MAT 211 Summer 2015 Homework 4 Solution Guide

Transcription

1 MAT 2 Summer 25 Homework 4 Solution Guide Due in class: June 22 nd Problem Which of the following subsets of P 2 are subspaces of P 2? If it is not a subspace, explain your reason; if it is a subspace, please find a basis for it a) {p(t) : p() = 2}; Answer No, because the zero polynomial is not in the set b) {p(t) : p(2) = } Answer Yes (You check the properties yourself) Let p(t) = a + bt + ct 2, then p(t) = a + 2b + 4c = Hence, a = (2b + 4c) Then p(t) = (2b + 4c) + bt + ct 2 = b(t 2) + c(t 2 4) The subspace is spanned by (t 2, t 2 4), and they are clearly linearly independent, hence (t 2, t 2 4) form a basis of the subspace Problem 2 Which of the following subsets of Mat (R) are subspaces of Mat (R)? If it is not a subspace, explain your reason; if it is a subspace, please find a basis for it a) The set of all diagonal matrices; b) The set of all matrices whose entries are all greater than or equal to Answer No, because it is not closed under scalar multiplication (take a matrix in the set then time it by ) Problem Find a basis for each of the following spaces and determine its dimension

2 [ ] a b a) The space of all matrices A = in Mat c d 2 2 (R) such that a + d = ; Answer This is Question in the midterm 2 Check the solution for midterm 2 b) The space of all 2 2 matrices A such that commute with B = c) The space of all matrices A such that commute with B = [ ] 2 a a 2 a Proof Let A = a 2 a 22 a 2 and A commutes with B Then AB = a a 2 a BA, ie, a a 2 a 2 a 22 a 2 a 2 a 22 = a a 2 a a a 2 Hence, we have ca = a 22 = a a 2 = a 2 a 2 = a = a 2 = Hence, a a 2 a A = a a 2 = a +a 2 +a a The basis is and the dimension is,,, 2

3 Problem 4 Find out which of the following transformations are linear For those that are linear, determine whether they are isomorphisms Explain your reason a) T (M) = M + I 2 from Mat 2 2 (R) to Mat 2 2 (R), where I 2 is the 2 2 identity matrix; Answer No, because while They are not equal T (M + M 2 ) = (M + M 2 ) + I 2 T (M ) + T (M 2 ) =(M + I 2 ) + (M 2 + I 2 ) =(M + M 2 ) + 2I 2 ([ ]) a b b) T = ad bc from Mat c d 2 2 (R) to R; Answer No, because ( [ ]) ([ ]) a b ka kb T k = T c d kc kd = (ka)(kd) (kb)(kc) = k 2 (ad bc) ([ ]) a b = k 2 T c d For any k,, the linearity does not hold [ ] f() f() c) T (f(t)) = from P f(2) f() to Mat 2 2 (R), where P is the space of all polynomials with degree smaller than or equal to Proof Yes, it is a linear transformation (Check the authenticity!) T is an isomorphism because A) dim(p ) = 4 = dim Mat 2 2 (R);

4 B) ker(t ) = {} If T (f(t)) is the zero matrix in Mat 2 2 (R), then f() = f() = f(2) = f() = Since ant nonzero polynomial in P has at most zeros, f can only be the zero function Problem 5 For which constants k is the linear transformation [ ] [ ] 2 T (M) = M M 4 k an isomorphism from Mat 2 2 (R) to Mat 2 2 (R)? Answer Given the standard basis B on Mat 2 2 (R) Suppose a [M] B = b c, d [ ] a b then M = c d [ ] [ ] [ ] [ ] 2 a b a b T (M) = 4 c d c d k [ ] [ ] 2a + c 2b + d a kb = 4c 4d kd [ ] a + c (2 k)b + d = 4c (4 k)d 4

5 Hence, a + c [T (M)] B = (2 k)b + d 4c (4 k)d = a + b 2 k + c 4 + d a = 2 k b 4 c 4 k d 4 k The matrix of T under the standard basis B is A = 2 k 4 4 k Then T is an isomorphism if and only if the matrix A is invertible, if and only if the rank of the matrix is 4 Hence, k 2 and k 4 Problem 6 Find the image and the kernel of the following linear transformations Write your answer as a span of vectors What is the rank and nullity of each linear transformation? Use your result to test the Rank-Nullity Theorem [ ] 2 a) T (M) = M from Mat (R) to Mat 2 2 (R); b) T (f(t)) = f(7) from P 2 to R; Answer Given P 2 the basis B = (, x, x 2 ), then for any polynomial a + bx + cx 2, then T (f(x)) = f(7) = a + 7b + 49c = [ 7 49 ] a b c 5

