308K. 1 Section 3.2. Zelaya Eufemia. 1. Example 1: Multiplication of Matrices: X Y Z R S R S X Y Z. By associativity we have to choices:


 Ezra Nash
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1 8K Zely Eufemi Section 2 Exmple : Multipliction of Mtrices: X Y Z T c e d f 2 R S X Y Z 2 c e d f 2 R S 2 By ssocitivity we hve to choices: OR: X Y Z R S cr ds er fs X cy ez X dy fz 2 R S 2 Suggestion : Write I nd multiply it on the right y the mtrix A c e Suggestion 2: Write I 2 nd multiply it on the left y the mtrix A c e 4 Let s multiply digonl mtrix nd n upper tringulr mtrix: d 22 m n p 22 XpR Sq Y pcr dsq ZpeR fsq 2 m n p m d n dp RpX cy ezq SpX dy f Zq d Then I A 2? f 2 d Then A 2 I 2? f 2 m n dp 22 Wht if we reverse the order of multipliction? m n p 22 d 22 m n m nd p pd m dn pd 22 Notice tht the results re different, so mtrix multipliction is not commuttive
2 2 Inverse of liner trnsformtion nd inverting mtrices Exmple : Multiply the following mtrices: c d 22 Given the mtrix A d c d p cq p q cd dp cq cp q d 22, cn you find mtrix B 22 such tht AB I 2? 22 2 In generl, given mtrix A nn, cn we find mtrix B such tht AB BA I n? No! Exmple: Try to find the inverse of Given the mtrix A d c d c Wht goes wrong? 22 Definition: The inverse of liner trnsformtion T : R m Ñ R n is nother liner trnsformtion denoted T : R n Ñ R m such tht T T pxq x nd T T pxq x Theorem: Not ll liner trnsformtions re invertile But if the inverse exists, it is unique! 22 A liner trnsformtion T : R n Ñ R m with mtrix A is invertile if nd only if (TFAE) (Liner Trnsformtion) T is oth injective nd surjective, forcing m n Yet it is enough to check n m nd either injective or surjective (Mtrix) The mtrix A hs oth linerly columns (injective) nd linerly independent rows (surjective) nd the columns spn R n (surjective), forcing it to e squre mtrix Yet it is enough to check it to e squre mtrix, nd tht it hs linerly independent columns (System) With coefficient mtrix A stisfies either for ll : Ax hs exctly one solution Enough to check the numer of eqution is equl to the numer of vriles, nd Ax hs exctly one solution 4 Properties: If A nd B re invertile: () pa q A (2) pabq B A () AC AD ñ C D The Method to find the inverse of mtrix using JordnGuss Elimintion: Find the inverse of the mtrix A 4 22 Step : Check tht A is squre mtrix: n 2 ecuse the dimensions is 2 2 Step 2: Write the prtitioned mtrix ra I n s: ra I 2 s {7 {7 24 {7 {7 4{7 {7 By () Sutrct 4 times row from row 2; (2) divide row y 7; () sutrct times row 2 from row Notice the mtrix we otin is of the form ri n A s {7 {7 Step : Red the mtrix A s the right hnd side squre: A 4{7 {7 6 Mtrices like ra I n s re clled Prtitioned Mtrices 2 7 But wht cn go wrong? Find the inverse of the mtrix B 6 2 rb I 2 s Notice tht we cnnot chieve the identity on the left squre! Why? The rows re not linerly independent, or the columns re not linerly independent!
3 4 Suspces Wht does single eqution represents in R 2? x y c represents line ( dimensionl oject in the 2dimensionl Eucliden spce) Wht does single eqution represents in R? x y cz d represents plne ( 2dimensionl oject in the dimensionl Eucliden spce) Wht does single eqution represents in R n for n 4? x 2 x 2 x n x n represents hyperplne It is n n dimensionl oject in the ndimensionl Eucliden spce 2 Wht other ojects of interest re contined in n Eucliden Spce tht do not include the whole spce? Suspces! Any Suspce S R n () must contin ; (2) S must e dditively close: if u, v P S ñ u sclrmultiplictively closed: if u P S, P R : r u P S v P S; S must e How to define suspces: A suspce my e the solution set of consistent system of equtions, provided tht is in the solution set A suspce my e the spn of some vectors (lwys) 4 Exmple: Determine if the set S of ll vectors of the form c where c is suspce of R Step : Check if P S R : Since p,, q T nd c, then P S Step 2: Check if S cn e spnned y some vectors: Notice tht since c, then c, so S Spnpp,, q T, p,, q T q ecuse c Step : Verify the definition of suspce: