MATH Linear Algebra Homework Solutions: #1 #6

Transcription

1 MATH 35-0 Linear Algebra Homework Solutions: # #6 Homework # [Cochran-Bjerke, L Solve the system using row operations to put the associated matrix in strictly triangular form and then back substitute 3x + x + x 3 0 x + x x 3 x x + x R 3 + R R R + 3R R Transforming the matrix back to a system of equations we get 3x + x + x 3 0 5x + x 3 6 x 3 Solving for each x value, we get So the values for x, x, x 3 are 5x + 6, x 3x + () + 0, 3x + 3 0, 3x 3, x x x x 3

2 [Glass, B Which of the following matrices are in row echelon form? Which are in reduced row echelon form? a) [ b) c) d) e) f) g) h) Properties of row echelon form All nonzero rows are above any row of all zero The leading coefficient (the first nonzero number from the left, also called the pivot) of a nonzero row is always strictly to the right of the leading coefficient of the row above it and the leading coefficient must be All entries in a column below a leading entry are zeroes The matrices in row echelon form are Properties of reduced row echelon form [ 3 4, , 0 0, 0 0 4, and It is in row echelon form Every leading coefficient is and is the only nonzero entry in its column The matrices in reduced row echelon form are 0 0, 0 0, and The remaining matrices are not in either form

3 3 [Gruenig, S For each of the systems of equations that follow, use Gaussian elimination to obtain a solution involving reduced row echelon form If there are infinite solutions, identify the free variables and describe all solutions (a) x x 3 x x 9 [ 3 9 R R R [ R R [ 3 0 R + R R [ (b) x 5 x x + 3x + x 3 x + x + x 3 3 3x + 4x + x R R R

4 R R R R 3 3R R R R R R R R 3 + R R Thus x 3 is a free variable Let x 3 t R, then x t + 5 x t 5 x 3 t (c) x x 3 x + x 5x + 8x

5 R R R R + R 3 R R R R + R R R + R 3 R This system of equations is inconsistent, so there is no solution 4 [Hoffman, L If compute the following quantities 6 8 (a) A (b) A + B A 0 and B 3, 4 5

6 3 (c) A 3B (d) (A) T (3B) T (e) AB (f) BA (g) A T B T (h) (BA) T [Li, Y For each pair of matrices, determine whether it is possible to perform the multiplication If it is possible, compute the product (a) (b) [ [ [ [ (c) (d) (e) Unable to compute because row number of first matrix is not equal to column matrix of second matrix [ [ [ [ [ [ Unable to compute because row number of first matrix is not equal to column matrix of second matrix 6

7 (f) [ [Parsons, C Let A and B be symmetric n n matrices For each of the following, determine whether the given matrix will be symmetric or non-symmetric It will be helpful to know that the (i, j)-entry of a matrix M AB, denoted m ij, can be computed with the sum m ij n a ik b kj, k where A is an m n matrix, B is an n r matrix, and M is an m r matrix Another hint is that A T A (a) C A + B We know A A T and B B T, or that a ij a ji and b ij b ji Then c ij a ij + b ij a ji + b ji c ji Therefore C C T, showing that C is symmetric (b) D A We know A A T, or that a ij a ji Then d ij Therefore D D T, and D is symmetric n k a ik a kj n k a ki a jk n k a jk a ki d ji (c) E AB We know that A A T and B B T Then E T (AB) T B T A T BA Since it may be that AB BA, then E E T, and E need not be symmetric Consider A, B 3 3, AB 4, BA (d) F ABA We know that A A T and B B T Then F T ((AB)A) T A T (AB) T A T B T A T ABA F Therefore F F T, and F is symmetric (e) G AB + BA We know that A A T and B B T Then G T (AB + BA) T (AB) T + (BA) T B T A T + A T B T BA + AB AB + BA G Therefore G G T, and G is symmetric (f) H AB BA We know that A A T and B B T Then H T (AB BA) T (AB + ( )(BA)) T (AB) T + (( )(BA)) T (AB) T + ( )(BA) T B T A T + ( )(A T B T ) BA AB Since it may be that AB BA BA AB, then H H T, and H need not be symmetric Consider A, B 3 3, AB 4, BA 0 4 0,

