nd re used, lthough on occsion it is convenient to use other vlues. The vlue of tht stisfies the inequlity < Principl rgument is clled the principl rgument nd is denoted by Arg( z). Two nonzero comple numbers in polr form re equl if nd only if they hve the sme modulus nd the sme principl rgument.

184 8. Polr Form nd DeMoivre s Theorem 5 Finding the Polr Form of Comple Number Find the polr form of ech of the comple numbers. (Use the principl rgument.). z i b. z i c. z i SOLUTION. Becuse nd b, then r, which implies tht r. From r cos nd b r sin, So, cos r nd nd b. Becuse nd b, then r, which implies tht r. So, cos nd sin b r r nd it follows tht So, the polr form is c. Becuse nd b, it follows tht r nd so z cos i sin.98.. z cos i sin. z cos.98 i sin.98. sin b r., The polr forms derived in prts (), (b), nd (c) re depicted grphiclly in Figure 8.8. Imginry is θ z = i Rel is i sin. z cos b. Imginry is θ z = + i Rel is z cos.98 i sin.98 Imginry is z = i θ Rel is c. z cos i sin Figure 8.8

185 6 Chpter 8 Comple Vector Spces Epress the comple number in stndrd form. z 8 cos i sin Converting from Polr to Stndrd Form SOLUTION Becuse cos nd sin, obtin the stndrd form z 8 cos i i. i sin 8 The polr form dpts nicely to multipliction nd division of comple numbers. Suppose you hve two comple numbers in polr form z nd z r cos i sin r cos i sin. Then the product of z nd is epressed s z z r r cos i sin cos i sin r r cos cos sin sin i cos sin sin cos. Using the trigonometric identities cos cos cos sin sin nd sin sin cos cos sin, you hve z z r r cos i sin. z This estblishes the first prt of the net theorem. The proof of the second prt is left to you. (See Eercise 75.) THEOREM 8. Product nd Quotient of Two Comple Numbers Given two comple numbers in polr form z nd z r cos i sin r cos i sin the product nd quotient of the numbers re s follows. z z r r cos i sin Product z z r r cos i sin, z Quotient LINEAR ALGEBRA APPLIED Elliptic curves re the foundtion for elliptic curve cryptogrphy (ECC), type of public key cryptogrphy for secure communictions over the Internet. ECC hs gined populrity due to its computtionl nd bndwidth dvntges over trditionl public key lgorithms. One specific vriety of elliptic curve is formed using Eisenstein integers. Eisenstein integers re comple numbers of the form z b, where i, nd nd b re integers. These numbers cn be grphed s intersection points of tringulr lttice in the comple plne. Dividing the comple plne by the lttice of ll Eisenstein integers results in n elliptic curve. thumb/shutterstock.com

186 8. Polr Form nd DeMoivre s Theorem 7 Theorem 8. sys tht to multiply two comple numbers in polr form, multiply moduli nd dd rguments. To divide two comple numbers, divide moduli nd subtrct rguments. (See Figure 8.9.) Imginry is Imginry is z z z z z θ + θ r r r r z θ θ Rel is r r r r z z θ θ θ θ Rel is To multiply z nd z : Multiply moduli nd dd rguments. Figure 8.9 To divide z nd z : Divide moduli nd subtrct rguments. Find z z nd z z for the comple numbers Multiplying nd Dividing in Polr Form nd SOLUTION Becuse nd re in polr form, pply Theorem 8., s follows. z multiply z z 5 divide z 5 cos i sin z cos z 5 z cos subtrct dd 6 i sin 6 i sin z cos i sin 6 subtrct dd cos 5 5 cos i sin 6. i sin REMARK Try verifying the division in Emple using the stndrd forms of z nd z. Use the stndrd forms of z nd z to check the multipliction in Emple. For instnce, z z 5 To verify tht this nswer is equivlent to the result in Emple, use the formuls for cos u v nd sin u v to obtin cos 5 cos 6 5 i 6 6 i i i. nd sin 5 sin i i i 6.

187 8 Chpter 8 Comple Vector Spces DEMOIVRE S THEOREM The finl topic in this section involves procedures for finding powers nd roots of comple numbers. Repeted use of multipliction in the polr form yields nd Similrly, nd z r cos i sin z r cos i sin r cos i sin r cos i sin z r cos i sin r cos i sin r cos i sin. z r cos i sin z 5 r 5 cos 5 i sin 5. This pttern leds to the net importnt theorem, nmed fter the French mthemticin Abrhm DeMoivre (667 75). You re sked to prove this theorem in Review Eercise 85. THEOREM 8.5 DeMoivre s Theorem If z rcos i sin nd n is ny positive integer, then z n r n cos n i sin n. Rising Comple Number to n Integer Power Find i nd write the result in stndrd form. SOLUTION First convert to polr form. For i, r which implies tht. By DeMoivre s Theorem, 96. So, nd i cos i sin. i cos i sin tn cos i sin 96cos 8 i sin 8 96 i

188 Recll tht consequence of the Fundmentl Theorem of Algebr is tht polynomil of degree n hs n zeros in the comple number system. So, polynomil such s p 6 hs si zeros, nd in this cse you cn find the si zeros by fctoring nd using the Qudrtic Formul. 6 Consequently, the zeros re ±, 8. Polr Form nd DeMoivre s Theorem 9 ± i, nd ± i. Ech of these numbers is clled sith root of. In generl, the nth root of comple number is defined s follows. Definition of the nth Root of Comple Number The comple number w bi is n nth root of the comple number z when z w n bi n. REMARK Note tht when k eceeds n, the roots begin to repet. For instnce, when k n, the ngle is n n n which yields the sme vlues for the sine nd cosine s k. DeMoivre s Theorem is useful in determining roots of comple numbers. To see how this is done, let w be n nth root of z, where w scos i sin nd z rcos i sin. Then, by DeMoivre s Theorem w n s n cos n i sin n nd becuse w n z, it follows tht s n cos n i sin n rcos i sin. Now, becuse the right nd left sides of this eqution represent equl comple numbers, equte moduli to obtin s n r, which implies tht s r, n nd equte principl rguments to conclude tht nd must differ by multiple of. Note tht r is positive rel number nd so s r n is lso positive rel number. Consequently, for some integer k, n which implies tht k n. k, n Finlly, substituting this vlue of into the polr form of w produces the result stted in the net theorem. THEOREM 8.6 The nth Roots of Comple Number For ny positive integer n, the comple number z rcos i sin hs ectly n distinct roots. These n roots re given by k nr cos n i sin where k,,,..., n. k n

189 Chpter 8 Comple Vector Spces n r π n π n The nth Roots of Comple Number Figure 8. Imginry is Rel is The formul for the nth roots of comple number hs nice geometric interprettion, s shown in Figure 8.. Becuse the nth roots ll hve the sme modulus (length) nr, they lie on circle of rdius nr with center t the origin. Furthermore, the n roots re eqully spced round the circle, becuse successive nth roots hve rguments tht differ by n. You hve lredy found the sith roots of by fctoring nd the Qudrtic Formul. Try solving the sme problem using Theorem 8.6 to get the roots shown in Figure 8.. When Theorem 8.6 is pplied to the rel number, the nth roots hve specil nme the nth roots of unity. + i Imginry is i i The Sith Roots of Unity Figure 8. + i Rel is Finding the nth Roots of Comple Number REMARK In Figure 8., note tht when ech of the four ngles 8, 58, 98, nd 8 is multiplied by, the result is of the form k. Determine the fourth roots of i. SOLUTION In polr form, i cos i sin so r nd. Then, by pplying Theorem 8.6, i cos k Setting k,,, nd, z cos i sin 8 8 z cos 5 5 i sin 8 8 z cos 9 9 i sin 8 8 z cos i sin 8 8 s shown in Figure 8.. cos k cos 8 5π 8 + i sin 5π 8 i sin i sin 8 Imginry is k. cos π 8 + i sin Rel is k π 8 cos 9π 8 + i sin Figure 8. 9π 8 cos π + i sin π 8 8

190 8. Eercises 8. Eercises Converting to Polr Form In Eercises, epress the comple number in polr form.. Imginry.. Imginry. Grphing nd Converting to Polr Form In Eercises 5 6, represent the comple number grphiclly, nd give the polr form of the number. (Use the principl rgument.) 5. i 6. i 7. i 5 8. i 9. 6i. i. 7.. i. i 5. i 6. 5 i Grphing nd Converting to Stndrd Form In Eercises 7 6, represent the comple number grphiclly, nd give the stndrd form of the number is cos i sin 7 7 cos i sin 5. cos i sin. 8 cos i sin 6. cos i sin. 5. 7cos i sin 6. 9cos i sin cos i sin 5 cos i sin is Rel is i Rel is Imginry is Imginry is i + i 6 6 cos 5 5 i sin 6 6 Rel is Rel is Multiplying nd Dividing in Polr Form In Eercises 7, perform the indicted opertion nd leve the result in polr form i sin cos cos i sin. [cos56 i sin56] cos5 i sin5... Finding Powers of Comple Numbers In Eercises 5, use DeMoivre s Theorem to find the indicted powers of the comple number. Epress the result in stndrd form. 5. i 6. i 6 7. i cos cos.5cos i sin.5cos i sin cos cos i sin cos6 i sin6 9cos i sin 5cos i sin i i 7 5 cos cos 5 i sin i sin cos i sin i sin 9 5 cos 6 5 cos i sin 5 i sin cos cos 6 cos 9 i sin i sin i sin Finding Squre Roots of Comple Number In Eercises 5 5, find the squre roots of the comple number. 5. i 6. 5i 7. i 8. 6i 9. i 5. i 5. i 5. i i sin 6 i sin 6

191 Chpter 8 Comple Vector Spces Finding nd Grphing nth Roots In Eercises 5 6, () use Theorem 8.6 to find the indicted roots, (b) represent ech of the roots grphiclly, nd (c) epress ech of the roots in stndrd form. 5. Squre roots: 6 cos i sin 5. Squre roots: 55. Fourth roots: 56. Fifth roots: 9 cos i sin 6 cos i sin cos 5 5 i sin Squre roots: 5i 58. Fourth roots: 65i 59. Cube roots: 5 i 6. Cube roots: i 6. Cube roots: 8 6. Fourth roots: 8i 6. Fourth roots: 6. Cube roots: Finding nd Grphing Solutions In Eercises 65 7, find ll the solutions of the eqution nd represent your solutions grphiclly i 66. 6i i 7. i 7. Electricl Engineering In n electric circuit, the formul V I Z reltes voltge drop V, current I, nd impednce Z, where comple numbers cn represent ech of these quntities. Find the impednce when the voltge drop is 5 5i nd the current is i. 7. Use the grph of the roots of comple number. () Write ech of the roots in trigonometric form. (b) Identify the comple number whose roots re given. Use grphing utility to verify your results. (i) Imginry is Rel is (ii) Imginry is Rel is 75. Proof When provided with two comple numbers z nd z r cos i sin r cos i sin, with z, prove tht z z r r cos i sin. 76. Proof Show tht the comple conjugte of z rcos i sin is z rcos i sin. 77. Use the polr forms of z nd z in Eercise 76 to find ech of the following. () zz (b) zz, z 78. Proof Show tht the negtive of z rcos i sin is z rcos i sin. 79. Writing () Let z rcos i sin cos 6 i sin 6. Sketch z, iz, nd zi in the comple plne. (b) Wht is the geometric effect of multiplying comple number z by i? Wht is the geometric effect of dividing z by i? 8. Clculus Recll tht the Mclurin series for e, sin, nd cos re e!!!... sin 5! 5! 7 7!... cos! 6! 6!.... () Substitute i in the series for e nd show tht e i cos i sin. (b) Show tht ny comple number z cn be epressed in polr form s z re i. bi (c) Prove tht if z re i, then z re i. (d) Prove the formul e i. True or Flse? In Eercises 8 nd 8, determine whether ech sttement is true or flse. If sttement is true, give reson or cite n pproprite sttement from the tet. If sttement is flse, provide n emple tht shows the sttement is not true in ll cses or cite n pproprite sttement from the tet. 8. Although the squre of the comple number bi is given by bi b, the bsolute vlue of the comple number z bi is defined s bi b. 8. Geometriclly, the nth roots of ny comple number z re ll eqully spced round the unit circle centered t the origin.

