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I was trying to solve a problem that asks me to show $p(Q^{-1}AQ) = Q^{-1}p(A)Q$ where $p(t)$ is an arbitrary polynomial $a_nt^n+...+a_1t + a_0$. I am wondering whether it is true that $(Q^{-1}AQ)^k = Q^{-1}A^kQ$ since if this is true then the problem can be solved easily. I cannot seem to prove it since I am not quite sure how to deal with a product of matrices raised to a power. If it's not true then is there any other way to solve the problem?

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    $\begingroup$ $A² = AA$, $A^3 = AAA$, ect. Try writing your product out for certain $k$ and see what happens. $\endgroup$ – Kaynex Jun 17 '18 at 0:49
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Suppose it is true for $n$, $(QAQ^{-1})^{n+1}=(QAQ^{-1})^n(QAQ^{-1})=QA^nQ^{-1}QAQ^{-1}=QA^{n+1}Q^{-1}$ and the proof follows recursively.

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    $\begingroup$ for the second equation the $(QAQ)$ term, is it supposed to be $(QAQ^{-1})$? $\endgroup$ – PsychoCom Jun 17 '18 at 1:02
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You can just open the power k and write $(Q^{-1}AQ)^{k} = (Q^{-1}AQ) *(Q^{-1}AQ)*...*(Q^{-1}AQ)$ where $Q^{-1}AQ$ appears k times. It immediately gives the result after canceling all $Q^{-1}Q$s.

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