8 and 0 0 AB BA [Rarden, E Let C be a non-symmetric n n matrix For each of the following, determine whether the given matrix must be symmetric or could be non-symmetric (a) A G + G T Start with A T [G + G T T Since transpose respects addition, we can distribute it A T G T + (G T ) T We know that, for any matrix, A R m n, (A T ) T A, giving A T G T + G G + G T A (b) Thus, A A T, and is symmetric B G G T Start with B T [G G T T Since transpose respects addition, we can distribute it B T G T (G T ) T We know that, for any matrix, A R m n, (A T ) T A, giving Thus, B B T, and is not symmetric [ 4 Assume G 0 B B T G T G B [ 4 0 B T [ 0 4 [ 0 4 B 4 0 [ (c) C G T G Start with C T [GG T T We know that [AB T B T A T, (A, B) R m n C T G T (G T ) T 8

9 We know that, for any matrix, A R m n, (A T ) T A, giving C T G T G C (d) Thus, C C T, and is symmetric D G T G GG T Start with D T [G T G GG T T We can distribute the transpose, giving D T (G T G) T (GG T ) T We know from above that (G T G) T G T G, and (GG T ) T fits the same form, giving D T G T G GG T D (e) Thus, D D T, and is symmetric E (I + G)(I + G T ) Start with E T [(I + G)(I + G T ) T Since (I + G) and (I + G T ) are just matrices, it fits the form (AB) T B T A T, giving E T (I + G T ) T (I + G) T Distribute the transpose, to get E T (I T + (G T ) T )(I T + G T ) I is symmetric, thus, I I T, We also know that (G T ) T G E T (I + G)(I + G T ) E (f) Thus, E E T, and is therefore symmetric F (I + G)(I G T ) Start with F T [(I + G)(I G T ) T Since (I + G) and (I + G T ) are just matrices, it fits the form (AB) T B T A T, giving F T (I G T ) T (I + G) T Distribute the transpose, to get F T (I T (G T ) T )(I T + G T ) 9

10 I is symmetric, thus, I I T We also know that (G T ) T G F T (I G)(I + G T ) F Thus, F F T Take for example: ([ [ ) ([ [ ) ([ ) ([ ) F F T [ 5 3 F [Bodvig, A Evaluate the following determinants by hand (a) det (b) det (c) det (3)( 3) (5)( ) 9 ( 0) (5)(4) ( 8)( ) [ [(3)(4)(5) + ()(5)() + ()()(4) [()()(4) + (3)(5)(4) + ()()(5) (d) (e) (f) det det det [(4)()( 4) + (3)()(5) + (0)(3)( ) [(0)()(5) + (4)()( ) + (3)(3)( 4) 4 ( 44) 58 [()()(3) + (3)( )() + ()(4)() [()()() + ()( )() + (3)(4)(3) [()(3)(6) + ( )()(5) + ()()() [()(3)(5) + ()()() + ( )()(6)

11 9 [Herman, A Let A and B be matrices Justify your answers (a) Does det (A + B) det (A) + det (B)? Let [ a b A c d [ e f and B g h Then det (A) + det (B) (a + e)(d + h) (c + g)(b + f) ad + ah + ed + eh cb cf gb gf, and det (A + B) ad bc + eh fg No, det (A + B) does not equal det (A) + det (B) (b) Does det (AB) det (A) det (B)? Note that Then and [ ae + cf be + df AB ag + ch bg + dh det (AB) (ae + bg)(cf + dh) (ce + dg)(af + bh) aedh + bgcf cebh dgaf, det (A) det (B) (ad bc)(eh gf) aedh dgaf cebh + bgcf Yes, det (AB) det (A) det (B) (c) Does det (AB)) det (BA)? Note that Then [ ae + cf be + df BA ag + ch bg + dh det (BA) (ae + cf)(bg + dh) (ag + ch)(be + df) aedh + cfbg agdf chbe, and Yes, det (AB) det (BA) (d) Does det (A T ) det (A)? Note that det (AB) aedh + bgcf cebh dgaf [ (A T a c ) b d

