# prove $[Av]_s = P^{-1} AP[v]_s$

Let $S= \{v_1,...,v_n\}$ be a basis for $\Bbb R^n$ and $P = [v_1\cdots v_n]$ where $v_j$ is column vector . Prove that for any square matrix A of order n and column vector V, $$[Av]_s = P^{-1} AP [v]_s$$

I don't even know how to start, can someone kindly give a hint? Thanks! Sorry for the formatting if it is wrong...

• Latex is much simpler as it seems on the first spot, you can learn it in 5 minutes. – user259412 Oct 27 '15 at 4:47
• @peterh sorry it was sent before I finish. Thx for ur comment – mshx Oct 27 '15 at 4:50

That looks like my homework (which is probably late by now). The hint states that $P$ is the transition matrix from $S$ to the standard basis $E$.
This means that for any column vector $u$:
$$P[u]_S=u$$
$[Av]_S=P^{-1}P[Av]_S=P^{-1}Av=P^{-1}AP[v]_S$