192 Answer Key Section cos i sin. 5. Imginry is Imginry is z = 6i 6 6 z = i Imginry is z = i 6 cos i sin. Imginry is 7cos i sin Rel is 8 cos i sin Rel is cos i sin Rel is Rel 6 is z = 7 6cos i sin. cos i sin 6 5. Imginry is 5 cos. i sin. 7. Imginry is θ = π i 9.. i 5 8 Imginry is Imginry is Imginry is z = i r = θ = 5 π θ = r =.75 5 i 8 z = + i 6 r = π Rel is Rel is Rel is Rel is Rel is

193 Answer Key. i Imginry is cos 6 i sin 6. cos i sin i i i cos i sin Squre roots: cos i sin i cos 5 i sin5 i 7. i cos i sin Squre roots: r = Imginry is θ = π 6 cos i sin 6 6 i cos i sin7 i 9. i cos 7 i sin7 Squre roots: 8 cos7 i sin i 8 cos5 i sin i Rel is θ = r = 7 Rel is cos i sin 9..5cos i sin 5. i cos i sin Squre roots: 6 cos i sin 6 5. () cos i sin 6 cos 7 7 i sin 6 6 (b) Imginry is (c) 55. () cos 7 i sin i 6 (b) (c) θ = 7 π 6 i i cos i sin cos 5 5 i sin 6 6 cos i sin cos i sin 6 6 θ = θ = 5π 6 π Imginry is i i i i 6 r = r = π θ = π θ = θ = 6 i Rel is Rel is π 6

194 Answer Key 57. () (b) (c) 59. () (b) (c) 6. () (b) 5 cos i sin 5 cos 7 7 i sin θ = θ = θ = π 5 π Imginry is π θ = (c) i i 6 θ= π 9 r = i π 9 r = 5 5 i Imginry is Imginry is r = 6 θ = 5 cos i sin cos i sin cos 6 6 i sin i.7.7i.8.i cos i sin cos i sin cos i sin 6 θ = 7π 6π 9 θ = Rel is Rel is Rel is 6. () 67. (b) (c) i i 65. cos i sin 8 cos 5 i sin5 8 8 cos 9 i sin9 8 8 cos i sin 8 8 θ = cos i sin cos i sin cos 5 5 i sin Imginry is θ = π cos i sin cos i sin cos i sin cos i sin Imginry is Imginry is 5π θ = 8 9π 8 θ = π π θ = π θ = θ= r = Rel is 8 r = π θ = 8 r = Rel is π θ = 5π θ = θ = π 8 Rel is

195 Answer Key 69. cos i sin 5 cos i sin 5 5 cos i sin cos 7 7 i sin 5 5 θ = π 5 θ = π Imginry is r = 5 cos 9 9 i sin 5 5 θ = π 5 Rel is 7. θ = 7π 5 cos i sin cos 7 7 i sin 6 6 cos i sin 6 6 Imginry is π θ = θ = 5 9π 5 θ = 7π 6 7. i 75. Proof 77. () r (b) cos i sin 79. () Imginry is iz = + i r = θ = Rel is π 6 z = + i Rel is z/i = i (b) Counterclockwise rottion of clockwise rottion of ; 8. True

196

197 8. Comple Vector Spces nd Inner Products. 8. Comple Vector Spces nd Inner Products. Objective: Represent vector in C n by bsis. Objective: Find the row spce nd null spce of mtri. Objective: Solve system of liner equtions. Objective: Find the Eucliden inner product, Eucliden norm, nd Eucliden distnce in C n A comple vector spce is vector spce in which the sclrs re comple numbers. The comple version of R n is the comple vector spce C n consisting of ordered n-tuples of comple numbers v = ( + b i, + b i, n + b n i) We lso represent vectors in C n by their coordinte mtrices reltive to the stndrd bsis e = (,,,, ), e = (,,,, ),, e n = (,,,, ), v = n bi b i bni As in of R n, the opertions of vector ddition nd sclr multipliction in C n re performed component by component. The dimension of C n is n. A bsis of subspce V of C n is ny linerly independent set of vectors tht spns V. If V hs dimension d, then ny linerly independent set of d vectors in V is bsis of V. Also, ny set of d vectors tht spns V is bsis of V. Emple: Is R subspce of the comple vector spce C over C? Solution: R is subset of C becuse R ={( + b i, + b i) where b = b = }. Moreover, p. 6

198 8. Comple Vector Spces nd Inner Products. Emple: Show tht S = {v, v, v } = {(i,, ), (, i, ), ( + i,, i)} is bsis for C. Solution: Becuse the dimension of C is, the set S will be bsis if it is linerly independent. So we must check tht Emple: Write u = ( i, 8 i, 5 i) s liner combintion of the vectors in the set S = {v, v, v } = {(i,, ), (, i, ), ( + i,, i)}, i.e. the sme set s in the previous emple. Solution: Emple: Find the dimension of nd bsis for S = spn{( i, i, 9 6i), ( i,, ), ( i, + i, 6 5i)}. Solution: S = the column spce of p. 6

199 8. Comple Vector Spces nd Inner Products. Emple: Find the rnk, nullity, nd bses for the row spce nd null spce of Solution: p. 65

200 8. Comple Vector Spces nd Inner Products. Emple: Solve Solution: Let u nd v be vectors in C n. The Eucliden inner product of u nd v is given by u v = u v * + u v * + + u n v n * = u v + u v + + u n v n Theorem 8.7 Properties of the Eucliden inner product Let u, v, nd w be vectors in C n nd let k be comple sclr. Then ) u v = (v u)* = v u Different from R n! ) (u + v) w = u w + v w ) (k u) v = k (u v) ) u (k v) = k*(u v) = k (u v) Different from R n! 5) u u (This lso sys tht u u is rel.) 6) u u = if nd only if u = p. 66

201 8. Comple Vector Spces nd Inner Products. The Eucliden norm of u in C n is u = u u The Eucliden distnce between u nd v in C n is d(u, v) = u v To find dot product u v on the TI-89, use Mtri Vector opsdotp(u,v) To find dot product u v in Mthemtic, use u.vconj which displys s u.v* To find dot product u v in PocketCAS, use dot(v,u) (PocketCAS tkes the comple conjugte of the first vector, rther thn the second.) To find norm v on the TI-89, use Mtri Normsnorm(v) To find norm v in Mthemtic, use Norm[v] To find norm v in PocketCAS, use lnorm(v) or norm(v) Emple: Let u = (i,, ), v = ( i,, ), nd w = ( + i, 5 + i, 6i). Find u Find w Find u v Find v w Find d(u, w) p. 67

202

203 8. Comple Vector Spces nd Inner Products 8. Comple Vector Spces nd Inner Products Recognize nd perform vector opertions in comple vector spces C n, represent vector in C n by bsis, nd find the Eucliden inner product, Eucliden norm, nd Eucliden distnce in C n. Recognize comple inner product spces. COMPLEX VECTOR SPACES All the vector spces studied so fr in the tet hve been rel vector spces becuse the sclrs hve been rel numbers. A comple vector spce is one in which the sclrs re comple numbers. So, if v, v,..., v m re vectors in comple vector spce, then liner combintion is of the form where the sclrs c re comple numbers. The comple version of R n, c,..., c is the comple vector spce C n m consisting of ordered n-tuples of comple numbers. So, vector in C n hs the form It is lso convenient to represent vectors in by column mtrices of the form As with R n, the opertions of ddition nd sclr multipliction in C n re performed component by component. Vector Opertions in Let v i, i nd u i, be vectors in the comple vector spce C. Determine ech vector.. b. c. v u iv v 5 iu SOLUTION. In column mtri form, the sum v u is b. Becuse i i 5i nd i i 7 i, iv i i, i c. c v c v v b i, b i,..., n b n i. v b i b i. n b n i. c m v m v u i i i i 7 i. 5i, 7 i. C n v 5 iu i, i 5 i i, 6i, 9 i 9 7i, i i, i C n

204 Chpter 8 Comple Vector Spces Mny of the properties of R n re shred by C n. For instnce, the sclr multiplictive identity is the sclr nd the dditive identity in C n is,,,...,. The stndrd bsis for C n is simply e,,,..., e,,,...,.. e n,,,..., which is the stndrd bsis for R n. Becuse this bsis contins n vectors, it follows tht the dimension of C n is n. Other bses eist; in fct, ny linerly independent set of n vectors in C n cn be used, s demonstrted in Emple. Verifying Bsis Show tht S v is bsis for C, v, v i,,, i, i,,,, i. SOLUTION Becuse the dimension of C is, the set v, v, v will be bsis if it is linerly independent. To check for liner independence, set liner combintion of the vectors in S equl to, s follows. c v c v c v,, c i,, c i, c i,,, c i,, c c i, c i, c i,, This implies tht c c i c i c i. So, c c c, nd v, v, v is linerly independent. Representing Vector in by Bsis Use the bsis S in Emple to represent the vector v, i, i. SOLUTION By writing C n REMARK Try verifying tht this liner combintion yields, i, i. v c v c v c v c c i, c i, c i, i, i you cn obtin c c i which implies tht c, c i i c i i c i i i, nd So, v iv v iv. c i i i.

205 8. Comple Vector Spces nd Inner Products 5 Other thn C n, there re severl dditionl emples of comple vector spces. For instnce, the set of m n comple mtrices with mtri ddition nd sclr multipliction forms comple vector spce. Emple describes comple vector spce in which the vectors re functions. The Spce of Comple-Vlued Functions Consider the set S of comple-vlued functions of the form f f if where f nd f re rel-vlued functions of rel vrible. The set of comple numbers forms the sclrs for S, nd vector ddition is defined by f g f if g ( i g f g if g. It cn be shown tht S, sclr multipliction, nd vector ddition form comple vector spce. For instnce, to show tht S is closed under sclr multipliction, let c bi be comple number. Then cf bif if f bf ibf f is in S. REMARK Note tht if u nd v hppen to be rel, then this definition grees with the stndrd inner (or dot) product in R n. The definition of the Eucliden inner product in C n is similr to the stndrd dot product in R n, ecept tht here the second fctor in ech term is comple conjugte. Definition of the Eucliden Inner Product in Let u nd v be vectors in C n. The Eucliden inner product of u nd v is given by u v u v u v u n v n. C n Finding the Eucliden Inner Product in Determine the Eucliden inner product of the vectors u i,, 5i nd v i, i,. SOLUTION u v u v u v u v i i i 5i i Severl properties of the Eucliden inner product C n C re stted in the following theorem. THEOREM 8.7 Properties of the Eucliden Inner Product Let u, v, nd w be vectors in C n nd let k be comple number. Then the following properties re true. u v w u w v w. u v v u.. ku v ku v. u kv ku v 5. u u 6. u u if nd only if u.

206 6 Chpter 8 Comple Vector Spces PROOF The proof of the first property is shown below, nd the proofs of the remining properties hve been left to you (see Eercises 59 6). Let Then u u, u,..., u n nd v u v u v u... v n u n v u v u... v n u n v u v u... v n u n u v u v u v. v v, v,..., v n. u n v n C n The Eucliden inner product in is used to define the Eucliden norm (or length) of vector in C n nd the Eucliden distnce between two vectors in C n. Definitions of the Eucliden Norm nd Distnce in The Eucliden norm (or length) of u in C n is denoted by u nd is u u u. The Eucliden distnce between u nd v is du, v u v. C n The Eucliden norm nd distnce my be epressed in terms of components, s follows (see Eercise 5). du, v u v u v... u n v n Finding the Eucliden Norm nd Distnce in Let u i,, 5i nd v i, i,.. Find the norms of u nd v. b. Find the distnce between u nd v. SOLUTION. u u u... u n u u u u v v v v 5 7 C n b. du, v u v, i, 5i 5 5 7

207 8. Comple Vector Spces nd Inner Products 7 COMPLEX INNER PRODUCT SPACES The Eucliden inner product is the most commonly used inner product in. On occsion, however, it is useful to consider other inner products. To generlize the notion of n inner product, use the properties listed in Theorem 8.7. C n Definition of Comple Inner Product Let u nd v be vectors in comple vector spce. A function tht ssocites u nd v with the comple number u, v is clled comple inner product when it stisfies the following properties.. u, v v, u. u v, w u, w v, w. ku, v ku, v. u, u nd u, u if nd only if u. A comple vector spce with comple inner product is clled comple inner product spce or unitry spce. A Comple Inner Product Spce Let u u nd be vectors in the comple spce C, u v v, v. Show tht the function defined by u, v u v u v is comple inner product. SOLUTION Verify the four properties of comple inner product, s follows.. v, u v u v u u v u v u, v. u v, w u v w u v w u w u w v w v w u, w v, w. ku, v ku v ku v ku v u v k u, v. u, u u u u u u u Moreover, u, u if nd only if u u. Becuse ll the properties hold, u, v is comple inner product. LINEAR ALGEBRA APPLIED Comple vector spces nd inner products hve n importnt ppliction clled the Fourier trnsform, which decomposes function into sum of orthogonl bsis functions. The given function is projected onto the stndrd bsis functions for vrying frequencies to get the Fourier mplitudes for ech frequency. Like Fourier coefficients nd the Fourier pproimtion, this trnsform is nmed fter the French mthemticin Jen-Bptiste Joseph Fourier (768 8). The Fourier trnsform is integrl to the study of signl processing. To understnd the bsic premise of this trnsform, imgine striking two pino keys simultneously. Your er receives only one signl, the mied sound of the two notes, nd yet your brin is ble to seprte the notes. The Fourier trnsform gives mthemticl wy to tke signl nd seprte out its frequency components. Eliks/Shutterstock.com