12 Then and Yes, det (A T ) det (A) det (A T ) ad bc, det (A) ad cb

13 Homework # [Ayyangar, A Let V R, so the vectors are real numbers To remind us that vectors are real numbers in this example, feel free to not use vector notation (use x instead of x) Define scalar multiplication as usual αx α x, and define addition, denoted by, by x y max {x, y} Is V a vector space? Clearly spell out why or why not (every property needs explanation) Assume nothing is clear to the reader and everything needs to be explained Also, use the names of the properties you are checking along with the numbers we use in class for the sake of clarity Since you need the practice, make sure you address all ten properties (a) Additive closure: If x and y are real numbers, then x y max {x, y} will also be a real number Therefore, additive closure works (b) Scalar multiplicative closure: If x is a real number, then αx α x will also be a real number Therefore, scalar multiplicative closure works (c) Commutative addition: If x and y are real numbers, x y max {x, y} and y x max {y, x} Since max {x, y}max {y, x}, commutative addition works (d) Associative addition: If x, y, and z are real numbers (x y) z max {max {x, y}, z}, and x (y z) max {x, max {y, z}} Since max {max {x, y}, z} max {x, max {y, z}} max {x, y, z}, associative addition works (e) Additive identity: Let x be a real number and y be another real number called the zero vector x y x if and only if y < x As x gets smaller and smaller, y must get even smaller to make x y x Since there isn t a fixed zero vector y such that x y x for all x, the additive identity does not work We could try 0, since x 0 max {x, } x, but R (f) Additive inverse: Since the additive identity does not work, the additive inverse does not work either (g) Distribution: Let x 3, y 4, and α 5 Then α (x y) 5( 3) 5 Also (α x α y) max {5, 0} 0 Since 5 is not equal to 0, distribution does not work (h) Distribution: Let x 3, α 4, and β 5 Then (α β) x (α x β x) max {, 5} 5 Since is not equal to 5, distribution does not work (i) Scalar multiplication associativity: Let α, β, and x be real numbers α (βx) αβ(x) Therefore, scalar multiplication associativity works (j) Scalar multiplication identity: Let x be a real number Since x multiplied by equals x, scalar multiplication identity works Therefore, V is not a vector space 3

14 [Cychosz, R Determine whether the following sets form a subspace of R 3 (a) S {[x, x, x 3 T : x + x 3 } i Check if it s empty Note that 0 0 S This is a non-empty set ii Check additive closure Take x, y S Then x + x 3, and y + y 3 Note that x y x + y x + y x + y x + y, x 3 y 3 x 3 + y 3 and x + y + x 3 + y 3 Thus x + y S As a specific example, consider , and 5 + ( 3) This is not a subspace (b) S {[x, x, x 3 T : x x x 3 } i Check if it s empty Note that S This is a non-empty set ii Check additive closure Take x, y S Then x y x + y x + y x + y x + y, x y x + y and since x + y x + y x + y, then additive closure works for this αx iii Check scalar multiplication closure Take α R Then αx αx, and αx αx αx αx Thus scalar multiplication closure works for this This is a subspace 4

15 3 [Diehl, K Determine whether the following sets form a subspace of R (a) The set of all lower triangular matrices You may need to find out what this means Try wwwmathworldcom i [ Nonempty 0 S 3 ii Additive Closure Take A, B S, then [ [ [ a 0 b 0 a + b A + B + 0 S a a b b a + b a + b iii Scalar Multiplicative Closure Take A S and α R, then [ [ a 0 αa 0 α A α S a a αa αa Then S is a subspace of R (b) The set of all matrices A such that a i [ Nonempty 5 S 3 ii Additive Closure Take A, B S, then [ [ [ a b a A + B + + b a a b b a + b a + b / S Then S is not a subspace of R (c) The set of all matrices B such that b 0 i [ Nonempty 0 S 3 ii Additive Closure Take A, B S, then [ [ [ 0 a 0 b 0 a A + B + + b S a a b b a + b a + b iii Scalar Multiplicative Closure Take A S and α R, then [ [ 0 a 0 αa α A α S a a αa αa Then S is a subspace of R 5

16 (d) The set of all symmetric matrices i [ Nonempty S 5 ii Additive Closure Take A, B S, then [ [ [ a a A + B b b + a + b a + b S a a b b a + b a + b iii Scalar Multiplicative Closure Take A S and α R, then [ [ a a α A α αa αa S a a αa αa Then S is a subspace of R (e) The set of all singular matrices The following theorem will be helpful Theorem An n n matrix A is singular if and only if det (A) 0 i [ Nonempty 0 0 S 0 0 ii Additive Closure Take A, B S, where A det(b) 0 Then Thus A + B / S Then S is not a subspace of R [ a and B [ b ([ ) a 0 det (A + B) det a 0 b b 0 4 [Dykstra, B Determine whether the following sets form a subspace of P 4 (a) The set of all polynomials in P 4 of even degree Note that det(a) 0 and 3 Take 0 P 4e and note that 0 0x 0, which is a polynomial of degree less than 4 and of even degree Thus, P 4e is nonempty Take x, y P 4e where x x + x + and y x + x Then x + y (x + x + ) + ( x + x ) x P 4e Since x + y P 4e, P 4e does not exhibit additive closure x, y P 4e Thus, P 4e is not a subspace of P 4 (b) The set of all polynomials p(x) P 4 such that p(0) 0 3 Take p P 4b where p(x) x Note that p(0) 0 Thus, P 4b is nonempty Take p, q P 4b and note that p(0) q(0) 0 Then (p + q)(0) p(0) + q(0) Thus, (p + q)(x) P 4b 6

