Showing geometric multiplicity is the same for two similar Matrices

[Muffins]

Homework Statement

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.

Homework Equations

A=S*B*S^-1
B=S^-1 *A*S

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!

 

[morphism]

Hint: S is invertible.

 

[Muffins]

So something like this?

We want to show that w is a linear combination of (Sv1, Sv2, ... ,Svg)
which says we have

c1*S*v1 + c2*S*v2 + ... + cg*S*vg = w

we can pull out the S...

S(c1v1 + c2v2 + ... + cgvg) = w

S^-1(S(c1v1 + c2v2 + ... + cgvg) = S^-1*w
(c1v1 + c2v2 + ... + cgvg) = S^-1 * w?

I tried going on a path similar to this last night but I wasn't seeing how this could show w is a linear combination of the S*v's... What am I still missing...?

 

[HallsofIvy]

I don't think you need to be that complicated. Saying that $\mu$ is an eigenvalue of A of multiplicity n basically means that $(A- \mu I)^n= 0$, where I and 0 are, of course, the identity and zero matrices of the same size as A. Now, if B is similar to A, B= SAS^-1 and A= S^-1BS for some invertible matrix, just as you say! Replace A in the previous equation by that and show that $(B- \mu I)^n= 0$ also.

Conceptually, of course, if A and B are similar matrices then they represent the same linear transformation written in different bases. "Eigenvalues" and "multiplicity of eigenvalues" are properties of the linear transformations not just specific representations so are the same for similar matrices.