Prove that similar matrices have the same geometric multiplicity

Question

I'm trying to prove in two cases where two similar matrices are diagonalizable or non-diagonalizable? If they are diagonalizable, obviously the statement holds. But what if they are not? Can non-diagonalizable matrices still be similar? If yes, show me an example please. Thank you.

Answer 1

An example to your request: I believe that $$ \begin{pmatrix}0&1\\0&0\end{pmatrix} $$ and $$ \begin{pmatrix}1&-1\\1&-1\end{pmatrix} $$ are similar, yet not diagonalizable.

Answer 2

Let $A,B\in M_n(\Bbb K)$ so that $\exists P\in GL_n(\Bbb K), A = P B P^{-1}$

$\forall \lambda \in \Bbb K, \gamma_A(\lambda)=\dim \operatorname{Ker} \left( A - \lambda I_n\right) = \dim \left\{X \in \Bbb K ^n, AX=\lambda X\right\}=\dim \left\{X \in \Bbb K ^n, P B P^{-1}X=\lambda X\right\} =\dim \left\{X \in \Bbb K ^n, B (P^{-1}X)=\lambda (P^{-1}X)\right\}=\dim \left\{X \in \Bbb K ^n, B X=\lambda X\right\}=\dim \operatorname{Ker} \left( B - \lambda I_n\right)=\gamma_B(\lambda)$