Diagonalization problem - Eigenvectors

Question

Considering the fallowing square matrix: A, B and P, (det P $ \neq0$), and the given expression:

$P^{-1}AP = B$

If $v$ is an eigenvector associated with eigenvalue $\lambda$ from B, which is eigenvector $w$ from A corresponds to the same eigenvector?

Answer 1

Since $v$ is an eigenvector associated with eigenvalue $\lambda$ for matrix $B$,

$$ P^{-1}APv = Bv = \lambda v$$

Then multiply by $P$ on the left.

$$ A(Pv) = \lambda (Pv)$$.

So, $Pv$ is an eigenvector associated with eigenvalue $\lambda$ for matrix $A$