208 8 Chpter 8 Comple Vector Spces 8. Eercises Vector Opertions In Eercises 8, perform the indicted opertion using u i, i, v i, i, nd w i, 6.. u. iw. iw. iv w 5. u iv 6. 6 iv iw 7. u iv iw 8. iv iw u Liner Dependence or Independence In Eercises 9, determine whether the set of vectors is linerly independent or linerly dependent. 9., i, i,. i, i,, i,,,, i,., i, i,, i, i,,,. i, i,, i,,,,, Verifying Bsis In Eercises 6, determine whether S is bsis for C n.. S, i, i,. S, i, i, 5. S i,,,, i, i,,, 6. S i,,,, i, i, i,, Representing Vector by Bsis In Eercises 7, epress v s liner combintion of ech of the following bsis vectors. () {i,,, i, i,, i, i, i} (b) {,,,,,,,, i 7. v,, 8. v i, i, 9. v i, i,. v i, i, i Finding Eucliden Inner Products In Eercises nd, determine the Eucliden inner product u v.. u i, i, i. u i, i, v i,, i v i,, i Properties of Eucliden Inner Products In Eercises 6, let u i, i, v i, i, w i,, nd k i. Evlute the epressions in prts () nd (b) to verify tht they re equl.. () u v. () u v w (b) v u (b) u w v w 5. () ku v 6. () u kv (b) ku v (b) ku v Finding the Eucliden Norm In Eercises 7, determine the Eucliden norm of v. 7. v i, i 8. v, 9. v 6 i, i. v i, i. v, i, i. v,,. v i, i, i, i. v, i, i, i Finding the Eucliden Distnce In Eercises 5, determine the Eucliden distnce between u nd v. 5. u,, v i, i 6. u i,, i, v i,, i 7. u i, i, i, v,, 8. u, i, i, v i, i, i 9. u,, v,. u,,, i, v i, i, i, Comple Inner Products In Eercises, determine whether the function is comple inner product, where u u, u nd v v, v.. u, v u u v. u, v u v u v. u, v u v 6u v. u, v u v u v Finding Comple Inner Products In Eercises 5 8, use the inner product u, v u v u v to find u, v. 5. u i, i nd v i, i 6. u i, i nd v i, i 7. u i, i nd v i, i 8. u i, nd v i, Finding Comple Inner Products In Eercises 9 nd 5, use the inner product u, v u v u v u v u v where nd v [ v v u [ u u u u ] v v ] to find u, v. 9. u i i v 5. v i u i i i i i i

209 8. Eercises 9 5. Let u b i, b i,..., n b n i. () Use the definitions of Eucliden norm nd Eucliden inner product to show tht (b) Use the results of prt () to show tht 5. Let v i,, nd v i, i,. If v z, z, z nd the set v is not bsis for C, v, v, wht does this imply bout z, z, nd z? 5. Let v i, i, i nd v,,. Determine vector such tht v is bsis for C, v, v. v u u u... u n. du, v u v u v... u n v n. 5. The comple Eucliden inner product of u nd v is sometimes clled the comple dot product. Compre the properties of the comple dot product in C n nd those of the dot product in R n. () Which properties re the sme? Which properties re different? (b) Eplin the resons for the differences. Properties of Comple Inner Products In Eercises 55 58, verify the sttement using the properties of comple inner product. 55. u, kv w ku, v u, w 56. u, 57. u, v u, v v, u 58. u, kv ku, v Proof In Eercises 59 6, prove the property, where u, v, nd w re vectors in C n nd k is comple number. 59. u v w u w v w 6. ku v ku v 6. u kv ku v 6. u u 6. u u if nd only if u. 6. Writing Let u, v be comple inner product nd let k be comple number. How re u, v nd u, kv relted? Finding Liner Trnsformtion In Eercises 65 nd 66, determine the liner trnsformtion T : C m C n tht hs the given chrcteristics. 65. T, i, T,, i 66. Ti, i, T, i, i Finding n Imge nd Preimge In Eercises 67 7, the liner trnsformtion T : C m C n is shown by Tv Av. Find the imge of v nd the preimge of w. 67. A w v i i i, i, i 68. w v i A i i,, A i i A i i, i i v i i,, v 5, w i i w i i i 7. Find the kernel of the liner trnsformtion in Eercise Find the kernel of the liner trnsformtion in Eercise 69. Finding n Imge In Eercises 7 nd 7, find the imge of v i, i for the indicted composition, where T nd T re the mtrices below. nd T [ i i T [ i i ] i i] 7. T T 7. T T 75. Determine which of the sets below re subspces of the vector spce of comple mtrices. () The set of symmetric mtrices. (b) The set of mtrices A stisfying A T A. (c) The set of mtrices in which ll entries re rel. (d) The set of digonl mtrices. 76. Determine which of the sets below re subspces of the vector spce of comple-vlued functions (see Emple ). () The set of ll functions f stisfying fi. (b) The set of ll functions f stisfying f. (c) The set of ll functions f stisfying fi fi. True or Flse? In Eercises 77 nd 78, determine whether ech sttement is true or flse. If sttement is true, give reson or cite n pproprite sttement from the tet. If sttement is flse, provide n emple tht shows the sttement is not true in ll cses or cite n pproprite sttement from the tet. 77. Using the Eucliden inner product of nd in u v u v u v... u v C n, u n v n. 78. The Eucliden norm of u in C n denoted by u is u u.

210 Answer Key Section 8.. i, 9 i. 8 i, 6 i 5. 5 i, 7. 9 i, i 9. Linerly dependent. Linerly independent. S is not bsis for C. 5. S is bsis for C. 7. (),, ii,, ii, i, i, i, i (b),,,,,,,, i 9. () i, i, ii,, ii, i, ii, i, i (b) i, i, i,, i,, i,, i. i. () (b) i i The epressions re equl. 5. () (b) i i The epressions re equl Not comple inner product. A comple inner product i 9. i 5. () nd (b) Proofs 5. z, z nd z cn be ny comple numbers. 55. u, kv w u, kv u, w ku, v u, w 57. u, v u, v v, u v, u v, u u v, u Proofs 65. T u i i u 67. i, i 69. i i 7. kert, t, ti, where t R (), (b), nd (d) re subspces. 77. Flse. u v u v u v... u n v n 5i,

211 Chpter 6: Liner Trnsformtions. 6. Introduction to Liner Trnsformtions. Chpter 6: Liner Trnsformtions. 6. Introduction to Liner Trnsformtions. Objective: Find the imge nd preimge of function Objective: Show tht function is liner trnsformtion. Recognize nd describe some common liner trnsformtions. A function T: V W mps vector spce V to W. V is the domin nd W is the codomin of S* T. If v V nd w W such tht T (v) = w then w is the imge of v under T. The set of ll imges of vectors in V is clled the rnge of T. The set of ll vectors in V such tht T (v) = w is the preimge of w. preimge domin v V T T T T imge rnge codomin w rnge W Let V nd W be vector spces. Then the function (or mp) T : V W is liner trnsformtion when the following two properties re true for ll v V nd w W nd for ny sclr c. And T()* ) T(u + v) = T(u) + T(v) ) T(cu) = c T(u) A liner trnsformtion is opertion preserving becuse pplying n opertion in V (before T ) gives the sme result s pplying the corresponding opertion in W (fter T ). ddition in V ddition in W sclr sclr multipliction multipliction in V in W T(u + v) = T(u) + T(v) T(cu) = c T(u) Emple: Show tht T : R R given by T(v, v ) = (v v, v + v ) is liner trnsformtion, find the imge of (, ), nd find the preimge of (, 8). [Note: techniclly, we should write T((v, v )) becuse v = (v, v ).] p. 69

212 Chpter 6: Liner Trnsformtions. 6. Introduction to Liner Trnsformtions. Solution: T(u + v) = T(u + v, u + v ) = ((u + v ) (u + v ), (u + v ) + (u + v )) = (u u, u + u ) + (v v, v + v ) = T(u) + T(v) T(cu) = T(cv, cv ) (cv cv, cv + cv ) (c(v v ), c(v + v )) c (v v, v + v ) =c T(u) Imge of (, ): T(, ) = ( + ( ), ( )) = (9, 7) Preimge of (, 8): T(v, v ) = (v v, v + v ) = (, 8) v v = v = v + v = 8 v = 6 (v, v ) = (5, ) S* Emple: Show tht f : R R given by f () = is not liner trnsformtion. Solution: f ( + y) = ( + y) = + y + y. Usully, this is not equl to f () + f (y) = + y. In prticulr, specific counteremple is f ( + ) =, but f () + f () =. Emple: Show tht f : R R given by f () = + is not liner trnsformtion. S* Solution: f ( + y) = + y +, which is never equl to f () + f (y) = + + y +. [However, f () = + is liner function, becuse its grph is stright line.] Two simple liner trnsformtions re the zero trnsformtion T : V W, T(v) = for ll v V nd the identity trnsformtion T : V V, T(v) = v for ll v V. Theorem 6. Properties of Liner Trnsformtions Let T: V W be liner trnsformtion, nd u nd v be vectors in V. Then ) T() = ) T( v) = T(v) ) T(u v) = T(u) T(v) ) If v = c v + c v + + c n v n then T(v) = T(c v + c v + + c n v n ) = c T(v ) + c T(v ) + + c n T(v n ) p. 7

213 Chpter 6: Liner Trnsformtions. 6. Introduction to Liner Trnsformtions. Proof: ) T() = ) T( v) = ) T(u v) = ) Prove T(c v + c v + + c n v n ) = c T(v ) + c T(v ) + + c n T(v n ) by induction: T(c v ) = c T(v ) by the second property of linerity Suppose T(c v + c v + + c n v n ) = c T(v ) + c T(v ) + + c n T(v n ). Then T(c v + c v + + c n v n + c n+ v n+ ) = T(c v + c v + + c n v n ) + T(c n+ v n+ ) by the first property of linerity = c T(v ) + c T(v ) + + c n T(v n ) + T(c n+ v n+ ) by supposition = c T(v ) + c T(v ) + + c n T(v n ) + c n+ T(v n+ ) by the second property of linerity Property of Theorem 6. tells us tht liner trnsformtion is completely determined by its ction on bsis for V. In other words, if {v, v,, v n } is bsis for V nd if T(v ), T(v ),, T(v n ) re known, then T(v) is determined for ny v V. Emple: Let T: R R be liner trnsformtion such tht T(,, ) = (,, ) T(,, ) = (, 5, 9) T(,, ) = (, 6, 5) Find T(, 7, ). Solution: T(, 7, ) = T(,, ) + 7 T(,, ) + T(,, ) = (,, ) + 7(, 5, 9) + (, 6, 5) = (7, 6, 5) p. 7

214 Chpter 6: Liner Trnsformtions. 6. Introduction to Liner Trnsformtions. p. 7 Emple: Let T: R R the function such tht T(v) = Av = v v Show tht T is liner function. Find T(, ) nd T(, ). Solution: For ny u nd v in R, T(u + v) = A(u + v) = Au + Av = T(u) + T(v). Also, for ny v in R nd ny sclr c, T(cv) = A(cv) = c(av) = ct(v). T(, ) = = 6 7 = (7,, 6) nd T(, ) = = 7 5 = (5,, 7) Notice tht T(, ) nd T(, ) re just the two columns of A. Theorem 6.: The Liner Trnsformtion Given by Mtri Let A be n mn mtri. The function T: R n R m the defined by T(v) = Av is liner trnsformtion. Vectors in R n re represented by n (column) mtrices nd vectors in R m re represented by m (column) mtrices. T(v) = n m mn m m n n n v n v v = m n mn m m n n n n v v v v v v v v v Emple: Rottion. We show tht the liner trnsformtion T: R R represented by the mtri A = ) cos( ) sin( ) sin( ) cos( rottes every vector in R counterclockwise by n ngle bout the origin.

