Verified answer

Recall that A is similar to B if and only if there is an invertible matrix S with S^(-1) A S = B. (Some books would have S A S^(-1) = B here, or put the S's with A instead of B. You should be able to prove that all of these choices are equivalent definitions of the concept of "similar to".)

Suppose that A is idempotent (that is, A^2 = A) and B is similar to A. Then there is an invertible matrix S with S^(-1) A S = B.

B^2 = BB = S^(-1) A S S^(-1) A S = S^(-1) A I A S = S^(-1) A^2 S = S^(-1) A S = B

showing that B is idempotent.

It is not true that if A is symmetric and B is similar to A that B must also be symmetric. For example consider A to be the 2x2 matrix with first row 0, 1 and second row 1, 0. Let S denote the 2x2 matrix with first row 1, 0 and second row 1, 1. You can check that S is invertible and that S^(-1) A S is the matrix with first row 1, 1 and second row 0, -1. Now A is similar to S^(-1) A S by definition, and A is symmetric, but S^(-1) A S is not.

a) true proof: if a matrix A is idempotent, then for some invertible matrix Q, A = Q^-1 * D * Q where D is a diagonal matrix containing only 1's and 0's on the diagonal. If B is similar to A, then there is a matrix P such that B = P^-1 * A * P expanding our earlier expression for A, we have B = Q^-1 * P^-1 * D * P * Q B = (PQ)^-1 * D * (PQ) Thus, B must be idempotent b) False see the other guy's example, he's right. In fact, any diagonalizable matrix is similar to a symmetric matrix.