Math 110 Homework Assignment 17 due date: Feb. 17, 2017 1. Suppose that T : R3 R2 is given (in the standard coordinates) by the matrix \u0014 \u0015 2 1 5 A= . 1 1 3 Let B be the basis B = [(1, 1, 1), (2, 0, 1), (3, 2, 1)] of R3 , and let A be the basis A = [(3, 5), (1, 2)] of R2 , find the matrix for T with respect to the new basis on both sides. 2. Let ~v1 = (1, 2), ~v2 = (2, 1), and let B be the basis B = [~v1 , ~v2 ] of R2 . Let T be the linear transformation from R2 to R2 given by T (~v1 ) = ~v1 and T (~v2 ) = ~0. (a) Write down the matrix for T in the new basis B. (You should be able to do this directly from the definition of T and the definition of \"writing the matrix of a linear transformation with respect to a basis\"). (b) Use this to write down the matrix for T in the standard basis. You might want to compare the answer for (b) with the answer for H6 3(a), with m = 2. Can you see why these are the same? Note: In part (b) you need to go from the matrix in B-basis form to the matrix in standard basis form, which is the reverse of what we did in class, so think for a bit to figure out which way the change of basis matrices should go. 3. We'll check in class that the determinant of a square matrix doesn't change when we change basis. The purpose of this question is to show that the trace of a matrix also doesn't change when we change basis. For an n n matrix A, the trace of A, tr(A) For instance, if 1 A= 2 6 is the sum of the numbers on the diagonal. 3 5 7 9 0 4 then tr(A) = 1 + 7 + 4 = 12. In the aij notation for the entries of a matrix, tr(A) = a11 + a22 + a33 + + ann . (a) If A and B are n n matrices, prove that tr(AB) = tr(BA) (the formula in the book for ij-th entry for a product of matrices may help). 1 (b) Suppose that A is an n m matrix and B an m n matrix. Then both products AB (an n n matrix) and BA (an m m matrix) are square matrices, so we can take their traces. Prove or disprove: tr(AB) = tr(BA) in this case. (c) If C is an n n matrix, and M an invertible n n matrix, prove that tr(C) = tr(M 1 CM). Suggestion: If you make the right choice of matrices A and B, part (c) will follow from part (a) with very little work. 4. Suppose that x1 , x2 , . . . , xn are numbers. The 1 1 1 x1 x2 x3 2 x2 x2 x23 1 A = x3 3 x2 x33 1 .. .. .. . . . n1 n1 n1 x3 x2 x1 n n matrix 1 xn x2n x3n .. .. . . n1 xn is called the Vandermonde matrix, and is surprisingly useful to know a few basic facts about it. Let's establish one of them now. If any of the two xi 's are equal to each other, then of course det(A) = 0 since we will have two repeated columns. What we'd like to show is that if all of the xi 's are different, then det(A) 6= 0, i.e., that A is an invertible matrix. (a) Explain why showing that det(A) 6= 0 is the same as showing that det(At ) 6= 0, where At means the transpose of A. [This is a very short answer]. (b) Suppose that At is invertible. Explain why this means that Ker(At ) = {~0}. (c) Conversely, suppose that Ker(At ) = {~0}. Explain why this means that At is invertible. (Suggestion: The rank-nullity theorem may help at one point.) (d) Explain why showing that det(At ) 6= 0 is the same as showing that the only vector in Ker(At ) is the zero vector. (e) If ~v = (c0 , c1 , . . . , cn1 ) is a vector in the kernel of At , and ~v is not the zero vector, explain how this would give you a polynomial of degree 6 n 1 with more than n 1 roots, which would be a contradiction. [Hint: write out what At~v = ~0 means.] Make sure that you answer carefully, for instance, in your explanation why is it important that all the xi 's be different? 2 \f\f\f\f