17 Take p P 4b and α R Then (αp)(0) α(p(0)) α0 0 Thus, (αp)(x) P 4b Since all three properties hold, P 4b is a proper subspace of P 4 5 [Hogan, J Determine whether the following sets form a subspace of C[, (a) The set of odd functions in C[, Recall that a function is odd if f( x) f(x) for all x in its domain, which is [, Answer: i Additive Closure Take f, g S Note (f +g)( x) f( x)+g( x) f(x) g(x) (f +g)(x) Therefore f + g S Thus this possesses additive closure ii Scalar Multiplicative Closure Take f S and α R Note (αf)( x) α(f( x)) α( f(x)) (αf)(x) Therefore αf S Thus this possesses Scalar Multiplicative Closure iii NonEmpty Note that x is a member of this set (b) The set of functions in C[, such that f( ) 0 and f() 0 Answer: i Additive Closure Take f, g S We know that f() 0, g() 0, f( ) 0, and g( ) 0 Thus when x, (f + g)() f() + g() Also when x, (f + g)( ) f( ) + g( ) Therefore (f + g) S ii Scalar Multiplicative Closure Take f S and α R Note (αf)(x) α(f(x)) Thus (αf)() α(f()) α(0) 0 Also (αf)( ) α(f( )) α(0) 0 Thus (αf)(x) S iii Nonempty An example is x (c) The set of functions in C[, such that f( ) 0 or f() 0 Answer: This is not a subspace as this does not possess additive closure We can take x, which is equal to zero when x, and x+, which is zero when x When these are added together we get (x + ) + (x ) x + x + x Thus () and ( ) Therefore this is not a subspace as (f + g) / S 7

18 6 [Mielke, M Let V be a vector space and let x, y, z V Prove the following: (a) If x + y x + z, then y z Proof Assume x + y x + z So, x + y + x x + z + x by additive inverse x + x + y x + x + z by commutative addition 0 + y 0 + z by additive inverse y z by additive identity (b) β0 0 for every β R Use the fact that β0 β(0 + 0) Proof We know β0 β(0 + 0) So, β0 β0 + β0 by distribution β0 + ( β0) β0 + β0 + ( β0) by additive inverse 0 β0 + 0 by additive inverse 0 β0 by additive identity (c) If αx 0 and α is nonzero, then x 0 Use (b) Proof Assume αx 0 and α 0 So, α (αx) α 0 (α α)x α 0 by scalar multiplication associativity x 0 by the 6(b) proof x 0 by scalar multiplicative identity 8

19 3 Homework #3 [Burch, A Let A R n n, and define the set C(A) to be the set of all matrices that commute with A, or C(A) {B : B R n n and AB BA} Show C(A) is a subspace of R n n Proof We need to check the three properties of a subspace (3) If we let I represent the identity matrix, we can say that, Therefore, C(A) is nonempty AI A IA () If we take B, B C(A) then, AB B A, AB B A, and which means, B + B C(A) () If we take B C(A) and α R then, This shows that αb C(A) Thus C(A) is a subspace of R n n A(B + B ) AB + AB B A + B A (B + B )A, (αb)a α(ba) α(ab) A(αB) [Wick, J In each of the following, determine the subspace of R consisting of all matrices that commute with the given matrix As an example, which you might want to work out for yourself so that you understand the process before trying the other problems, consider ([ ) {[ } 0 b 0 C : b 0 0 b, b R (a) [ Note that [ [ 0 0 a b 0 c d and [ [ a b 0 0 c d 0 Since the matrices commute, [ 0 0 a b [ 0 0, a b [ b 0 d 0 [ b 0, d 0 then b 0, a d, and c can equal anything Therefore, ([ ) {[ } 0 0 b 0 C : b 0 b b, b R 9