6 Hence, the matrix of T under the basis is A = [ 7 49 ] The image is span{}, which is indeed all the real numbers (Note that T maps P 2 to R) The kernel of A is 7 49 ker(a) = span,, namely, ker(t ) = span ( 7 + x, 49 + x 2) It is clear that dim(im(t )) =, dim(ker(t )) = 2 and dim(im(t )) + dim(ker(t )) = = dim(p 2 ) c) T (M) = M [ ] 2 [ ] 2 M from Mat 2 2 (R) to Mat 2 2 (R) Problem 7 Determine whether the following vectors in the given linear space are linearly independent or not Explain your answer and show all your work a) The polynomials f(t) = 7+t+t 2, g(t) = 9+9t+4t 2 and h(t) = +2t+t 2 in the linear space P 2 ; Answer Given P 2 the standard basis B = (, x, x 2 ), then 7 9 [f] B =, [g] B = 9, [h] B = 2 4 Since 7 9 rref 9 2 = I, 4 The coordinate vectors are linearly independent Hence, f, g and h are linearly independent in P 2 6

7 b) The matrices [ ], [ ] 2 4 in the linear space Mat 2 2 (R), [ ] 2 5 7, [ ] (Hint: you can first give a basis to the linear space, then you just need to show whether the coordinate vectors are linearly independent or not) Answer The same idea as a) The matrices turn out to be linearly dependent Problem 8 Let T be a linear transformation from Mat 2 2 (R) to Mat 2 2 (R) defined by [ ] 2 T (M) = M 6 Find the matrix of T with respect to the following basis of Mat 2 2 (R) ([ ] [ ] [ ] [ ]) B =,,, 2 2 Answer (Note that the basis hereisnot the standard basis) given any a M Mat 2 2 (R), suppose [M] B = b c, then d [ ] [ ] [ ] [ ] [ ] a + c b + d M = a + b + c + d = 2 2 2c a 2d b Then [ ] [ ] a + c b + d 2 T (M) = 2c a 2d b 6 [ ] (a + c) + (b + d) 2(a + c) + 6(b + d) = (2c a) + (2d b) 2(2c a) + 6(2d b) [ ] (a + b) + (c + d) (2a + 6b) + (2c + 6d) = 2(c + d) (a + b) 2(2c + 6d) (2a + 6b) [ ] [ ] = (a + b) + (c + d) + (2a + 6b) 2 7 [ ] + (2c + 6d) [ ] 2

8 Hence, a + b a [T (M)] B = 2a + 6b c + d = 2 6 b c = 2 6 [M] B 2c + d 2 6 d 2 6 Therefore, the matrix of T with respect to the given basis B is Problem 9 Let V be the space of all upper triangular 2 2 matrices Consider the linear transformation [ ] a b T = ai c 2 + bp + cp 2 [ ] 2 from V to V, where I 2 is the 2 2 identity matrix, P = and P 2 represents the product of P with itself a) Find the matrix A of T with respect to the basis ([ ] [ ] [ ] ) B =,, Answer First, we have [I 2 ] B =, [P ] B = 2, [ ] 8 and P 2 = hence 9 [P 2 ] B = 8 9 8

9 Let M = [ ] a a b, then [M] c B = b c [T (M)] B = a[i 2 ] B + b[p ] B + c[p 2 ] B = a + b 2 + c 8 9 = 2 8 [M] B 9 Hence, the matrix of T under the basis B is b) Find bases of the image and kernel of T, and thus determine the rank of T Proof hence, im(a) = span, 2, im(t ) = span ([ ] [ ]) 2, Rank(T ) = dim(im(t )) = 2 ker(a) = span 4, hence, ker(t ) = span ([ ]) 4 9

10 Problem In the plane V defined by the equation 2x + x 2 2x =, consider the bases 2 U = ( a, a 2 ) = 2, 2 2 and B = ( b, b 2 ) = 2, 2 a) Find the change of basis matrix S from B to U; Answer Since b = a, b 2 = a + a 2, we have [ ] [ ] [ b ] U =, [ b 2 ] U = Hence, the change of basis matrix S from B to U is S = [[ b ] U [ ] [ ] b 2 ] U = b) Find the change of basis matrix from U to B; Answer The matrix is S = [ ] c) Write an equation relating the matrices [ a a 2 ], [ b b2 ], and S = S B U Answer [ ] b b2 = [ ] a a 2 S