215 Chpter 6: Liner Trnsformtions. 6. Introduction to Liner Trnsformtions. r cos( ) cos( ) sin( ) r cos( ) T = r sin( ) sin( ) cos( ) r sin( ) cos( )cos( ) sin( )sin( ) = r sin( )cos( ) cos( )sin( ) r cos( ) = r sin( ) T(.5v) T(u + v) u + v T(u) T(v) u v.5v Observe tht T is liner trnsformtion. It preserves vector ddition nd sclr multipliction. Emple: projection. The liner trnsformtion T: R R represented by the mtri A = projects vector v = (, y, z) to T(v) = (, y, ). In other words, T mps every vector in R to its orthogonl projection in the -y plne. Emple: Trnspose. We show tht the liner trnsformtion T: M m,n M n,m given by T(A) = A T is liner trnsformtion. If A nd B re mn mtrices, then T(A + B) = (A + B) T = A T + B T = T(A) + T(B). If A is n mn mtri nd c is sclr, then T(cA = (ca) T = c(a T ) = c T(A). p. 7

216 Chpter 6: Liner Trnsformtions. 6. Introduction to Liner Trnsformtions. Emple: The Differentil Opertor. Let C[, b], lso written s C [, b], be the set of ll functions whose derivtives re continuous on the intervl [, b]. We show tht the differentil opertor D defines liner trnsformtion from C[, b] into C[, b]. d Using opertor nottion, we write D ( f ) = [ f ], where f is in C[, b]. d From clculus, we know tht for ny functions f nd g in C[, b], d d d D ( f + g) = [ f + g] = [ f ] + [ g] = D ( f ) + D ( g) C[, b] d d d nd for ny function f in C[, b] nd for ny sclr c, d d D (c f ) = [c f ] = c [ f ] = c D ( f ) C[, b] d d Emple: The Definite Integrl Opertor. Let P be the vector spce of ll polynomil functions. b We show tht the definite integrl opertor T: P R defined by T(p) = p ( ) d is liner trnsformtion. From clculus, we know tht for ny polynomils p nd q in P, b T(p + q) = [ p q]( ) d = [ p ( ) q( )] d = p ( ) d + q ( ) d = T(p) + T(q) nd for ny polynomil p in P nd for ny sclr c, b b b T(cp) = [ cp ]( ) d = cp ( ) d = c p( ) d = ct(p). b b b p. 7

217 Chpter 6: Liner Trnsformtions. 6. The Kernel nd Rnge of Liner Trnsformtion. 6. The Kernel nd Rnge of Liner Trnsformtion. Objective: Find the kernel of liner trnsformtion. Objective: Find bsis for the rnge, the rnk, nd the nullity of liner trnsformtion. Objective: Determine whether liner trnsformtion is one-to-one or onto. Objective: Prove bsic results bout one-to-one nd/or onto liner trnsformtions. Objective: Determine whether vector spces re isomorphic. S* Let T: V W be liner trnsformtion. Then the set of ll vectors v in V such tht T(v) = is the kernel of T nd is denoted by ker(t). Remember tht for ny liner trnsformtion, T() =, so ker(t) lwys contins. Emple Find the kernel of the projection T: R R defined by T(, y, z) = (, y, ). Solution: The solution to T(, y, z) = (,, ) is =, y =. So ker(t) = {(,, z): z R} Emple Find the kernel of the zero trnsformtion T: V W defined T(v) =. Solution: T(v) = is true for ll v V so ker(t) = V Emple Find the kernel of the identity trnsformtion T: V V defined T(v) = v. Solution: T(v) = mens v = so ker(t) = {}. Emple Find the kernel of the liner trnsformtion T: R R defined by T(,, ) = (5 6, 8 ). Solution: T(,, ) = (, ) yields the system rref 5 which hs the solution = t, = t, = t so ker(t) = {( t, t, t): t R} 5 5 Theorem 6.: The Kernel of T: V W is Subspce of V The kernel of liner trnsformtion T: V W is subspce of the domin V. Proof: ker(t) so ker(t) is not empty. If u, v ker(t), then T(u) = nd T(v) = so T(u + v) = T(u) + T(v) = so u + v ker(t) p. 75

218 Chpter 6: Liner Trnsformtions. 6. The Kernel nd Rnge of Liner Trnsformtion. If c is sclr nd v ker(t), then T(v) = so T(cv) = ct(v) = so cv ker(t) Emple Find bsis for the kernel of the liner trnsformtion T: R 5 R defined by T() = A, where Solution: ker(t) = the nullspce of A (i.e. the solution spce of A = ). S* S* Let T: V W be liner trnsformtion. Then the set of ll vectors w in W tht re imges of vectors in V is the rnge of T, nd is denoted by rnge(t). Tht is, rnge(t) = {T(v): v V} kernel T domin T rnge codomin p. 76

219 Chpter 6: Liner Trnsformtions. 6. The Kernel nd Rnge of Liner Trnsformtion. Theorem 6.: The Rnge of T: V W is Subspce of W The rnge of liner trnsformtion T: V W is subspce of the codomin W. Proof: rnge(t) becuse T() =, so rnge(t) is not empty. If T(u) nd T(v) re vectors in rnge(t), then u nd v re vectors in V. V is closed under ddition, so u + v V, so T(u + v) rnge(t). T(u + v) = T(u) + T(v), so T(u) nd T(v) rnge(t) implies tht T(u) + T(v) rnge(t). Let c be sclr nd T(v) rnge(t) Then v V. Since V is closed under sclr multipliction, cv V, so T(cv) rnge(t). T(cv) = ct(v), so Let c be sclr nd T(v) rnge(t) implies tht ct(v) rnge(t). so T(cv) = ct(v) = so cv ker(t) Emple Find bsis for the rnge of the liner trnsformtion T: R 5 R defined by T() = A, where S* Solution: rnge(t) = the column spce of A. Let T: V W be liner trnsformtion. The dimension of the kernel of T is clled the nullity S* of T nd is denoted by nullity(t), the dimension of the rnge of T is clled the rnk of T nd is denoted by rnk(t). p. 77

220 Chpter 6: Liner Trnsformtions. 6. The Kernel nd Rnge of Liner Trnsformtion. Theorem 6.5: The Sum of Rnk nd Nullity S* Let T: V W be liner trnsformtion from the n-dimensionl vector spce V into vector spce W. Then rnk(t ) + nullity(t ) = n i.e. dim(rnge(t )) + dim(kernel(t )) = dim(domin(t )) Emple Find the rnk nd nullity of the liner trnsformtion T: R 8 R 5 defined by T() = A, where Solution: S* A function T: V W is one-to-one (injective) if nd only if every vector in the rnge of T hs single (unique) preimge. An equivlent definition is tht T is one-to one if nd only if T(u) = T(v) implies tht u = v. S* A function T: V W is onto (surjective) if nd only if the rnge of T is W (the codomin of T ). And equivlent definition is tht tht T is one-to one if nd only if every element in W hs preimge in V. Emple: T: R R defined by T(, y) = (, y, ) is one-to-one but not onto (points with z hve no preimge). Emple: The projection T: R R defined by T(, y, z) = (, y) is onto but not one-to-one. p. 78

221 Chpter 6: Liner Trnsformtions. 6. The Kernel nd Rnge of Liner Trnsformtion. Emple: The projection T: R R defined by T(, y) = ( y, 6y) = ( y) (, ) is neither one-to-one (becuse there re mny different (, y) pirs tht give the sme vlue of y ) nor onto (becuse points with y re not in the rnge of T). Emple: The trnsformtion T: R n R n defined by T(v) = Av, where A is n invertible mtri, is one-to-one nd onto. T is one-to-one becuse T(u) = T(v) implies Au = Av. Since A is invertible, we cn pre-multiply by A to obtin A Au = A Av or u = v. T is onto becuse the rnk of n invertible nn mtri (nmely, A) is n. So we lso hve rnk(t ) = n. By Theorem 6.7 (below), T is onto becuse its rnk equls the dimension of its codomin. Theorem 6.6: One-to-One Liner Trnsformtions Let T: V W be liner trnsformtion. Then T is one-to-one if nd only if ker(t ) = {}. Proof S* If T is one-to-one, then T(v) = hs only one solution, nmely, v =. Thus, ker(t ) = {}. Conversely, suppose ker(t ) = {} nd T(u) = T(v). Becuse T is liner trnsformtion, T(u v) = so u v = becuse Therefore, p. 79

222 Chpter 6: Liner Trnsformtions. 6. The Kernel nd Rnge of Liner Trnsformtion. Theorem 6.7: Onto Liner Trnsformtions Let T: V W be liner trnsformtion, where W is finite dimensionl. Then T is onto if nd only rnk(t ) = dim(w). Proof Let T: V W be liner trnsformtion. S* If T is onto, then W is equl to the rnge of T, so rnk(t ) = dim(rnge(t)) = dim(w). Let dim(w) = n. If rnk(t ) = n, then there re n linerly independent vectors T(v ), T(v ),, T(v n ) in the rnge of T. Since the rnge of T is in W, the vectors T(v ), T(v ),, T(v n ) re linerly independent W. By Thm.. (n linerly independent vectors in n n-dimensionl spce form bsis), the vectors T(v ), T(v ),, T(v n ) form bsis for W. So ny vector w W cn be written s liner combintion w = c T(v ) + c T(v ) + + c n T(v n ) = T(c v + c v + + c n v n ), which is in the rnge of T. Therefore, T is onto. Theorem 6.8: One-to-One nd Onto Liner Trnsformtions Let T: V W be liner trnsformtion, where vector spces V nd W both hve dimension n. Then T is one-to-one if nd only if it is onto. Proof If T is one-to-one, then by Thm 6.6, ker(t ) = {}, nd dim(ker(t )) =. By Thm 6.5, S* dim(rnge(t)) = Therefore, by Thm 6.7, T is onto. Conversely, if T is onto, then dim(rnge(t)) = dim(w) = n so by Thm 6.5, dim(ker(t)) = Therefore, ker(t ) = {}, nd by Thm 6.6, T is one-to-one. p. 8

223 Chpter 6: Liner Trnsformtions. 6. The Kernel nd Rnge of Liner Trnsformtion. Emple: Consider the liner trnsformtions Let T: V W represented by T() = A. Find the rnk nd nullity of T, nd determine whether T is one-to-one or onto. (Notice tht ll of the mtrices re in row-echelon form.) ) A = 5 b) A = 5 c) A = d) A = 5 dim(codomin) m dim(domin) n dim(rnge) r = rnk(t) dim(kernel) = nullity(t) One-to-one? Onto? ) b) c) d) A function tht is both one-to-one (injective) nd onto (surjective) is clled bijective. A bijective function is invertible. A liner trnsformtion T: V W tht is one-to-one nd onto is clled n isomorphism. If V nd S* W re vector spces such tht there eists n isomorphism from V to W, then V nd W re sid to be isomorphic to ech other. p. 8

224 Chpter 6: Liner Trnsformtions. 6. The Kernel nd Rnge of Liner Trnsformtion. Theorem 6.9: Isomorphic Spces nd Dimension Two finite-dimensionl vector spces V nd W re isomorphic if nd only if they re of the sme dimension. Proof Assume V is isomorphic to W, nd V hs dimension n. Then there eists liner trnsformtion T: V W tht is one-to-one nd onto. Becuse T is one-to-one, dim(ker(t)) =, so dim(rnge(t)) = Moreover, becuse T is onto, dim(w) = so V nd W re of the sme dimension. Conversely, ssume V nd W both hve dimension n. Let {v, v, v n } be bsis for V, nd S* {w, w, w n } be bsis for W. Then n rbitrry vector v V cn be uniquely written s v = c v + c v + + c n v n. Define the liner trnsformtion T: V W by T(v) = c w + c w + + c n w n. Then T is one-to-one: If T(v) =, so ker(t ) = {}, nd by Thm 6.6, T is one-to-one. Also, T is onto: Since the bsis {w, w, w n } is contined in the rnge of T, which is contined in W. Thus, rnk(t ) = n nd Thm 6.7, T is onto. p. 8

225 Chpter 6: Liner Trnsformtions. 6. Mtrices for Liner Trnsformtions. 6. Mtrices for Liner Trnsformtions. Objective: Find the stndrd mtri for liner trnsformtion. Objective: Find the stndrd mtri for the composition of liner trnsformtions nd find the inverse on n invertible liner trnsformtion. Objective: Find the mtri for liner trnsformtion reltive to nonstndrd bsis. Objective: Prove results in liner lgebr using linerity. Recll Theorem 6., Prt : If v = c v + c v + + c n v n then T(v) = T(c v + c v + + c n v n ) = c T(v ) + c T(v ) + + c n T(v n ) This sys tht liner trnsformtion T: V W is completely determined by its ction on bsis of V. In other words, if {v, v,, v n } is bsis of V nd T(v ), T(v ),, T(v n ) re given, then T(v) is determined for ny v in V. This is the key to representing T by mtri. Recll tht the stndrd bsis for R n in column vector nottion is S = {e, e,, e n } =,,, Theorem 6.: Stndrd Mtri for Liner Trnsformtion Let T: R n R m be liner trnsformtion such tht for the stndrd bsis vectors e i of R n, n T(e ) =, T(e ) =,, T(e n ) = n. m m mn S* Then the mn mtri whose n columns correspond to T(e i ) n A = n m m mn is such tht T(v) = Av for every v in R n. A is clled the stndrd mtri for T. Proof Choose ny v = v e + v e + + v n e n in R n. Then T(v) = T(v e + v e + + v n e n ) = v T(e ) + v T(e ) + + v n T(e n ) p. 8

226 Chpter 6: Liner Trnsformtions. 6. Mtrices for Liner Trnsformtions. p. 8 S On the other hnd, in column vector nottion, v = v n v v Av = mn m m n n v n v v = n mn m m n n n n v v v v v v v v v = v m + v m + + v n mn n n = v T(e ) + v T(e ) + + v n T(e n ). Emple Find the stndrd mtri for the liner trnsformtion T(, y) = ( y, y, y ). Solution:. T: R R. Emple Find the stndrd mtri A for the liner trnsformtion T tht is the reflection in the line y = in R, use A to find the imge of the vector v = (, ) nd sketch the grph of v nd its imge T(v). Solution: A = [T(e ), T(e )] = T(v) = T(, ) =