20 (b) [ 0 (c) Note that [ [ a b 0 c d [ a + c b + d c d and [ [ [ a b a a + b c d 0 c c + d Since the matrices commute, [ [ a + c b + d a a + b, c d c c + d then a + c a and c + d d means c 0, and b + d a + b means a d Therefore, ([ ) {[ } b b C : b 0 0 b, b R [ Note that [ [ a b c d, [ a + c b + d, a + c b + d and [ [ [ a b a + b a + b c d c + d c + d Since the matrices commute, [ [ a + c b + d a + b a + b, a + c b + d c + d c + d then a + c a + b c + d, and b + d a + b c + d, which means c b and a d Therefore, ([ ) {[ } b b C : b b b, b R 3 [Mollet, B Let A R Determine whether the following sets are subspaces of R (a) S {B R : BA O}, where O is the matrix of all zeros Solution: i Nonempty We [ need to ensure that a matrix exists within the subspace For example, the matrix 0 0 S 0 0, since OA O Thus S is nonempty ii Additive Closure Next, we need to check that additive closure holds Take B, C S So BA O and CA O So (B + C)A BA + CA O + O O Therefore (B + C) S 0

21 iii Scalar Multiplicative Closure Next, we need to check that scalar multiplicative closure holds Take α R Then Therefore αb S So S is a subspace of R (b) S {B R : AB BA} Solution: [ a a Counterexample: Take A a a (αb)a α(ba) αo O, B [ 0, and C 0 0 A(B + C) AI IA (B + C)A [ 0 0 So B + C I and 0 Therefore additive closure fails because A(B +C) (B +C)A Therefore S is NOT a subspace of R (Scalar multiplicative closure fails when α 0, and the nonempty condition fails if A I or A O) (c) S 3 {B R : AB + B O} Solution: i Nonempty We need to[ ensure that a matrix exists within the subspace that is nonempty For example, 0 0 the matrix S 0 0 3, since AO + O O Thus S 3 is nonempty ii Additive Closure Next, we need to check that additive closure holds Take B, C S So AB + B O and AC + C O So A(B + C) + (B + C) AB + AC + B + C (AB + B) + (AC + C) O + O O Therefore (B + C) S 3 iii Scalar Multiplicative Closure Next, we need to check that scalar multiplicative closure holds Take α R Then So αb S 3 Therefore S 3 is a subspace of R A(αB) + (αb) α(ab + B) α(o) O

22 4 [Steffen, D Given 3 x x u 6 v, 6 5 answer the following questions Justify your answers by writing out the necessary linear combination or explaining why no linear combination is possible (a) Is u Span (x, x )? Take α x + α x u Then, α 3 + α Then, There is no solution so a linear combination is not possible and u / Span (x, x ) (b) Is v Span (x, x )? Consider ( 6) 9 3x + ( )x ( 8) v ( 4) 5 Therefore, v Span (x, x ) 5 [Kreher, E Determine whether the following are spanning sets for R Justify your answers {[ [ } 4 (a), 3 6 [ x Take R, and consider x which yields the system, [ x x [ 4 x 3 6 x [ [ 4 α + α 3, 6 [ 4 x 0 0 3/x + x Since 3/x + x 0, we have a spanning set for the line, but not for R

23 {[ [ [ } (b),, 3 4 [ x Take R, and consider x which yields the system, [ x x [ x 3 4 x [ [ [ α + α + α 3 3, 4 [ 0 /7 ( 3x + x )/7 0 0/7 (x + x )/7 Since α 3 is a free variable, there are infinite solutions, making it a spanning set for R 6 [Janis, J Determine whether the following sets are spanning sets for P 3 Justify your answers (a) {, x, x } Take ax + bx + c P 3 Consider the linear combination ax + bx + c α () + α (x ) + α 3 (x ) ax + bx + c α + x α + x α 3 α 3 ax + bx + c (α + α 3 )x + α α 3 This system of equations can be solved using the augmented matrix 0 a 0 c b 0 a 0 c b There is only a solution to this system if b 0, but in general, b 0 for P 3 Thus this is not a spanning set for P 3 (b) {x +, x +, x } Take ax + bx + c P 3 Consider the linear combination ax + bx + c α (x + ) + α (x + ) + α 3 (x ) ax + bx + c xα + α + xα + α + x α 3 α 3 ax + bx + c x α 3 + x(α + α ) + α + α α 3 This system of equations can be solved using the augmented matrix 0 0 a 0 0 b + c + a 0 b 0 0 c + b a c 0 0 a We now know α b + c + a α c + b a α 3 a Since there is a solution for each α, then this is a spanning set for P 3 3

