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Let $n$ be a positive integer and $\mathbb{F}$ be a field. Suppose $A$ $\in$ $M_{n\times n}(\mathbb{F})$ and $P$ is and invertible matrix, such that $P \in M_{n\times n}(\mathbb{F})$. If $f$ is any polynomial over $\mathbb{F}$, prove that

$$f(P^{-1}AP) = P^{-1}f(A)P.$$

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Have you tried induction on the degree of $f$? – ajotatxe Apr 20 '14 at 0:11
No, I will think in the solution using induction. – ViKaN Apr 20 '14 at 0:36
No need for induction. Just write out an arbitrary polynomial and use algebraic properties. See the hint I gave below. – Seth Apr 20 '14 at 1:27

Hint: $(P^{-1}AP)^m = P^{-1}A^mP$.

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I think I have the proof, let show you, I will be a little bit formal.

Let $f \in \mathbb{F}$[$X$]. Such that $f = \sum_{i=0}^{n}c_ix^i$. For $n,c_i\in \mathbb{F}$.

Let $\mathcal{A}$ be a linear algebra, and we define $f(\alpha) = \sum_{i=0}^{n}c_i\alpha^i$, for all $\alpha \in \mathcal{A}$.

First we will show that $(P^{-1}AP)^m = P^{-1}A^mP.$ This mean the following,

$(P^{-1}AP)^m$ = $(P^{-1}AP)\cdot(P^{-1}AP)\ldots(P^{-1}AP)$ m times.

But as the multiplication of matrices is associative, we get that following:

$(P^{-1}AP)^m = P^{-1}A(PP^{-1})A(PP^{-1})\ldots(PP^{-1})AP$ $= P^{-1}A^mP$ ----- ($\ast$).

So by the definition of our funtion we get:

$$f(P^{-1}AP) = \sum_{i=0}^{n}c_i(P^{-1}AP)^i$$ by ($\ast$) we can state the following: $$=\sum_{i=0}^{n}c_iP^{-1}A^iP$$ $$=P^{-1}(\sum_{i=0}^{n}c_iA^i)P$$ $$=P^{-1}f(A)P$$ As we wanted to proof. $\blacksquare$

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Good job, this was what I had in mind. – Seth Apr 20 '14 at 1:38

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