If p,, cq T, pe, f, gq T P S, then c e f g ñ p eq p fq pc gq ñ p,, cq T pe, f, gq T P S And if r P R nd p,, cq T P S, then c ñ r r rc ñ rp,, cq T P S Two importnt suspces from Liner Trnsformtion: The Kernel of the trnsformtion nd the Rnge of the trnsformtion 2 Exmple: Consider the liner trnsformtion S : R Ñ R with mtrix 2 In other words Spx, y, zq T px 2y, 2x y, q The Kernel is the set of ll vectors px, y, zq T in the domin R tht re mpped to, ie: x 2y, 2x y ñ KerpSq tp,, zq T P R u The Rnge is the set of ll vectors p,, q in the codomin R tht re imges of vectors in the domin, ie: x 2y, 2x y ñ Spx 2, y 2, zqt p,, q T S : R Ñ R KerpSq Ñ tu R Ñ RngepSq ImpSq 6 Note: S : R n Ñ R m is injective if nd only if KerpSq tu And S : R n Ñ R m is Surjective if nd only if RngepSq R m the codomin of S
4 4 42 Bsis nd Dimension If we were interested in suspces, which cn e written s the spn of set, then we should e interested in how to find the miniml set enough to spn it Recll tht if u P Spnpv,, v k q then Spnpu, v,, v k q Spnpv,, v k q, ie: u ecomes n unnecessry vector to descrie the spn of the set 2 Bsis of suspce S is set of vectors of B t u,, u m P R n u such tht () B spns S (descries the suspce), nd (2) B is linerly independent (minimlity) Stndrd sis of R n : e p,,, q T, e 2 p,,, q T,, e n p,,, q T The dimension of R n is surprise surprise n! 4 How to find the sis of suspce S descried s the spn of given set of vectors: Exmple: Suspce S Spnp u p2,, q T, u 2 p,, q T, u p, 9, q T q Step : Write mtrix whose rows re the listed vectors: 2 9 Step 2: Apply GussElimintion: Step : Collect the nonzero row nd write them s column vectors They re the sis of the given suspce S Spn Conclusion: The spn of the rows of two equivlent mtrices is the sme Moreover, if we pply JordnGuss elimintion insted, we lso get nother sis, which is simpler, ut with more work to e performed TRUE nd Flse: Given two equivlent mtrices (one cn e trnsform to the other y elementry opertions) A mn nd B mn, oth of them without zero rows; if the rows of A re linerly dependent, then so re the rows of B 6 TRUE or Flse: All sis of given suspce S hve the sme numer of vectors This numer is clled the dimension of S, Indeed, fter performing JordnGuss elimintion on ny sis, we will otin the sme numer of nonzero rows with the sme leding vriles 7 How to uild sis from scrtch: Exmple: S tp,, cq T : c u If the suspce is, then the only possile sis is t u: Not the cse ecuse S contins p,, 4q Pick nonzero element of S (ny!): sy p,, 4q T Strt your sis list with this elements: B tp,, 4q T u Step : Pick nother element of S which is not in the spn in the list: p2, 6, 8q T no ecuse it is in the spn of the sis list But p,, q T is not Then B tp,, 4q T, p,, q T u 4 Step 2: Check if ll the elements of S re in the Spn of the sis: p,, q T p,, cq T where c p,, q T p,, q T Repet steps nd 2 s mny times s necessry until you find tht the sis list spns the suspce In this exmple the sis is tp,, q T, p,, q T u 4
5 8 Remrk: A smller list won t e le to spn the suspce S, nd if this is the result, it is ecuse we terminted the lgorithm erlier due to misscheck And igger list is not linerly independent, nd if this is the result, it is ecuse we continued the lgorithm longer due to misscheck 9 Theorem: If B hs m elements of suspce S of dimension m, in order for B to e sis of S, it is enough to check tht B is linerly independent or B spns S Theorem: Suspces S S 2 R m, then dimps q dimps 2 q m Moreover, if S S 2, then the suspces re equl if nd only if the dimensions re equl Find n exmple of four vectors u, u 2, u, u 4 P R 2 such tht if we remove ny two of them, the remining 2 re sis 2 True or Flse: If S nd S 2 re suspces of R n of the sme dimension, then S S 2 Given set of m vectors in R n, with m n, when cn we complete it to e sis? 4 Given set of m vectors in R n, with m n, when cn we reduce it to e sis? Find the simplest sis you cn get for S Spnpp, 2,, 4q T, p, 2,, 2q T, p, 2,, 2q T, p,,, q T, p, 6, 7, 8q T q Wht is the dimension of S?