24 4 Homework #4 [Westlund, J Determine whether the following vectors are linearly independent in R 3 (a) (b) (c) 0 0,, 0 0 We must show the constants equal zero Consider the linear combination 0 c 0 + c + c When the matrix is reduced to row echelon form we get They are linearly independent because all the variables equal zero 4,, 4 Consider the linear combination 4 c + c + c When the matrix is reduced to row echelon form we get There are infinite solutions Therefore the vectors are linearly dependent 0, 3 We must show the constants equal zero 0 c + c 0 3 When the matrix is reduced to row echelon form we get The vectors are linearly independent because c c 0 4

25 [Waddell, T Determine whether the following vectors are linearly independent in P 3 (a) x +, x +, x Take c, c, c 3 R and consider c (x + ) + c (x + ) + c 3 (x ) 0 Distribute c, c, c 3 so Combine like terms c x + c + c x + c + c 3 x c 3 0 c 3 x + (c + c )x + (c + c c 3 ) 0 Rewrite zero as a linear combination of P 3 Set term coefficients to zero and c 3 x + (c + c )x + (c + c c 3 ) 0x + 0x + 0 c 3 0, c + c 0, c + c c 3 0 Create an augmented matrix and get it into reduced row echelon form , 0 which then reduces down to So c c c 3 0 The result fits the definition of linear independence, therefore the vectors are linearly independent in P 3 (b) x +, x Take c, c R and consider c (x + ) + c (x ) 0 Distribute c and c and rewrite zero as a linear combination Combine like terms Set term coefficients to zero and c x + c + c x c 0x + 0x + 0 c x + c x + (c c ) 0x + 0x + 0 c 0, c 0, c c 0 Since c c 0 the definition for linear independence has been met and therefore the vectors are linearly independent in P 3 5

26 Please note that part (b) and part (a) have different spans, but can still be linearly independent 3 [Schram, W Let x, x, and x 3 be linearly independent vectors in R n, and let y x x, y x 3 x, y 3 x 3 x Are y, y, and y 3 linearly independent? Justify your answer Consider the linear combination Plugging in the forms for y, y, and y 3 yields Combining like terms gives C y + C y + C 3 y 3 0 C (x x ) + C (x 3 x ) + C 3 (x 3 x ) 0 x ( C 3 C ) + x (C C ) + x 3 (C + C 3 ) 0 This can be written as a system of equations given by C 3 C 0 C C 0 C + C 3 0 Then the equations are put into a matrix to make sure that all solutions are zero to verify linear independence Since , then one of the variables is a free variable, and there are infinitely many solutions Thus variables are not restricted to zero, and the vectors y, y, y 3 are linearly dependent 4 [Deschamp, B Let A be an m n matrix Prove that if A has linearly independent column vectors, then N(A) {0} Hint: for any x R n, Ax x a + x a + + x n a n, where a i is the i-th column of A Proof We need to prove the only vector in N(A) is the zero vector Consider x N(A) We need to show x 0 Since x N(A), we know Ax 0 From the hint we know we can rewrite Ax as the linear combination Ax x a + x a + + x n a n Thus x a + x a + + x n a n 0 We know the columns of A are linearly independent, and so the only solution to the linear combination is x x x n 0 Since the x i are the entries of x, then x 0 Therefore N(A) {0} 6

27 5 [Molash, D Let S and T be subspaces of a vector space V Prove that their intersection, S T {v : v S and v T }, is a subspace of V Take x, y S T, and let α R (a) Nonempty Since S and T are subspaces, then 0 S, and 0 T Thus 0 S T, and S T is nonempty (b) Additive closure Within the subspace S, by the definition of a subspace, x+y S, whenever x, y S Similarly, within the subspace T, x + y T, whenever x, y T Since x, y S T, then x, y S and x, y T, and thus x + y S and x + y T Thus x + y S T, and S T has additive closure (c) Scalar multiplicative closure For the subspace S, the definition states that αx S, whenever α R and x S Similarly for the subspace T, αx T, whenever α R and x T Since x S T, then x S and x T From above we know αx S and αx T Thus αx S T, α R, and S T has scalar multiplicative closure As a result, S T is a subspace of V 7

28 5 Homework #5 [Hammer, E Find a basis for the subspace S of R 4 consisting of all vectors of the form a + b a b + c b, c where a, b, c R What is the dimension of S? Justify your answers First, we will start with finding the span Using the form a + b a b + c b, c we can find that the span is a + b a b 0 0 a b + c b a 0 + b b + c 0 a 0 + b + c 0 c 0 0 c 0 0 And lastly to find a basis we must find out if our three vectors are linearly independent Consider 0 0 a 0 + b + c Using an augmented matrix with the augmented portion being the zero vector, we can use row reduction to find if they are, in fact, linearly independent Note that Since a 0, b 0, and c 0, then we find that these vectors are linearly independent We find that these vectors are linearly independent and span, so we have found a basis 8

