PROBLEM 5 (a) T We have B = S' A S where S is some invertible matrix and S' is the inverse of A. Therefore B is the product of invertible matrices, and so is itself invertible. (b) T Pythagoras: 3*3 + 4*4 = 5*5 so the dot product of x with y is zero. (c) T we have v = Au for some vector u. Therefore v.w = Au.w = u.Aw (because A is symmetric), and Aw=0 (w is in the kernel of A), so u.Aw = 0. (d) F The columns of an orthogonal matrix are unit vectors; therefore all the entries have absolute value at most 1. (e) F Similar matrices have the same determinant, but det(3A) = 27 det(A), so det(A)=0 and A is not invertible. (f) F Any real eigenvalue of an orthogonal matrix is either 1 or -1 because ||Av|| = ||v|| for all vectors v. (This is Example 4 of 7.1 in the textbook, page 299.) (g) T The sum of the dimensions of the two eigenspaces is n (where A is an n-by-n matrix), so the geometric multiplicities add up to at least n, hence exactly n. (h) F If Av = 2v then 0 = (AA - 4A + 3I) v = (2*2 - 4 + 3) v = 3v, so v=0. (i) F For example, A could be the non-diagonalizable matrix [ 1 1 0 ] [ 0 1 0 ] [ 0 0 2 ] (j) F Once we have n eigenvalues of an n-by-n matrix, we have them all, and the trace is just their sum. Here n=6 and all the eigenvalues are positive, so the trace must be positive as well.