Similar matrices have the same eigenvalues with the same geometric multiplicity

$B = P^{-1}AP \ \Longleftrightarrow \ PBP^{-1} = A$. If $Av = \lambda v$, then $PBP^{-1}v = \lambda v \ \Longrightarrow \ BP^{-1}v = \lambda P^{-1}v$. so, if $v$ is an eigenvector of $A$, with eigenvalue $\lambda$, then $P^{-1}v$ is an eigenvector of $B$ with the same eigenvalue. So, every eigenvalue of $A$ is an eigenvalue of $B$ and since you can interchange the roles of $A$ and $B$ in the previous calculations, every eigenvalue of $B$ is an eigenvalue of $A$ too. Hence, $A$ and $B$ have the same eigenvalues.

Geometrically, in fact, also $v$ and $P^{-1}v$ are the same vector, written in different coordinate systems. Geometrically, in fact, also $A$ and $B$ are matrices associated to the same endomorphism. So, they have the same eigenvalues, eigenvectors and geometric multiplicities.