Prove that $P^{-1}v$ is an eigenvector of $B$ corresponding to the eigenvalue $\lambda$.

Question

I'm having a difficutly proving the following:

Let $A,B \in M_{n\times n} (\mathbb{R})$ be similar matrices. Let $P \in M_{n\times n} (\mathbb{R})$ be an invertible matrix satisfying $P^{-1}AP = B.$ Let $v$ be an eigenvector of $A$ corresponding to the eigenvalue $\lambda$. Prove that $P^{-1}v$ is an eigenvector of $B$ corresponding to the eigenvalue $\lambda$.

I've tried looking the the common characteristic polynomial, but didn't get anywhere.

Answer 1

In fact, we can verify the result by direct calculation: since $P^{-1}AP=B$, we then have $$A=PBP^{-1}.$$ Noticing that $\lambda v=Av=PBP^{-1}v$, we have by right-multiplying $P^{-1}$ on both sides $$\lambda P^{-1}v=B(P^{-1}v),$$ which shows the result.

Answer 2

$BP^{-1}v=P^{-1}Av=P^{-1}\lambda v=\lambda P^{-1}v$...

Answer 3

\begin{aligned}P^{-1}AP & = B\newline \implies P^{-1}APP^{-1} &= BP^{-1} \newline \implies P^{-1}A &= BP^{-1} \newline \implies P^{-1}A v &= BP^{-1} v \newline \implies \lambda P^{-1} v &= BP^{-1} v\newline \implies \lambda (P^{-1} v) &= B(P^{-1} v) \end{aligned} The final line implies $P^{-1} v$ is an eigenvector of $B$, with eigenvalue $\lambda$.