6 4: Row nd Column Spces The Row Spce of mtrix A mn denoted y rowpaq is the suspce of R n spnned y the m rows of A red s elements of R n 2 The Column Spce of mtrix A mn denoted y colpaq is the suspce of R m spnned y the n columns of A red s elements of R n The Null Spce of mtrix A nm denoted y NullpAq is the set of solutions to Ax The dimension of the null spce is denoted N ullitypaq dimpn ullpaqq 4 Guss Elimintion nd the sis of the row nd column spce of A: 2 7 Exmple: Given Mtrix A Find sis for rowpaq nd colpaq: 8 4 Step : Find the Echelon form B of the Mtrix A using GussElimintion: A B 8 4 Description: The Echelon form hs two nonzero rows, whose leding entries re in the first column nd in the second column These re clled Pivot Columns Moreover, if you kept trck of the opertions performed, we relize tht r r r 2 Step 2: The nonzero rows of the Echelon form B spn rowpaq $ '& rowpaq Spn 2 7, '% Indeed, s stted in the description r ws liner comintion of r nd r 2, nd r 2 ws exchnged y simpler vector r 2 r Step : The columns in the originl mtrix A corresponding to the pivot columns of the echelon form B, in this cse the first nd second, spn colpaq: $ & colpaq Spn % 2, 2 Even if you don t need to compute it, notice tht c c c 2 nd c 4?? Step 4: Add column of s to mtrix B nd find the solution set to the system of liner equtions understnding this new mtrix s the ugmented mtrix of homogeneous system 2 7 rb s Then x r nd x 4 s re free vriles nd x 2 r nd x r 9s{ Therefore, the solution set is: $ r '& 9 s 9, $ r s r r / '& 9, s : r, s P R Spn, / '% / '% / s This lst suspce is the null, nd thus nullitypaq 2 Remrk: Notice tht oth hve the sme dimension: dimprowpaqq 2 dimpcolpaqq, yet rowpaq R 4 nd colpaq R This is lwys the cse, the dimension of oth the column nd row spces re the sme, clled rnk of the mtrix A Remrk: Notice tht rnkpaq 2 nd nullitypaq 2 nd the second dimension of A (the numer of columns) is 4 It is lwys the cse tht A mn ñ rnkpaq nullitypaq n s, / /,  6
7 6 PopQuiz Consider the liner trnsformtion T : R Ñ R 2 descried y T 2 trnsformtion S : R 2 Ñ R 2 with mtrix B Find the mtrix A tht represents T Is T injective? Is T surjective? Is T invertile? Find KerpT q nd NullpAq Find RngepT q nd ColpAq x y z 4x y z 2y z And consider the liner Is S injective? Is S surjective? Is S invertile? If yes, cn you find S? Do it in two wys descried in lecture! Wht is the mtrix C ssocited to S T? Is S T injective? Is S T surjective? Is S T invertile? TrueFlse The union (s sets) of two suspces of R n is lwys suspce of R n The intersection (s sets) of two suspces of R n is lwys suspce of R n Given vector n nd mtrix A nm The set of solutions to the system Ax is not suspce of R m Let S nd S 2 e suspces of R n, nd define S to e the set of ll vectors of the form s s 2 is in S 2 Then S is suspce of R n s 2 where s is in S nd THE BIG THEOREM: Let T : R m Ñ R n with mtrix A r,, m s where,, m P R n Recll tht RnkpAq NullitypAq m nd RnkpAq n T is injective or onetoone ô KerpT q ô NullitypAq ô t,, m u is linerly independent set ô ll systems Ax hve t most one solution It requires m n T is surjective or onto ô RnkpAq n ô Spnp,, m q R n ô ll systems Ax hve t lest one solution It requires m n In the cse tht n m: Injective ô Surjective ô t,, n u is sis of R n 7
8 7 : Determinnts The determinnt is single numer ssocited to squre mtrix Nonsqure mtrices don t hve determinnt! 2 The determinnt of A pq is The determinnt of A 22 is d c c d c 4 The determinnt of A d e f is ei fg cdh fh di ceg g h i The mtrix M ij paq of the mtrix A nn is the pn q pn q mtrix otined y deleting the ith row nd the jth column of A The determinnt of M ij paq is clled the Minor of ij 6 The Cofctor C ij C ij paq of the mtrix A nn is p q i j detpm ij paqq 7 The Cofctor expnsions of the Determinnt: The Determinnt Formul expnded on the ith row: detpa nn q i C i i2 C i2 in C in The Determinnt Formul expnded on the jth column: 8 Importnt Exmples where the formul gets : detpi n q detpa nn q j C j 2j C 2j nj C nj The determinnt of digonl mtrix is the product of the digonl entries The determinnt of tringulr mtrix is the product of the digonl entries The determinnt of the trnspose of mtrix is the sme s the determinnt of the mtrix: detpa T q detpaq The determinnt of mtrix with row or column of zeros is zero detpabq detpaq detpbq 9 The Invertiility test of determinnts: A mtrix A nn is invertile if nd only if detpaq Moreover, if A is invertile, then detpa q detpaq In generl, it is flse tht the determinnts of two equivlent mtrices (through sequence of elementry opertions) re equl The row opertion of exchnging two rows chnges the determinnt y fctor, ie: If B is the mtrix otined from A, then detpbq detpaq The row opertion of multiplying row y constnt c chnges the determinnt y multiplying it y the constnt c, ie: If B is the mtrix otined from A, then detpbq c detpaq The row opertion of dding constnt multiple of row to nother row does not chnge the determinnt Exmple: Compute the determinnt of the following mtrix using the cofctor expnsion in the rd column 2 4 det p q det 7 7 p q det Alterntively, we could hve computed using the cofctor expnsion in the 2nd row: 2 4 det p q2 p q det 7 p q 2 4 pq det Show the formul for n using the cofctor expnsion nd the formul for n 2 8
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