Same Eigenvalues?

Let $v$ be an eigenvector of $A$ with corresponding eigenvalue $\lambda$ , ie $Av=\lambda v$ .

We have $M^{-1} A=BM^{-1}$ ; so $BM^{-1} v = \lambda M^{-1} v$ ; hence $M^{-1} v$ is an eigenvector of $B$ with corresponding eigenvalue $\lambda$ , so all eigenvalues of $A$ are also eigenvalues of $B$ (and vice-versa).

Alternatively, the characteristic polynomial of $A$ is $\chi_A(t)=|tI-A|$ . The characteristic polynomial of $B$ is

$\chi_B(t)=|tI-B|=|tI-M^{-1} AM|=|M^{-1}| \cdot |tM-AM| = |M^{-1}|\cdot |M| \cdot |t-A| = |tI-A| = \chi_A(t)$

so again we see the eigenvalues are the same.

The matrices $A$ and $B$ represent the same linear operator in a different basis (other terms for this are that $A$ and $B$ are similar, or conjugate matrices).