# Applying a polynomial onto $S^{-1}JS$.

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 $$f(SJS^{-1})= Sf(J)S^{-1}$$?

• 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 $\alpha f(x)$? Aug 14, 2020 at 17:32
• Hint : $(SJS^{-1})^2 = (SJS^{-1})(SJS^{-1})=SJ(S^{-1}S)JS^{-1} = SJJS^{-1} = SJ^2S^{-1}$. Aug 14, 2020 at 17:36

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.