29 [Hill, G The vectors x, x 5, x 3 3, x 4 7, x span R 3 Pare down the set {x, x, x 3, x 4, x 5 } to form a basis of R 3 Justify your answer In order to find a basis for R 3, we must find a set of vectors that both spans and is linearly independent over R 3 First, let s see whether the set of vectors is linearly independent We can do this by assembling c x + c x + c 3 x 3 + c 4 x 4 + c 5 x 5 0 into one matrix and row reducing, yielding Looking at the row-reduced matrix, we see that several of the vectors have redundant information Since c 3 and c 4 are free variables, the vectors x 3 and x 4 are redundant, and by eliminating them we should achieve linear independence In the interest of clarity, however, we check the three vectors, yielding Thus c c c 5 0 Therefore, these three vectors are linearly independent However, we next have to show that the remaining vectors span R 3 It was proven in class that any set of n linearly independent vectors, where n is the dimension of the vector space V, spans V In this case, the vector space is R 3, a vector space with a dimension of 3 Since we have a linearly independent set of exactly 3 vectors, we know that they also span R 3 Thus, these vectors form a basis We also find that the dimension of this subspace is 3 3 [Nienhueser, A Let S be the subspace of P 3 consisting of all polynomials p(x) such that p(0) 0, and let T be the subspace of all polynomials q(x) such that q() 0 Find bases for (a) S The general form for P 3 is p(x) c x + c x + c 0 Then for S, p(0) 0 c 0 Thus c 0 0, and p(x) c x + c x With this we can determine that a spanning set of S is {x, x} We need to check linear independence Consider c x + c x 0 0x + 0x Then c 0 and c 0 Therefore the vectors are linearly independent With these two pieces of information, span and linearly independence, we can determine that x and x is a basis for S 9

30 (b) T The general form is again p(x) c x + c x + c 0 Then for T, p() 0 c + c + c 0 For this to be true, then c 0 (c + c ) Thus p(x) c x + c x (c + c ) c (x ) + c (x ) Meaning a spanning set of T is {x, x } Next we need to prove linear independence If then c (x ) + c (x ) 0, c x + c x (c + c ) 0x + 0x + 0, and c 0, c 0, and c c 0 Thus c 0, c 0, and the vectors are linearly independent With this set of information, span and linearly independence, we can determine that {x, x } is a basis for T (c) S T The general form is again p(x) c x + c x + c 0 Then for S T, p(0) 0 c 0 Thus c 0 0 Then p(x) c x + c x Also, p() 0 c + c, and thus c c Thus the span of S T is c x c x c (x x) In other words, {x x} is a spanning set for S T Since there is only one nonzero vector in our set, this means the set is linearly dependent With this we know that {x x} is a basis for S T 4 [Parsons, C Recall an early example of a vector space: {[ } x V : x, x R, x > 0, x where for α R and x, y V, [ αx α x x α [ x + y and x y x y Do the vectors [ [, 4 form a basis for V? Solution: We check the vectors in three stages (a) Confirm the zero vector Recall 0 x 0 [ 0 Take x [ x + 0 x [ x x [ x Then x x Also consider α R Then α 0 [ α 0 α [

31 (b) Check linear independence Take α, α R Consider [ [ [ 0 α α 4 [ [ [ α α 0 α 4 α [ [ α α 0 α 4 α [ [ α α 0 α +α Then we get α α 0 and α +α, or log ( α +α ) log (), which yields α +α 0 Using row reduction yields [ [ 0 0 0, and α α 0 Therefore these vectors are linearly independent (c) Check if they span V Take α, α R Consider [ [ α α 4 [ α α α +α Then x α α, or α x + α Also, x α +α, or log (x ) α + α Then Then [ x x [ x log (x ) (x + α ) + α 3α log (x ) x x α 3 (log (x ) x ) α x + 3 (log (x ) x ) 3 x + 3 log (x ) As a check, and α α 3 x + 3 log (x ) 3 log (x ) + 3 x x, α + α 3 x + ( 3 log (x ) + 3 log (x ) ) 3 x 3 x + 3 log (x ) + 3 log (x ) 3 x log (x ) Then α +α log (x ) x Therefore the vectors span V Since the vectors are linearly independent and they span V, they form a basis for V 3

