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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...

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  • $\begingroup$ Latex is much simpler as it seems on the first spot, you can learn it in 5 minutes. $\endgroup$
    – peterh
    Commented Oct 27, 2015 at 4:47
  • $\begingroup$ @peterh sorry it was sent before I finish. Thx for ur comment $\endgroup$
    – mshx
    Commented Oct 27, 2015 at 4:50

1 Answer 1

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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 $u$:

$$P[u]_S=u$$

Hence:

$[Av]_S=P^{-1}P[Av]_S=P^{-1}Av=P^{-1}AP[v]_S$

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  • $\begingroup$ haha, yes it's homework. Anyway, I summit alr. ;P $\endgroup$
    – mshx
    Commented Oct 27, 2015 at 15:12

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