227 Chpter 6: Liner Trnsformtions. 6. Mtrices for Liner Trnsformtions. S* The composition of T : R n R m with T : R m R p is defined by T(v) = T (T (v)) where v is vector in R n. This composition is denoted by T T. Theorem 6.: Compostion of Liner Trnsformtions S* Let T : R n R m with T : R m R p be liner trnsformtions with stndrd mtrices A nd A, respectively. The composition T: R n R m, defined by T(v) = T (T (v)), is liner trnsformtion. Moreover, the stndrd mtri A for T is given by the mtri product A = A A. Proof To show tht T is liner trnsformtion, let u nd v be vectors in R n nd let c be sclr. Then T(u + v) = T (T (u + v)) = T (T (u) + T (v)) = T (T (u)) + T (T (v)) = T(u) + T(v) Also, T(cv) = T (T (cv)) = T (ct (v)) = ct (T (v)) = ct(v) To show tht A A is the stndrd mtri for T, T(v) = T (T (v)) = T (A v) = A (A v) = (A A )v using the ssocitive property of mtri multipliction Composition is not commuttive becuse mtri multipliction is not commuttive. We cn generlize to compositions of three or more liner trnsformtions. For emple, T = T T T is defined by T(v) = T (T (T (v))) nd is represented by A A A. p. 85

228 Chpter 6: Liner Trnsformtions. 6. Mtrices for Liner Trnsformtions. Emple Given the liner trnsformtions T : R R with T : R R defined by T = nd T = Find the stndrd mtri for T = T S T. Solution: A liner trnsformtion T : R n R n is invertible if nd only if there eists liner trnsformtion T : R n R n such tht for every v in R n, T (T (v)) = v nd T (T (v)) = v T is the inverse of T. When the inverse eists, it is unique, nd we denote it by T. Theorem 6. Eistence of n Inverse Trnsformtion Let T: R n R n be liner trnsformtion with stndrd mtri A. Then the following conditions re equivlent. ) T is invertible. ) T is n isomorphism. ) A is invertible. If T is invertible with stndrd mtri A, then the stndrd mtri for T is A. Emple Given the liner trnsformtions T: R R defined by T(,, ) = (, +, + + ). Show tht T is invertible, nd find its inverse. Solution: S p. 86

229 Chpter 6: Liner Trnsformtions. 6. Mtrices for Liner Trnsformtions. Let V be n n-dimensionl vector spce with bsis B = {v, v,, v n }nd W be n m- dimensionl vector spce with bsis B = {w, w,, w n }. Let T: V W is liner trnsformtion such tht [T(v )] B =, [T(v )] B = m m,, [T(v n )] B = n n mn, i.e. T(v ) = w + w + + m w m T(v ) = w + w + + m w m T(v n ) = n w + n w + + mn w m Then the mn mtri whose n columns correspond to [T(v i )] B n A = n m m mn is such tht [T(v)] B = A[v] B for every v in V. A is clled the mtri of T reltive to the bses B nd B. Emple Let D : P P be the differentil opertor tht mps polynomil p of degree or less to its derivtive p. Find the mtri for D using the bses B = {,,, } nd B = {,, }. Solution: The derivtives of the bsis vectors in B re [D ()] B = [] B = [() + + ] B = S [D ()] B = p. 87

230 Chpter 6: Liner Trnsformtions. 6. Mtrices for Liner Trnsformtions. p. 88

231 Chpter 6: Liner Trnsformtions. 6. Mtrices for Liner Trnsformtions. Emple Let B = B = {, cos(), sin(), cos(), sin()} nd let D : spn(b) spn(b) be the differentil opertor tht mps function f of to its derivtive f. Find the mtri for D using the bses B nd B. Solution: The derivtives of the bsis vectors in B re [D ()] B = [] B = [() + cos() + sin() + cos() + sin()] B = ; S p. 89

232

233 Chpter 6: Liner Trnsformtions. 6. Trnsition Mtrices nd Similrity. 6. Trnsition Mtrices nd Similrity. Objective: Find nd use mtri for liner trnsformtion reltive to bsis. Objective: Use properties of similr mtrices nd digonl mtrices. In 6. we sw tht the mtri for liner trnsformtion T: V W depends on the choice of bsis B for V nd B for W. Oftentimes, we cn find bses tht give very simple mtri for given liner trnsformtion. In this section, we will consider liner trnsformtions T: V V nd we will use the sme bsis for the domin s for the codomin. We will find mtrices reltive to different bses. (In Chpter 7, we will lern how to find the bses tht give us simple mtrices.) The mtri for T: V V reltive to the bsis B = {v, v,, v n } is A, where [T()] B = A[] B The mtri for T: V V reltive to the bsis B = {v, v,, v n } is A, where [T()] B = A[] B The trnsition mtri from B to B is P BB = [ v ] B [ v ] B [ v n ] B, so P[] B = [] B The trnsition mtri from B to B is P, so P A [] B = [] B [] B [T()] B So we hve [T()] B = A[] B nd we lso hve [T()] B = P BB A P BB [] B Therefore, P P [] B A [T()] B S A = P AP 6 Emple: Let T: R R which hs the stndrd mtri A = for T reltive to the bsis B = {(,, ), (, 7, ), (,, )}. S Solution: A is the mtri reltive to the stndrd bsis S Find the mtri A 6 P BS = p. 9

234 Chpter 6: Liner Trnsformtions. 6. Trnsition Mtrices nd Similrity. p. 9 Digonl mtrices: Observe tht A = = () () ) ( nd A k+ = k k k () () ) ( = () () ) ( k k k. Also, A T = A. If D = d n d d is digonl mtri nd none of the digonl elements re zero, then D = d n d d / / / Two squre mtrices A nd B re similr if nd only there is n invertible mtri P such tht B = P AP. For emple, the mtri representtions of liner trnsformtion reltive to two different bses re similr. Theorem 6:: Properties of Similr Mtrices Let A, B, nd C be squre mtrices. Then ) A is similr to A. ) If A is similr to B, then B is similr to A. ) If A is similr to B nd B is similr to C, then A is similr to C.

235 Chpter 6: Liner Trnsformtions. 6.5 Applictions of Liner Trnsformtions. 6.5 Applictions of Liner Trnsformtions. Objective: Identify liner trnsformtions defined by reflections, epnsions, contrctions, or shers in R. Objective: Use liner trnsformtion to rotte figure in R. Reflections in R. Reflection in y-is Horizonl reflection T(, y) = (, y) Reflection in -is Verticl reflection T(, y) = (, y) The mtri A for Tv = Av is A = [ Te Te ] p. 9

236 Chpter 6: Liner Trnsformtions. 6.5 Applictions of Liner Trnsformtions. Reflection in the line y = T(, y) = (y, ) T(, y) = (y, ) The mtri A for Tv = Av is A = [ Te Te ] Epnsions nd Contrctions in R. Horizonl epnsion Verticl epnsion T(, y) = (k, y) for k > T(, y) = (, ky) for k > p. 9

237 Chpter 6: Liner Trnsformtions. 6.5 Applictions of Liner Trnsformtions. Horizonl contrction Verticl contrction T(, y) = (k, y) for < k < T(, y) = (, ky) for < k < The mtri A for Tv = Av is A = [ Te Te ] p. 95

238 Chpter 6: Liner Trnsformtions. 6.5 Applictions of Liner Trnsformtions. Shers in R. Horizonl sher T(, y) = (, y) Verticl sher T(, y) = (, y) The mtri A for Tv = Av is A = [ Te Te ] p. 96

239 Chpter 6: Liner Trnsformtions. 6.5 Applictions of Liner Trnsformtions. Rottion in R. T(, y) = ( cos( ) y sin( ), sin( ) + y cos( )) p. 97

240 Chpter 6: Liner Trnsformtions. 6.5 Applictions of Liner Trnsformtions. Rottion in R. The mtri A for Tv = Av is A = [ Te Te Te ] Rottion bout -is by /6 rd = counterclockwise (right-hnd rule). p. 98

241 Chpter 6: Liner Trnsformtions. 6.5 Applictions of Liner Trnsformtions. Rottion bout y-is by /6 rd = Rottion bout y-is by /6 rd = The mtri A for Tv = Av is A = [ Te Te Te ] p. 99

242

243 Chpter 7: Eigenvlues nd Eigenvectors. 7. Eigenvlues nd Eigenvectors. Chpter 7: Eigenvlues nd Eigenvectors. 7. Eigenvlues nd Eigenvectors. Objective: Prove properties of eigenvlues nd eigenvectors. Objective: Verify eigenvlues nd corresponding eigenvectors. Objective: Find eigenvlues nd corresponding eigenspces. Objective: Use the chrcteristic eqution to find eigenvlues nd eigenvectors. Objective: Use softwre to find eigenvlues nd eigenvectors. Recll the emple A = from Section 6.5. Then Ae = nd Ae = = = e = = e Let =.5. Then A = =.5 = Let =. Then A =.5 =.5. 5 = Given squre mtri A, if we hve nonzero vector nd sclr ( lmbd ) such tht S* A = then is n eigenvector of A with the corresponding eigenvlue. Eigen is Germn for prticulr. Emple: hs eigenvectors =.5.5, =.5.5 with eigenvlues =, =. p.

244 Chpter 7: Eigenvlues nd Eigenvectors. 7. Eigenvlues nd Eigenvectors. Emple Show tht eigenvlue is n eigenvector of..6, nd find the corresponding S* Solution: Theorem 7.: Eigenvectors of Form Subspce If A is n nn mtri with n eigenvlue, then the set of ll eigenvectors of, together with the zero vector is subspce of R n. S* Proof: {: is n eigenvector of } {} = {: A = } {: A = } is not empty becuse it contins the zero vector. {: A = } is closed under sclr multipliction becuse for ny vector {: A = } nd for ny sclr c, A(c) = c(a) = c() = (c) {: A = } is closed under ddition becuse for ny vectors, {: A = }, A( + ) = A + A = + = ( + ) To find n eigenvlue of mtri A, we strt with the eqution A =. A = A = I = I A = (I A) I A is singulr det(i A) = p.

245 Chpter 7: Eigenvlues nd Eigenvectors. 7. Eigenvlues nd Eigenvectors. Theorem 7.: Eigenvlues nd Eigenvectors of Mtri Let A be n nn mtri. ) An eigenvlue of A is sclr such det(i A) =. ) The eigenvectors of A corresponding to re the nonzero solutions of (I A) = i.e. the nullspce of I A. Proof: S* A hs n eigenvector with eigenvlue if nd only if if nd only if A I is singulr (so its determinnt is zero) nd is in the nullspce of A I. det(i A) = is clled the chrcteristic eqution of A. The chrcteristic polynomil det(i A) is n n th -degree polynomil in. The Fundmentl Theorem of Algebr sttes tht n n th -degree polynomil hs ectly n roots (i.e. n n th -degree polynomil eqution hs ectly n solutions). In other words, we will lwys hve det(i A) = ( ) ( ) ( n ) where the n constnts,,, n re the eigenvlues of A. Note tht some i my be comple or some i my be repeted, for emple =. Emple Find the eigenvlues nd corresponding eigenvectors of A =. Solution by hnd: = det( p.

246 Chpter 7: Eigenvlues nd Eigenvectors. 7. Eigenvlues nd Eigenvectors. so = = Eigenvector for = Eigenvector for = Emple Find the eigenvlues nd corresponding eigenvectors of A =. Solution: = det( so = = p.