32 6 Homework #6 [Steffen, D Consider the following bases for R : E {e, e }, U {u, u }, and W {w, w }, where [ [ [ [ 3 4 u, u, w 5, w 3 (a) Find the transition matrix from U to E (b) Find the transition matrix from E to U (c) Convert x S EU S UE ([ ) det 5 [ 3 to [x U x U S EU x E (d) Find the transition matrix from W to U S UE [ [ u u 5 [ 5 [ [ 5 3 E [ 5 [ 9 8 U [ 5 We know S S W U S UE S W E [ 5 UE and S W E [ Then [ [ [ S W U [ (e) Convert [x W to [x 4 U W [ [ [ 4 45 x U S W U x W x U W U As a check [ 4 W and [ U [ [ 6 [ 4 3 [ 5 [ 3, 0 E [ 3 0 E 3

33 [Waddell, T Given v [, v 6 [, S 4 V U [ 4 find vectors u and u such that S is the transition matrix from V {v, v } to U {u, u } To begin with, note that and S V U S EU S V E, S V E [, 6 4 as shown in class Therefore we have [4 [ S EU, 6 4 V U V E which can also be written as [ 4 V U [ x x x x EU [ 6 4 V E Writing out the matrix multiplication we get the system of equations and which then can be put into an augmented matrix After doing row reduction we get the form meaning x + 6x 4, x + 4x, x + 6x, x + 4x, S EU [ 5 0 We know the columns of S UE will be u and u, which means we need to find S UE Then [ S UE S 5 EU [ [ [ [ 0 Therefore u and u 5 33,

34 3 [Li, Y Let E {, x} and F {x, x + } be bases for P (a) Find the transition matrix from F to E We know that for any polynomial in P [, p(x) ax + b We find that E is the standard basis b For p(x) ax + b, we assign the vector We also find that for F, a x + x [ and The transition matrix is x + + x S F E [ [ (b) Find the transition matrix from E to F We can compute the transition matrix directly [ S EF S F E 4 [ [ [ 3 (c) Convert x to [x E Check that your answer is correct F We need to compute S F E x F x E [ From previous problem, we know that S F E It is stated in the problem that [ 3 x Hence we get that F x E S F E x F [ [ 3 F [ 4 4 We then get the polynomial p(x) 4x + 4 We can check our answer We take the basis F, and rewrite x E in terms of F, giving 3(x ) (x + ) 6x 3 x 4x 4 The answers match

35 4 [Hogan, J Several collections of vectors exist that are used as bases for various vector spaces involving functions Two such collections are the Laguerre polynomials, L, and the Hermite polynomials, H Below are the first four polynomials for each collection, which each form a basis for P 4 L {, t, 4t + t, 6 8t + 9t t 3 } H {, t, + 4t, t + 8t 3 } (a) Prove that one of these collections forms a basis for P 4 You need not prove that both collections are bases Solution: Put H into the linear combination c () + c (t) + c 3 ( + 4t ) + c 4 ( t + 8t 3 ) 0 c + tc c 3 + 4t c 3 tc 4 + 8t 3 c 4 0 (8c 4 )t 3 + (4c 3 )t + (c c 4 )t + (c c 3 ) 0 Now that the linear combination has been combined into like terms it is time to split them into the system of equations Finally, use row reduction to solve, yielding c 4 0 4c 3 0 c c 4 0 c c Thus c c c 3 c 4 0, and this system is linearly independent Since the dimension of P 4 is 4, then by a theorem we also have span Thus this is a basis for P 4 (b) Find the transition matrix from L to H Solution: We knows S HE and S LE

36 We compute S HE Finally 5 S LH S HE S LE Note we first found the transition matrices from H to E and from L to E Next we inverted the S HE in order to be able to go from E to H Finally, we multiplied the inverted matrix and S LE matrix to go from L to H (c) Convert 3 + t 6t to your choice of L or H Note that the given polynomial is written in the standard basis for P 4 Solution: If we convert to L, we start by encoding the polynomial in a vector using the form a at 3 + bt + ct + d b c d Then Then the conversion is done by computing t 6t x E 6 3 S LE x E Interpreting the vector as a polynomial yields If we convert to H, we compute Interpreting the vector as a polynomial yields ( 4t + t ) 3 ( t) 8 S HE x E () + 3(t) 6( + 4t ) L H E 36