Suppose $f$ is a polynomial with complex coefficients and let $A$ be an $n \times n$ matrix. Further, let $S$ be the invertible matrix s.t. $SJS^{-1} = A$ and $J$ is in Jordan Canonical form. How might I prove that $f(SJS^{-1})= Sf(J)S^{-1}$?

  • $\begingroup$ Can you see why if $f(x) = x^n$? If it works for $f(x)$ and $g(x)$, can you see why it works for $f(x)+g(x)$ and for $\alpha f(x)$? $\endgroup$ Aug 14, 2020 at 17:32
  • $\begingroup$ Hint : $(SJS^{-1})^2 = (SJS^{-1})(SJS^{-1})=SJ(S^{-1}S)JS^{-1} = SJJS^{-1} = SJ^2S^{-1}$. $\endgroup$ Aug 14, 2020 at 17:36

1 Answer 1


By induction, one can see that if $A=S^{-1}JS$, then $A^n=S^{-1}J^nS$. Then, for any polynomial $p(x)=a_0+a_1x+\ldots + a_nx^n$, $a_n\neq0$,

$$ \begin{align} p(A):&=a_0I+a_1A+\ldots + a_n A^n=a_0(S^{-1}IS) +a_1(S^{-1}JS)+\ldots + a_n(S^{-1}J^nS)\\ &=S^{-1}(a_0I + a_1J +\ldots + a_nJ^n)S\\ &=S^{-1}p(J)S \end{align} $$

This is independent on whether $J$ is a Canonical Jordan matrix or not. Of course, it is very useful when $J$ is such a cononical matrix for in that situation $J^n$ can be easily computed.


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