1. The problem statement, all variables and given/known data Suppose A and B are similar matrices, and that (mu) is an eigenvalue of A. We know that (mu) is also an eigenvalue of B, with the same algebraic multiplicity(proved in class) Suppose that g is the geometric multiplicity of (mu), as an eigenvalue of B. Show that (mu) has geometric multiplicity g as an eigenvalue of A. 2. Relevant equations A=S*B*S^-1 B=S^-1 *A*S 3. The attempt at a solution We know that g = dim(EigenSpace of (mu) for (B)), so this tells us that the space has a basis (v1, v2, ... , vg) From our previously worked problem, we know that if A and B are similar, an v is an eigenvector for B, then S*v is an eigenvector for B, so it follows that (Sv1, Sv2, ... , Svg) are eigenvectors for A. Now to verify that this is a basis of A, we need to show that they are linearly independent and span (EgienSpace of (mu) for A). So, (c1*S*v1 + c2*S*v2 + ... + cg*S*vg) = 0 vector, factor out an S and we get S*(c1*v1 + c2*v2 + ... + cg*vg) = 0 vector, and because the v's form a basis they're linearly independent, forcing the c's to be 0. Here's where my troubles hit. I'm entirely confused as to how to show (S*v1, ... S*vg) span the eigenspace(mu) for A. My professor suggested to take an arbitrary vector w, which is an element of the eigenspace(mu) for A, and show that the w is a linear combination of the S*v's.. but I'm at a loss on how to logically show that. Any help will be greatly appreciated guys!!!!!! Thank you mucho!