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Let $n$ be a positive integer and $F $ a field. Suppose $A$ is an $n \times n$ matrix over $F$ and $P$ is an invertible $n \times n$ matrix over $F$. If $f$ is any polynomial over $F$, prove that $f(P^{-1}AP)=P^{-1}f(A)P$.

My attempt: By definition of $F[x]\times M_n(F)\to M_n(F)$, we have $f(P^{-1}AP)$ $=\sum_{i=0}^n f_i\cdot (P^{-1}AP)^i$. We claim $\forall k\geq 0$, $(P^{-1}AP)^k=P^{-1}A^kP$. Base case: $k=0$. Then $(P^{-1}AP)^0$ $=I_n$ and $P^{-1}A^0P$ $=P^{-1}I_n P$ $=P^{-1}P$ $=I_n$. Thus $(P^{-1}AP)^0=P^{-1}A^0P$. Inductive step: suppose $(P^{-1}AP)^k=P^{-1}A^kP$, for some $k\in \Bbb{N}$. Then $(P^{-1}AP)^{k+1}$ $=(P^{-1}AP)^k\cdot (P^{-1}AP)$ $=P^{-1}A^kP \cdot (P^{-1}AP)$ $=P^{-1}A^k (P\cdot P^{-1})AP$ $=P^{-1}A^k I_n AP$ $=P^{-1} A^{k+1}P$. Thus $(P^{-1}AP)^{k+1}$ $=P^{-1} A^{k+1}P$. By principle of mathematical induction $(P^{-1}AP)^k=P^{-1}A^kP$, $\forall k\ge 0$. So $\sum_{i=0}^n f_i\cdot (P^{-1}AP)^i$ $= \sum_{i=0}^n f_i\cdot (P^{-1}A^iP)$. Since $M_n(F)$ is linear algebra over $F$, we have $\sum_{i=0}^n f_i\cdot (P^{-1}A^iP)$ $= \sum_{i=0}^n P^{-1}(f_i\cdot A^i)P$ $= (\sum_{i=0}^n P^{-1}(f_i\cdot A^i))P$ $=P^{-1}(\sum_{i=0}^nf_i\cdot A^i)P$ $=P^{-1}f(A)P$. Here is definition of linear algebra over $F$. Is my proof correct?

user264745
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    That's really hard to read. Could you perhaps format it better? Space, paragraphs, line breaks, displayed equations, etc.... – Arturo Magidin Sep 07 '22 at 18:44
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    Yes, this is correct. If you wish to improve it, in addition to Arturo's vital suggestion, you could also make the definition of $f$ in terms of the $f_i$s a little more explicit. The $f_i$s are not introduced anywhere before they are used. – Theo Bendit Sep 07 '22 at 19:18

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First note that $(P^{-1}xP)^n=P^{-1}x^n(A)P$. This is obvious by inspection, it follows basically from the fact that multiplication is associative and $P^{-1}P=I$. The general result now follows:

$$P^{-1}f(A)P=P^{-1}(a_0+a_1A+\dotsb +a_nA^n)P=$$ $$P^{-1}a_0P+P^{-1}a_1AP+\dotsb +P^{-1}a_nA^nP=$$ $$a_0+a_1(P^{-1}AP)+a_2(P^{-1}AP)^2+\dotsb +a_n(P^{-1}A^nP)=f(P^{-1}AP)$$

Sebastiano
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