247 Chpter 7: Eigenvlues nd Eigenvectors. 7. Eigenvlues nd Eigenvectors. Eigenvector for = Eigenvector for = Emple Find the eigenvlues nd corresponding eigenvectors of A = Solution: = det( so = = = Eigenvector for = p. 5

248 Chpter 7: Eigenvlues nd Eigenvectors. 7. Eigenvlues nd Eigenvectors. Using rref we find which hs solution = E vector for = = p. 6

249 Chpter 7: Eigenvlues nd Eigenvectors. 7. Eigenvlues nd Eigenvectors. Using rref we find which hs solution Note tht this eigenvlue, which hs multiplicity two, hs two linerly independent eigenvectors. Emple Find the eigenvlues nd corresponding eigenvectors of A = b where b. Solution: = det( so = = Eigenvector for = Eigenvector for = Note tht this eigenvlue hs multiplicity two but hs only one linerly independent eigenvector. Using softwre to find eigenvlues nd eigenvectors. Let A = nd B =. TI-89: Use nd Mtri eigvl(a) Mtri eigvc(a) p. 7

250 Chpter 7: Eigenvlues nd Eigenvectors. 7. Eigenvlues nd Eigenvectors. The eigenvectors will be the columns of the mtri returned by eigvc, corresponding in order to the eigenvlues from eigvl. The eigenvectors re normlized (unit vectors). However, eigvl nd eigv will not work with comple eigenvlues nd eigenvectors. In this cse you must find the chrcteristic polynomil using Mtri det( Mtri identity()-b) Then use the Algebr menu to solve: ComplecSolve(^+=,) or fctor : ComplecFctor(^+) Use rref to find the eigenvectors: Mtri rref( Mtri identity()-b) Mthemtic: On the Bsic Mth Assistnt Plette, under Bsic Commnds More use Eigenvlues[] nd Eigenvectors[]. For emple, the eigenvlue 5 corresponds to the eigenvector (, ). Mthemtic will hndle rel nd comple eigenvlues nd eigenvectors. However, if you wnt the chrcteristic polynomil, you cn use Det[*IdentityMtri[]-] Use the Fctor[ ], Simplify[ ], nd Solve[lhs ==rhs,vr] funtions s pproprite. To find the eigenvectors the long wy, you cn use MtriForm[RowReduce[ ]] Theorem 7.: Eigenvlues of Tringulr Mtrices If A is n nn tringulr mtri, then its eigenvlues re the entries on the min digonl. S* Proof: If A is n nn tringulr mtri, then I A is tringulr mtri with,,, nn on the digonl. Since I A is tringulr, its determinnt is the product of the entries on the min digonl ( )( ) ( nn ) by Theorem.. Therefore, the eigenvlues of A re the entries on the min digonl:,,, nn. p. 8

251 Chpter 7: Eigenvlues nd Eigenvectors. 7. Digonliztion. 7. Digonliztion. Objective: Prove properties of eigenvlues nd eigenvectors. Objective: Find the eigenvlues of similr mtrices, determine whether mtri is digonlizble, nd find mtri P such tht P AP is digonl. Objective: Find, for liner trnsformtion T: V V bsis B for V such tht the mtri for T reltive to B is digonl. An nn mtri A is digonlizble when there eists n invertible mtri P such tht D = P AP is digonl mtri. Recll from Section 6. tht two squre mtrices A nd B re similr if nd only there is n invertible mtri P such tht B = P AP. So we cn lso sy tht A is digonlizble when it it similr to digonl mtri D. Theorem 7.: Similr Mtrices Hve the Sme Eigenvlues If A nd B re similr nn mtrices, then they hve the sme eigenvlues. S* Proof: Becuse A nd B re similr, there eists n invertible mtri P such tht B = P AP, so I B = P IP P AP = P (I A)P = P (I A) P = P (I A) P = (I A) Since A nd B hve the sme chrcteristic polynomil, they must hve the sme eigenvlues. Theorem 7.5: Condition for Digonliztion An nn mtri A is digonlizble if nd only if it hs n linerly independent eigenvectors. Proof: First, ssume A is digonlizble. Then there eists n invertible mtri P such tht D = P AP is digonl mtri. Then PD = AP. Let the column vectors of P be p, p,, p n nd the min digonl entries of D be,,, n. Now PD = [ p p p n ] = [ p p n p n ] n nd AP = A[ p p p n ] = [ Ap Ap Ap n ] p. 9

252 Chpter 7: Eigenvlues nd Eigenvectors. 7. Digonliztion. Since PD = AP, we hve Ap i = i p i for ech of the n column vectors of P. Since P is invertible, the n column vectors re lso linerly independent. So A hs n linerly independent eigenvectors. Conversely, ssume A hs n linerly independent eigenvectors p, p,, p n with corresponding eigenvlues,,, n. Let P be the mtri whose columns re the n eigenvectors: P = [ p p p n ]. Then AP = A[ p p p n ] = [ Ap Ap Ap n ] = [ p p n p n ] = [ p p p n ] n Thus, AP = PD, where D = is digonl mtri. Since the p i re linerly n independent, P is invertible, so D = P AP nd A is digonlizble. Emple Digonlize A = by finding n invertible mtri P nd digonl mtri D such tht D = P AP. Use the chrcteristic polynomil to find the eigenvlues nd the reduced row-echechelon function of your softwre to find the eigenvectors. Solution: = det( = so = = = p.

253 Chpter 7: Eigenvlues nd Eigenvectors. 7. Digonliztion. Eigenvector for = Eigenvector for = Eigenvector for = p.

254 Chpter 7: Eigenvlues nd Eigenvectors. 7. Digonliztion. Emple: Digonlize A = by finding n invertible mtri P nd digonl mtri D such tht D = P AP. Use eigvl nd EigVc on the TI-89, or Eigenvlues[ ] nd Eigenvectors[ ] in Mthemtic, or eigenvlues nd eigenvectors in PocketCAS. Solution: TI-89: Mtri eigvl(a) Mtri eigvc(a)p P^-*A*P S* Mthemtic: Eigenvlues[] MtriForm[p=Trnspose[Eigenvectors[]]] (Note tht Mthemtic gives the trnspose of P.) MtriForm[Inverse[p]..p] PocketCAS eigenvlues() [,, ] p:=eigenvectors() p^-**p p.

255 Chpter 7: Eigenvlues nd Eigenvectors. 7. Digonliztion. Emple: Try to digonlize A = Use eigvl nd EigVc on the TI-89, or Eigenvlues[ ] nd Eigenvectors[ ] in Mthemtic, or eigenvlues nd eigenvectors in PocketCAS. Solution: TI-89: Mtri eigvl(a) Mtri eigvc(a)p P^(-)*A*P Error: Singulr Mtri Mthemtic: Eigenvlues[] MtriForm[p=Trnspose[Eigenvectors[]]] (Note tht Mthemtic gives the trnspose of P.) MtriForm[Inverse[p]..p] PocketCAS eigenvlues() [,, ] p:=eigenvectors() Not digonlizble t eigenvlue p.

256

257 Chpter 7: Eigenvlues nd Eigenvectors. 7. Symmetric Mtrices nd Orthogonl Digonliztion. 7. Symmetric Mtrices nd Orthogonl Digonliztion. Objective: Recognize, nd pply properties of, symmetric mtrices. Objective: Recognize, nd pply properties of, orthogonl mtrices. Objective: Find n orthogonl mtri P tht orthogonlly digonlizes symmetric mtri A. A squre mtri A is symmetric when A = A T. Theorem 7.7: Properties of Symmetric Mtrices If A is symmetric nn mtri, then S* ) A is digonlizble. ) All eigenvlues of A re rel. ) If is n eigenvlue of A with multiplicity k, then hs k linerly independent eigenvectors. Tht is, the eigenspce of hs dimension k. The proof will be deferred until Ch. 8.5 Theorem 7.9: Property of Symmetric Mtrices Let A be n nn symmetric mtri. if nd re distinct eigenvlues of A, then their corresponding eignevectors nd re orthogonl. Proof: Let nd re distinct eigenvlues of A with corresponding eignevectors nd, so A = nd A =. Recll tht we cn write = T. Then S* ( ) = ( ) = (A ) = ( ) Therefore, ( )( ) =. Since, we must hve = so nd re orthogonl. p. 5

258 Chpter 7: Eigenvlues nd Eigenvectors. 7. Symmetric Mtrices nd Orthogonl Digonliztion. p. 6 Emple: Find the eigenvlues nd eigenvectors of. Are the eigenvlues rel? Wht re the dimensions of the eigenspces? Find the ngles between the eigenvectors. Solution: = det( ) (using the clcultor to tke the determinnt nd fctor it.) =, =, =. All eigenvlues re rel. Solve ( I ) = using rref. Write =. = t is free vrible. We solve for = t nd = t. The solution is t t t = t so =. The eigenspce of = (multiplicity = ) is one-dimesionl. Solve ( I ) = nd Solve ( I ) = using rref. (Note =.) = s nd = t re free vribles. We solve for = s t. The solution is t s t s = s + t so = nd =. is the Greek letter i

259 Chpter 7: Eigenvlues nd Eigenvectors. 7. Symmetric Mtrices nd Orthogonl Digonliztion. The eigenspce of = (multiplicity = ) is two-dimensionl. = (,, ) (,, ) = so is orthogonl to. = (,, ) (,, ) = so is orthogonl to. = (,,) (,,) (,,) (,,) = = so the ngle between nd is rccos( ) =. Alterntive method: Use eigvl nd eigvc on the clcultor to obtin = = nd = with eigenvectors.8697 p =.88, p = , p = We cn clculte p = p = p =, p p =, p p =.98, nd p p = 6.8 5, so the eigenvectors re lmost orthonorml. A squre mtri P is clled orthogonl when it is invertible nd P = P T. Theorem 7.8: Property of n Orthogonl Mtri S* An nn mtri P is orthogonl if nd only if its column vectors form n orthonorml set. Proof: Write P = [ p p p n ] in terms of its column vectors. Then T T T p p p p p p p n T T T P T = p nd P T P = p p p p p p n. T T T p n p n p p n p p n p n Thus, P T P = I if nd only if p i = p i p i = for ech i nd p i p j = whenever i j. Therefore, P T = P if nd only if its column vectors form n orthonorml set. p. 7

260 S* Chpter 7: Eigenvlues nd Eigenvectors. 7. Symmetric Mtrices nd Orthogonl Digonliztion. Becuse of this theorem, it would hve been better if orthogonl mtrices hd been clled orthonorml mtrices. but we re stuck with the nme orthogonl. Theorem 7.: Fundmentl Property of n Symmetric Mtrices Let A be rel nn mtri. Then A is orthogonlly digonlizble nd hs rel eigenvlues if nd only if A is symmetric. Proof: Suppose A is orthogonlly digonlizble nd rel, so there eists rel, orthogonl mtri P such tht D = P AP is digonl nd rel. Then = PDP = PDP T so A T = (PDP T ) T = (P T ) T D T P T = PDP T = A Conversely, Theorems 7.7 nd 7.9 together tell us tht ll symmetric mtrices re orthogonlly digonlizble with rel eigenvlues. An nn symmentric mtri hs n eigenvlues, counting multiplicity. Becuse the dimension of ech eigenspce equls the multiplicity of the corresponding eigenvlue, we hve n linerly independent eigenvectors, so the mtri is digonlizble (Theorem 7.5). Eigenvectors with different eigenvlues re orthogonl. Within ech eigenspce, we cn use the Grm-Schmidt process to find orthonorml eigenvectors. The set of ll of the orthonorml eigenvectors from ll of the eigenspces mke up the columns of n orthogonl mtri P, such tht is digonl. D = P AP Emple (continued): Digonlize using ) the chrcteristic eqution nd rref ) eigvl nd eigvc on the TI-89 ) Eigenvlues[ ] nd Eigenvectors[ ] in Mthemtic ) eigenvlues nd eigenvectors in PocketCAS p. 8

261 Chpter 7: Eigenvlues nd Eigenvectors. 7. Symmetric Mtrices nd Orthogonl Digonliztion. p. 9 Solution: ) From the first prt of the emple, =, =, =. Solve ( I ) = using rref. = t is free vrible. We solve for = t nd = t. The solution is t t t = t so =. Solve (, I ), = = s nd = t re free vribles. We solve for = s t. The solution is t s t s = s + t so = nd =. Normlize p = =. Grm-Schmidt: p = = u = ( p )p = = / / ; p = u u = 6 so D = nd P = 6 / / 6 / / / 6 / / / Check tht P T P = I nd multiply P AP = P T AP to check tht it gives you D.

262 Chpter 7: Eigenvlues nd Eigenvectors. 7. Symmetric Mtrices nd Orthogonl Digonliztion. ) TI-89: Use eigvl nd eigvc to obtin D = nd P = Check P T P =..E.98.E. 6.8E E 5. so P is lmost orthogonl. By looking t the rtios between components, it ppers tht p = unitv([-,,]) = 6 / 6 / 6 6 / 6 p = unitv([,,]) = / / / p = unitv([,-,]) = / / Try new P = 6 / 6 / 6 6 / 6 / / / / / p gives Check tht P T P = I nd multiply P AP = P T AP to check tht it gives you D. ) Mthemtic: Eigenvlues[] yields {,-,-} so D = Eigenvlues[] gives {{,,},{-,,},{-,,}} so =, =, = Wrning: MtriForm[Eigenvlues[]] gives which hs the eigenvectors in the rows (insted of in the columns)! p.

263 Chpter 7: Eigenvlues nd Eigenvectors. 7. Symmetric Mtrices nd Orthogonl Digonliztion. Orthogonlize[Eigenvlues[]] gives the orthonorml bsis / / / 6 so P = / / / / / 6 Wrning: MtriForm[Orthogonlize[Eigenvlues[]]] gives P T, not P! Check tht P T P = I nd multiply P AP = P T AP to check tht it gives you D. ) PocketCAS: Computing M T M shows in the off-digonl elements tht the vectors re orthogonl, nd the digonl elements give the norm squred of ech vector. so D = eigenvectors( ) gives orthogonl, but unnormlized obtin P T. eigenvectors. grmschmidt( ) will construct n orthonorml bsis from the rows of mtri, so use grmschmidt(trnspose(p)) to / / / 6 so P = / / 6 / / / 6 Check tht P T P = I nd multiply P AP = P T AP to check tht it gives you D. p.

