Assignment 2: Solutions, comments, marking schemeA marking scheme will appear at the end of this file some time after it is posted. For now, I just want to post some solutions and say some stuff.First some general comment: This is an honours course, and one is supposed to do things carefully in such a course. This assignment is the first place we are proving things involving properties of matrix algebra and related facts, so it seems to me we should do that carefully. Thus, below, I will check things like sizes, and check whether sums or products are defined, etc. First, in problem 1, let an nxn matrix A be given. 1.(a) (18.(a), 2.1) A - A^T is skew-symmetric. Pf: A^T is again nxn, so A - A^T is defined. Now, (A - A^T)^T = (A + (-1)A^T)^T = A^T + (-1)A^TT = A^T + (-1)A = A^T - A = (easy but long using stuff from p. 39 or the defn. of the negative of a matrix ) - (A - A^T). // 1.(b) (18.(b), 2.1) First, some comments: I noticed that part (c) had an answer in the back, but did not give myself the extra few seconds to realize that that answer would spoil part (b) if people looked at it. The purpose of doing 17 in class and assigning 18.(a) was to get people to do (b) with NO further "help". Even so, I essentially spoiled (b) with the hint given in class. Now... I have the impression that people looked at the answer to (c) in the back of the text and somehow decided that it answers (b). It doesn't. (b) asks us to show that THERE ARE matrices S, W with certain properties. (c) asks us to show that there is no choice about the S and the W in (b), i.e., that if we have ALREADY SHOWN that there is such an S and such a W, then we have found the only ones having those properties. Answer to (b): From class, we know that A + A^T is symmetric (problem 17 of 2.1). Call it C. Let S be (1/2)C. Then S^T = (Theorem 2, p. 41) (1/2) C^T = (1/2)C = S. So S is symmetric. From (a), D = A - A^T is skew-symmetric. Let W denote (1/2)D. Then W^T = (1/2)D^T = (1/2)(-D) = (1/2)((-1)D) = (Theorem 1, property 7) ((1/2)(-1)) D = ((-1)(1/2)) D = -((1/2)D) = - W. So W is skew-symmetric. And S + W = (1/2) (A + A^T) + (1/2) (A - A^T) = (use several properties of matrix algebra -- 5, 1, 6, 8, 3 of Theorem 1, at least) ... = A. 2. (20.(a) of 2.2) Suppose AX = 0 for every nx1 fellow X. Fix an i between 1 and n inclusive. Let X have a 1 in row i and 0's in all other rows. Then AX = column i of A. So column i of A is 0. Since this holds for all such i's, A is 0. 2. (37.(a) of 2.2) Let A, B be nxn. Then A+B is defined and is nxn. One checks that all the sums and products written below are defined. (A + B)^2 = (A+B)(A+B) = (Theorem 1, part 4) A(A+B) + B(A+B) = (Theorem 1, part 3, twice) AA + AB + BA + BB = A^2 + AB + BA + B^2. Now, suppose AB = BA. Then (A+B)^2 = A^2 + AB + AB + B^2 = A^2 + (1)AB + (1)AB + B^2 = A^2 + (1+1)AB + B^2 = A^2 + 2AB + B^2. For the implication in the other direction, we basically need a cancellation law for matrix addition: If U, V, W are matrices of the same size, and U + W = V + W, then U = V. Proof: Suppose the first equation holds for such matrices. Then U = U + 0 = U + (W + (-W)) = (U + W) + (-W) = (V + W) + (-W) = ... = V. // So we can cancel matrices in sums, on the right. By commutativity of matrix addition, we can cancel them on the left too. Now suppose (A+B)^2 = A^2 + 2 AB + B^2. Then from the expression for the left side given at the start of this solution, we can cancel A^2 on the left, and then B^2 on the right. And then write 2 AB as (1+1) AB = 1AB + 1AB = AB + AB, and finally cancel an AB on either left or right of what is left after the two cancellations in the expression referred to above, to get that AB = BA. // |