let J(A) be the Jordan form of A. and let f be some polynomial. is it true that $\det(xI-f(A))=\det(xI-f(J(A))$

Question

I tried a couple of examples and it turned out to be true, but I couldn't prove it..

Answer 1

Note that $A = S J(A) S^{-1}$, where $S$ is the appropriate matrix changing the basis; therefore $$ A^k = S \big(J(A)\big)^k S^{-1}$$ $$f(A) = S f(J(A)) S^{-1}$$and \begin{align} \det (xI - f(A)) &= \det \big(x S S^{-1} - S f(J(A)) S^{-1}\big) = \\ &= \det(S)\det \big(x I - f(J(A))\big)\det( S^{-1}) =\\&= \det \big(x I - f(J(A))\big) \end{align}