264 Chpter 7: Eigenvlues nd Eigenvectors. 7. Symmetric Mtrices nd Orthogonl Digonliztion. Emple: Digonlize. p.

265 Chpter 7: Eigenvlues nd Eigenvectors. 8.5 Unitry nd Hermitin Mtrices. 8.5 Unitry nd Hermitin Mtrices. Objective: Find the conjugte trnspose (Hermitin conjugte) of comple mtri A. Objective: Determine if mtri A is unitry. Objective: Find the eigenvlues nd eigenvectors of Hermitin mtri, nd digonlize Hemitin mtri. In order to prove the theorems in Section 7. bout rel symmetric mtrices, we must consider comple Hermitin mtices. The conjugte trnspose (or Hermitin conjugte) of comple mtri A, denoted by A or A H or A*, is given by A = (A ) T = A. T Tht is, Tke the comple conjugte of ech of entry of A, nd trnspose the mtri. Notice tht if A is rel (ll entries re rel), then A = A T. A is Hermitin if nd only if A = A. Notice tht if A is rel, then Hermitin is the sme s symmetric. To tke the Hermitin conjugte A on the TI-89, use Mtri T To tke the Hermitin conjugte A in Mthemtic, use ct To tke the Hermitin conjugte A in PocketCAS, use trn() A is unitry if nd only if A = A. Notice tht if A is rel, then unitry is the sme s orthogonl. Emple: Clssify the following mtrices s Hermitin, unitry, both, or neither. i ) A = : A i = i. A = A so A is Hermitin. AA = I so A is lso unitry. i ) B = : B =. B B so B is not Hermitin. BB = I so B is unitry. i ) C = : C i = 7 i 5. C = C so C is Hermitin. 7 i 5 CC = I so C is not unitry. p.

266 Chpter 7: Eigenvlues nd Eigenvectors. 8.5 Unitry nd Hermitin Mtrices. ) D = : D = i. D D so D is not Hermitin. D D = i unitry. I so D is not Theorem 8.8: Properties of the Hermitin Conjugte (Comple Trnspose) If A, B, nd C comple mtrices with dimensions mn, mn, nd np respectively, nd k is comple number, then the following properties re true. S* ) (A ) = A ) (A + B) = A + B ) (ka) = k A ) (AC) = C A Theorem 8.9: Unitry Mtrices An nn comple mtri U is unitry if nd only if its row vectors form n orthonorml set in C n using the Eucliden inner product. Proof: Recll tht the Eucliden inner product of u nd v in C n is given by u v = u v + u v + + u n v n Note tht if we write u nd v s row vectors, then u v = u v u = u u nd u is orthogonl to v if nd only if u v =. u Write A = u in terms of its row vectors. Then u n u u u u u u n A T = [ u u u n ] nd AA = u u u u u u n. u n u u n u u n u n Thus, AA = I if nd only if u i = u i u i = for ech i nd u i u j = whenever i j. Therefore, A = A if nd only if its row vectors form n orthonorml set. Theorem 8.9.: A is unitry if nd only if A is unitry. p.

267 Chpter 7: Eigenvlues nd Eigenvectors. 8.5 Unitry nd Hermitin Mtrices. Corollry 8.9.: Unitry Mtrices An nn comple mtri A is unitry if nd only if its column vectors form n orthonorml set in C n using the Eucliden inner product. Theorem 8.: The Eigenvlues of Hermitin Mtri If H is Hermitin mtri, then its eigenvlues re rel. Proof: Let be n eigenvlue of H nd let v be its corresponding eigenvlue, so Hv = v. Then (v Hv) = v H (v ) = v Hv Now v Hv = v (Hv) = v (v) = (v v) = v So (v Hv) = v = v becuse v is rel. Since v nd v = v, must be rel. p. 5

268

269 Chpter 8 Comple Vector Spces 8.5 Unitry nd Hermitin Mtrices Find the conjugte trnspose A* of comple mtri A. Determine if mtri A is unitry. Find the eigenvlues nd eigenvectors of Hermitin mtri, nd digonlize Hermitin mtri. CONJUGATE TRANSPOSE OF A MATRIX Problems involving digonliztion of comple mtrices nd the ssocited eigenvlue problems require the concepts of unitry nd Hermitin mtrices. These mtrices roughly correspond to orthogonl nd symmetric rel mtrices. In order to define unitry nd Hermitin mtrices, the concept of the conjugte trnspose of comple mtri must first be introduced. Definition of the Conjugte Trnspose of Comple Mtri The conjugte trnspose of comple mtri A, denoted by A*, is given by A* A T where the entries of A re the comple conjugtes of the corresponding entries of A. Note tht if A is mtri with rel entries, then A* A T. To find the conjugte trnspose of mtri, first clculte the comple conjugte of ech entry nd then tke the trnspose of the mtri, s shown in the following emple. Determine A* for the mtri A 7i i SOLUTION 7i A i i. A* A T 7i Finding the Conjugte Trnspose of Comple Mtri i 7i i i i i Severl properties of the conjugte trnspose of mtri re listed in the following theorem. The proofs of these properties re strightforwrd nd re left for you to supply in Eercises 7 5. THEOREM 8.8 Properties of the Conjugte Trnspose If A nd B re comple mtrices nd k is comple number, then the following properties re true.. A** A. A B* A* B*. ka* ka*. AB* B*A*

270 8.5 Unitry nd Hermitin Mtrices UNITARY MATRICES Recll tht rel mtri A is orthogonl if nd only if A A T. In the comple system, mtrices hving the property tht A A* re more useful, nd such mtrices re clled unitry. Definition of Unitry Mtri A comple mtri A is unitry when A A*. Show tht the mtri A is unitry. SOLUTION Begin by finding the product AA*. AA* i i Becuse A i i AA* i i I A Unitry Mtri i i i i i i it follows tht A* A. So, A is unitry mtri. Recll from Section 7. tht rel mtri is orthogonl if nd only if its row (or column) vectors form n orthonorml set. For comple mtrices, this property chrcterizes mtrices tht re unitry. Note tht set of vectors v, v,..., v m in C n ( comple Eucliden spce) is clled orthonorml when the sttements below re true.. v i, i,,..., m. v i v j, i j The proof of the net theorem is similr to the proof of Theorem 7.8 presented in Section 7.. THEOREM 8.9 Unitry Mtrices An n n comple mtri A is unitry if nd only if its row (or column) vectors form n orthonorml set in C n.

271 Chpter 8 Comple Vector Spces The Row Vectors of Unitry Mtri REMARK Try showing tht the column vectors of A lso form n orthonorml set in C. Show tht the comple mtri A is unitry by showing tht its set of row vectors forms n orthonorml set in C. A i 5i 5 SOLUTION Let r, r, nd be defined s follows. Begin by showing tht r, r, nd re unit vectors. r 5i r 5, i i, 5 5 r i i r i i i i r 5i 55 5i i 5 6 Then show tht ll pirs of distinct vectors re orthogonl. r r i i i i i r r 5i 5 i i 5i 5 r r i 5i 5 i i 5 65 i i i 5 r, i,, i i 5 5 i 65 i 5 r i, r i 65 So, r, r, r is n orthonorml set. i,, 5 i 5 i 5 5 i 5 5 i i 5 5

272 HERMITIAN MATRICES 8.5 Unitry nd Hermitin Mtrices A rel mtri is symmetric when it is equl to its own trnspose. In the comple system, the more useful type of mtri is one tht is equl to its own conjugte trnspose. Such mtri is clled Hermitin fter the French mthemticin Chrles Hermite (8 9). Definition of Hermitin Mtri A squre mtri A is Hermitin when A A*. As with symmetric mtrices, you cn recognize Hermitin mtrices by inspection. To see this, consider the mtri A. A i c c i The conjugte trnspose of A hs the form A* A T i b b i i b b i If A is Hermitin, then A A*. So, A must be of the form A b b i Similr results cn be obtined for Hermitin mtrices of order n n. In other words, squre mtri A is Hermitin if nd only if the following two conditions re met.. The entries on the min digonl of A re rel.. The entry ij in the ith row nd the jth column is the comple conjugte of the entry in the jth row nd the ith column. ji b b i d d i c c i d d i c c i d d i. b b i d. Which mtrices re Hermitin? i. b. i i i i c. i i d. i i Hermitin Mtrices i SOLUTION. This mtri is not Hermitin becuse it hs n imginry entry on its min digonl. b. This mtri is symmetric but not Hermitin becuse the entry in the first row nd second column is not the comple conjugte of the entry in the second row nd first column. c. This mtri is Hermitin. d. This mtri is Hermitin becuse ll rel symmetric mtrices re Hermitin. i

273 Chpter 8 Comple Vector Spces REMARK Note tht this theorem implies tht the eigenvlues of rel symmetric mtri re rel, s stted in Theorem 7.7. One of the most importnt chrcteristics of Hermitin mtrices is tht their eigenvlues re rel. This is formlly stted in the net theorem. THEOREM 8. The Eigenvlues of Hermitin Mtri If A is Hermitin mtri, then its eigenvlues re rel numbers. PROOF Let be n eigenvlue of A nd let b i v b i i. n b n be its corresponding eigenvector. If both sides of the eqution Av v re multiplied by the row vector v*, then v*av v*v v*v b b... n b n. Furthermore, becuse v*av* v*a*v** v*av it follows tht v* Av is Hermitin mtri. This implies tht v* Av is rel number, so is rel. To find the eigenvlues of comple mtrices, follow the sme procedure s for rel mtrices. Find the eigenvlues of the mtri A. i i A i i i i SOLUTION The chrcteristic polynomil of A is I A i i So, the chrcteristic eqution is re, 6, nd. Finding the Eigenvlues of Hermitin Mtri i i i i i i i i i i i i 6, nd the eigenvlues of A

274 8.5 Unitry nd Hermitin Mtrices 5 TECHNOLOGY Some grphing utilities nd softwre progrms hve built-in progrms for finding the eigenvlues nd corresponding eigenvectors of comple mtrices. To find the eigenvectors of comple mtri, use procedure similr to tht used for rel mtri. For instnce, in Emple 5, to find the eigenvector corresponding to the eigenvlue substitute the vlue for into the eqution to obtin i i i i, i i i Solve this eqution using Guss-Jordn elimintion, or grphing utility or softwre progrm, to obtin the eigenvector corresponding to, which is shown below. v i i i i i i Eigenvectors for 6 nd cn be found in similr mnner. They re nd respectively. i 6 9i LINEAR ALGEBRA APPLIED i i 5, v v v v v v. Quntum mechnics hd its strt in the erly th century s scientists begn to study subtomic prticles nd light. Collecting dt on energy levels of toms, nd the rtes of trnsition between levels, they found tht toms could be induced to more ecited sttes by the bsorption of light. Germn physicist Werner Heisenburg (9 976) lid mthemticl foundtion for quntum mechnics using mtrices. Studying the dispersion of light, he used vectors to represent energy levels of sttes nd Hermitin mtrices to represent observbles such s momentum, position, nd ccelertion. He noticed tht mesurement yields precisely one rel vlue nd leves the system in precisely one of set of mutully eclusive (orthogonl) sttes. So, the eigenvlues re the possible vlues tht cn result from mesurement of n observble, nd the eigenvectors re the corresponding sttes of the system following the mesurement. Let mtri A be digonl Hermitin mtri tht represents n observble. Then consider physicl system whose stte is represented by the column vector u. To mesure the vlue of the observble A in the system of stte u, you cn find the product u*au u u u u u u u u u. u u u Becuse A is Hermitin nd its vlues long the digonl re rel, u*au is rel number. It represents the verge of the vlues given by mesuring the observble A on system in the stte u lrge number of times. Jezper/Shutterstock.com

275 6 Chpter 8 Comple Vector Spces Just s rel symmetric mtrices re orthogonlly digonlizble, Hermitin mtrices re unitrily digonlizble. A squre mtri A is unitrily digonlizble when there eists unitry mtri P such tht P AP is digonl mtri. Becuse P is unitry, P P*, so n equivlent sttement is tht A is unitrily digonlizble when there eists unitry mtri P such tht P* AP is digonl mtri. The net theorem sttes tht Hermitin mtrices re unitrily digonlizble. THEOREM 8. Hermitin Mtrices nd Digonliztion If A is n n n Hermitin mtri, then. eigenvectors corresponding to distinct eigenvlues re orthogonl.. A is unitrily digonlizble. PROOF To prove prt, let v nd v be two eigenvectors corresponding to the distinct (nd rel) eigenvlues nd. Becuse Av v nd Av v, you hve the equtions shown below for the mtri product Av * v. Av * *A* * * v v v v v v Av v * Av * v * v * v v v v * v So, v *v v *v v *v v becuse *v nd this shows tht v nd v re orthogonl. Prt of Theorem 8. is often clled the Spectrl Theorem, nd its proof is left to you. v The Eigenvectors of Hermitin Mtri The eigenvectors of the Hermitin mtri shown in Emple 5 re mutully orthogonl becuse the eigenvlues re distinct. Verify this by clculting the Eucliden inner products v v, v v, nd v v. For emple, v v i i6 9i i i6 9i i 6 9i i 8. The other two inner products v v nd v v cn be shown to equl zero in similr mnner. The three eigenvectors in Emple 6 re mutully orthogonl becuse they correspond to distinct eigenvlues of the Hermitin mtri A. Two or more eigenvectors corresponding to the sme eigenvlue my not be orthogonl. Once ny set of linerly independent eigenvectors is obtined for n eigenvlue, however, the Grm-Schmidt orthonormliztion process cn be used to find n orthogonl set.

