# Showing that similar matrices have the same minimal polynomial

I am in the process of proving the title.

The hint says, for any polynomial $f$, we have $$f(P^{-1}AP) = P^{-1}f(A)P.$$ A is an $n \times n$ matrix over $F$ while $P$ is an invertible matrix such that the above matrix multiplication $P^{-1}AP$ makes sense.

Why is this true?

Thank you.

• Hint: What is $(P^{-1}AP)^n$? – user60589 Nov 30 '15 at 20:27
• @user60589 It is $P^{-1}A^nP$ – user247618 Nov 30 '15 at 20:30
• Can you use this with the fact that $f$ is polynomial? – user60589 Nov 30 '15 at 20:34
• That would be useful if the OP wanted to prove the hint - but I think the task is to prove that similar matrices have the same minimal polynomial – WW1 Nov 30 '15 at 20:38
• @user60589 aha! got it. Thanks. Also thank you WW1 but I think I know how to proceed from here! – user247618 Nov 30 '15 at 20:44

if $f$ is a polynomial then you have: $$f(x)=a_nx^n+...+a_1x+a_0$$ Then you have $$f(P^{-1}AP)=a_n(P^{-1}AP)^n+...+a_1(P^{-1}AP)+a_0I$$ which is $$f(P^{-1}AP)=a_n(P^{-1}APP^{-1}AP...P^{-1}AP)+...+a_1(P^{-1}AP)+a_0P^{-1}IP$$ or $$f(P^{-1}AP)=P^{-1}a_nA^nP+...+P^{-1}a_1AP+a_0P^{-1}IP$$ which finally gives $$f(P^{-1}AP)=P^{-1}(a_nA^n+...+a_1A+a_0I)P=P^{-1}f(A)P$$