276 8.5 Unitry nd Hermitin Mtrices 7 Digonliztion of Hermitin Mtri Find unitry mtri P such tht P* AP is digonl mtri where i i A i i i i. SOLUTION The eigenvectors of A re shown fter Emple 5. Form the mtri P by normlizing these three eigenvectors nd using the results to crete the columns of P. So, v, i, 5 7 v i, 6 9i, v i, i, P 7 i 7 7 Try computing the product P* AP for the mtrices A nd P in Emple 7 to see tht * AP P i i i i 5. where, 6, nd re the eigenvlues of A. You hve seen tht Hermitin mtrices re unitrily digonlizble. It turns out tht there is lrger clss of mtrices, clled norml mtrices, tht re lso unitrily digonlizble. A squre comple mtri A is norml when it commutes with its conjugte trnspose: AA* A*A. The min theorem of norml mtrices sttes tht comple mtri A is norml if nd only if it is unitrily digonlizble. You re sked to eplore norml mtrices further in Eercise 56. The properties of comple mtrices described in this section re comprble to the properties of rel mtrices discussed in Chpter 7. The summry below indictes the correspondence between unitry nd Hermitin comple mtrices when compred with orthogonl nd symmetric rel mtrices. Comprison of Symmetric nd Hermitin Mtrices A is symmetric mtri A is Hermitin mtri (rel) (comple). Eigenvlues of A re rel.. Eigenvlues of A re rel.. Eigenvectors corresponding. Eigenvectors corresponding to distinct eigenvlues re to distinct eigenvlues re orthogonl. orthogonl.. There eists n orthogonl. There eists unitry mtri P mtri P such tht such tht P T AP P*AP is digonl. is digonl.

277 8 Chpter 8 Comple Vector Spces 8.5 Eercises Finding the Conjugte Trnspose In Eercises, determine the conjugte trnspose of the mtri.. i i. i 5 i i. 5 i 6. i Finding the Conjugte Trnspose In Eercises 5 nd 6, use softwre progrm or grphing utility to find the conjugte trnspose of the mtri Identifying Unitry Mtrices In Eercises 6, determine whether A is unitry by clculting AA*... A i i A i i i i i i 5. i i i i i i i A A i 5 5 i i i i i i. A i. A i i 5 5 i i i i i i i i i i 5 i i i i i 6 i i Non-Unitry Mtrices In Eercises 7, eplin why the mtri is not unitry. 7. i i A 8. A i 9. A i i i i i i i 6 i i i 6. A i i Row Vectors of Unitry Mtri In Eercises 7, () verify tht A is unitry by showing tht its rows re orthonorml, nd (b) determine the inverse of A. i i A 8. A Identifying Hermitin Mtrices In Eercises 6, determine whether the mtri is Hermitin.. i. i i i. i i A A i i 5 i 6 6 i i i i i Finding Eigenvlues of Hermitin Mtri In Eercises 7, determine the eigenvlues of the mtri A A A i i i i i 9. A i. i. A i i i i i 5 i 6 i i i i i i A i i i i

278 8.5 Eercises 9. Finding Eigenvectors of Hermitin Mtri In Eercises 6, determine the eigenvectors of the mtri in the indicted eercise.. Eercise 7. Eercise 5. Eercise 6. Eercise 8 Digonliztion of Hermitin Mtri In Eercises 7, find unitry mtri P tht digonlizes the mtri A A i A i i i 9... i A i i A i A i A. Show tht A I n is unitry by computing AA*.. Let z be comple number with modulus. Show tht the mtri A is unitry. A z iz i i i 6 i z iz i i i. Consider the following mtri. A i i i i i i () Is A unitry? Eplin. (b) Is A Hermitin? Eplin. (c) Are the row vectors of A orthonorml? Eplin. (d) The eigenvlues of A re distinct. Is it possible to determine the inner products of the pirs of eigenvectors by inspection? If so, stte the vlue(s). If not, eplin why not. (e) Is A unitrily digonlizble? Eplin. Unitry Mtrices In Eercises 5 nd 6, use the result of Eercise to determine, b, nd c such tht A is unitry A 6 i A 5 b c b c Proof In Eercises 7 5, prove the formul, where A nd B re n n comple mtrices. 7. A** A 8. A B* A* B* 9. ka* ka* 5. AB* B*A* 5. Proof Let A be mtri such tht A* A O. Prove tht ia is Hermitin. 5. Show tht deta deta, where A is mtri. Determinnts In Eercises 5 nd 5, ssume tht the result of Eercise 5 is true for mtrices of ny size. 5. Show tht deta* deta. 5. Prove tht if A is unitry, then. 55. () Prove tht every Hermitin mtri A cn be written s the sum A B ic, where B is rel symmetric mtri nd C is rel nd skew-symmetric. (b) Use prt () to write the mtri A i s the sum A B ic, where B is rel symmetric mtri nd C is rel nd skew-symmetric. (c) Prove tht every n n comple mtri A cn be written s A B ic, where B nd C re Hermitin. (d) Use prt (c) to write the comple mtri A i i i s the sum A B ic, where B nd C re Hermitin. 56. () Prove tht every Hermitin mtri is norml. (b) Prove tht every unitry mtri is norml. (c) Find mtri tht is Hermitin, but not unitry. (d) Find mtri tht is unitry, but not Hermitin. (e) Find mtri tht is norml, but neither Hermitin nor unitry. (f) Find the eigenvlues nd corresponding eigenvectors of your mtri in prt (e). (g) Show tht the comple mtri i i i deta is not digonlizble. Is this mtri norml?

279 Answer Key Section 8.5. i i. 5 i i i 5 i 6 i i i i i i 5. i i i 7. A is not unitry becuse it is singulr. 9. A is not unitry becuse it is not squre mtri. i. AA* i I So, A is not unitry.. AA* I So, A is unitry. 5. AA* So, A is unitry. 9. () r i, i, r i, i r, r, r r (b) A i i i i I i r 5, 5 i, r 5, 5 i 7. () r, r, r r (b) A i 5 i. A is Hermitin becuse A A*.. A is not Hermitin becuse the entry, on the min digonl, is not rel number. So, A A*. 5. A is not Hermitin becuse the mtri is not squre i i. v, i 5. v,, v, i v i,, v 6 i, i, P i P i i P A A* AA* Therefore A* A nd I n is unitry. 5. A 7 5. Proofs i 55. () A A A i A A i (b) (c) A A A * (d) i i i i i i i I n i A A * i i i i

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281 Chpter 7: Eigenvlues nd Eigenvectors. 7. Applictions of Eigenvlues nd Eigenvectors. 7. Applictions of Eigenvlues nd Eigenvectors. Objective: Model popultion growth using n ge trnsion mtri nd ge distribution vector, nd find stble ge distribution vector. Objective: Use mtri eqution to solve system of first-order liner differentil equtions. To model popultion growth of popultion with lifespn of L yers, we prtition the lifespn into n equl-size clsses st ge clss nd ge clss i th ge clss n th ge clss L L L, n, n n ( i ) L il, n n ( n ) L, L n The ge distribution vector is = where i is the number of individuls in the i th ge clss. n Let p i be the probbility tht fter L/n yers, member of the i th ge clss will survive to become member of the (i + ) th ge clss. (Note tht p i for i =,,, n, nd p n =.) Let b i be the verge number of offspring produced by member of the i th ge clss. (Note tht p i for i =,,, n.) b p The ge trnsition mtri or Leslie mtri is A = b p b b p n n bn If i is the ge distribution vector t specific time, then the ge distribution vector L/n yers lter is i+ = A i Emple: 8% of popultion of mice survives the first yer. Of tht 8%, 5% survives the second yer. The mimum lifespn is three yers. The number of offspring for ech member of the popultion is in the first yer, 6 in the second, nd in the third yer. The popultion now consists of members in ech ge clss. How mny members will there be in ech ge clss in one yer? In two yers? p. 7

282 Chpter 7: Eigenvlues nd Eigenvectors. 7. Applictions of Eigenvlues nd Eigenvectors. Solution: = nd A = After one yer, = = for ge ge ge After two yers, = = for ge ge ge Emple: A certin type of lizrd hs mimum lifespn of two yers. Only 8% of lizrds survive from their first yer to their second yer. The verge number of offspring for ech member of the popultion is.5 in the first yer, nd in the second yer. Find stble ge distribution vector for this popultion. Solution: A =. Using the TI-89, we find eigenvlues with correspronding eigenvectors A negtive eigenvlue (nd n eigenvector with some positive nd some negtive entries) does not mke sense. nd A = must both hve ll non-negtive entries, becuse the entries represent the number of individuls in ech ge clss. We therefore use the eigenvector for = The number of individuls in ech clss should be whole number, so try = Now check: A = The rtio of the ge clsses is stble t (first yer) : (second yer) = p. 8

283 Chpter 7: Eigenvlues nd Eigenvectors. 7. Applictions of Eigenvlues nd Eigenvectors. p. 9 A system of first-order liner differentil equtions hs the form n nn n n n n n n n y y y y y y y y y y y y with initil conditions y n C n C y C y () () () Where ech i y is function of t nd dt dy y i i. This is liner system becuse the y = y n y y is liner trnsformtion of y = y n y y. y (t) = Ay(t), where A is mtri of constnts. These re first-order differentil equtions becuse they contin first (nd not higher) derivtives. Emple: Solve the system y y y y y y y y y y y y with initil conditions () () () C y C y C y Solution: From clculus, we know tht the generl solution to y (t) = ky(t) is y(t) = Ce kt. So the solution to the system (with digonl mtri) is t C e t y ) ( t C e t y ) ( t C e t y ) ( Emple: Solve the system y y y y y y y y y y with initil conditions () () () y y y

284 Chpter 7: Eigenvlues nd Eigenvectors. 7. Applictions of Eigenvlues nd Eigenvectors. p. Solution: y (t) = Ay(t), where A = nd y() = Try substitution y(t) = Pw(t), where P is some invertible constnt mtri. Then y (t) = Pw (t) nd P y (t) = w (t) so y (t) = A y(t) P y (t) = P A y(t) w (t) = P A(Pw(t)) w (t) = (P AP)w(t) If is P AP digonl, i.e. P AP = D =, then w(t) = t t t e B e B B e for unknown constnts B, B, B. Using the TI-89 to find the eigenvlues nd eigenvectors of A, we find P = nd D = So w(t) = t t e B B B e, y(t) = Pw(t) = P t t e B B B e, nd y() = = P B B B, so B B B = P = Then y(t) = Pw(t) = P t t e.. e = t t t e e e = t t t e e e 6 6 Other softwre my yield other P nd D mtrices, but w(t) will lwys be the sme.

285 Chpter 7: Eigenvlues nd Eigenvectors. 7. Applictions of Eigenvlues nd Eigenvectors. Emple: Solve the system y y y y y y 5y y y with initil conditions y () y () y () Solution: y (t) = Ay(t), where A = nd y() = Try substitution y(t) = Pw(t) y (t) = A y(t) Pw (t) = A Pw(t) w (t) = (P AP) w(t) We wnt digonl P AP =, so w(t) = for unknowns B, B, B. Using the TI-89 to find the eigenvlues nd eigenvectors of A, we find P = nd D = So w(t) =, y(t) = y() = B = P B, so B B B B = Then y(t) = Pw(t) = P Use Epnd on the TI to obtin y(t) = p.

286 Chpter 7: Eigenvlues nd Eigenvectors. 7. Applictions of Eigenvlues nd Eigenvectors. An emple from continuum mechnics of symmetric mtri is the stress tensor T =. The figure uses the nottion tht T: R R is the liner trnsformtion ( T e i ) Te ij is the force per unit re on the pln perpendiculr to e i, in the e j direction. ii is is pressure or norml stress. ij is shering stress for i j. T is symmetric becuse of conservtion of ngulr momentum. i Emple: Let T =. Use the TI-89 to digonlize T: P = nd D = 6 The mutully orthogonl vectors p = , p = = , nd p = ( p i ) T is prllel to p i nd where re clled the principl directions, where the stress vector there re no sher stresses d ij for i j. The three stresses =, =6, nd =, re clled principl stresses. p.

287 Chpter 7: Eigenvlues nd Eigenvectors. 7. Applictions of Eigenvlues nd Eigenvectors. Emple (multivrible clculus): The Hessin mtri H of function of severl vribles is symmetric. If f (, y): R R is (not necessrily liner) function of two vribles, then f f H = y hs orthogonl eigenvectors. Ner criticl point, the level curves of f f y y f (, y) = constnt re ellipses if the eigenvlues of H hve the sme sign, stright lines if one eigenvlue is zero, or hyperbols if the eigenvlues hve opposite signs. See the tet for more discussion. f (, y) = 8 y + 7y 5 Eigenvlues = 5, 5 Eigenvectors =, f (, y) = + y + y + + y + 5 Eigenvlues =, Eigenvectors =, f (, y) = 7 + y 7y 75 Eigenvlues = 5, 5 Eigenvectors =, p.