# eigenvalues of a matrix under a polynomial

Let $A\in M_{n}(\mathbb{C})$, and suppose $A$ has eigenvalues $\lambda_{1},\lambda_{2},...,\lambda_{n}$ counting multiplicities. Let $f(\cdot)$ be a polynomial. How can we show that $f(A)$ has eigenvalues $f(\lambda_{1}),f(\lambda_{2}),\cdots,f(\lambda_{n})$ counting multiplicities?

I can show that if $Ax_{1}=\lambda x_{1}$ then $f(A)x_{1}=f(\lambda) x_{1}$, but this does not take multiplicities into account. Any suggestions will be much appriciated

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Use trigonalisation on $A$ to reduce to the case of upper triangular matrices. –  Joel Cohen Sep 28 '11 at 20:42
In addition to what Joel states, there is a well-known theorem that states: if B=(P^-1AP), then (P^-1)f(A)P=f(B), so you can draw conclusions about f(A) from the triangular matrix it is similar to. –  josh Sep 28 '11 at 20:55

Let $K$ be an algebraically closed field, let $A$ be an $n$ by $n$ matrix with coefficients in $K$, let $X$ be an indeterminate, let $f$ be in $K[X]$, let $L$ be the set of eigenvalues of $A$, let $M$ be the set of eigenvalues of $f(A)$, and for $\lambda\in L,\mu\in M$ let $E_\lambda,F_\mu$ be the respective generalized eigenspaces.

As the OP noticed, we have $f(L)\subset M.$

For $\mu$ in $M$ put $$S_\mu:=\bigoplus_{\lambda\in f^{-1}(\mu)}E_\lambda\subset K^n.$$ To answer the question, it suffices to show

$$(1)\quad F_\mu=S_\mu\quad\forall\ \mu\in M.$$

This will imply in particular $f(L)=M$.

We claim

$$(2)\quad E_\lambda\subset F_{f(\lambda)}\quad\forall\ \lambda\in L.$$

We prove that (2) implies (1). By (2) we have $S_\mu\subset F_\mu$ for all $\mu$ in $M$. As we also have $$\sum_{\mu\in M}\ \dim F_\mu=n=\sum_{\mu\in M}\ \dim S_\mu,$$ we get (1).

We're left with proving (2). Let $\lambda$ be in $L$. As $A$ and $f(A)$ preserve $E_\lambda$, they induce endomorphisms $A_\lambda$ and $f(A)_\lambda=f(A_\lambda)$ of $E_\lambda$. Let $\nu$ be an eigenvalue of $f(A)_\lambda$. It suffices to show $\nu=f(\lambda)$. We can assume that $f$ is non constant and monic. We have $$f(X)-\nu=(X-a_1)\cdots(X-a_d)$$ for some $a_1,\dots,a_d$ in $K$, and thus
$$f(A)_\lambda-\nu=f(A_\lambda)-\nu=(A_\lambda-a_1)\cdots(A_\lambda-a_d).$$ This endomorphism being singular and $\lambda$ being the only eigenvalue of $A_\lambda$, one of the $a_i$ is equal to $\lambda$, yielding $\nu=f(\lambda)$, as required.

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@Pierre-YvesGaillard: I think the crucial step is to show that whatever property you want is continuous on $M_n(\mathbb{C})$. I admit I didn't think through the argument for this case --- but $f$ is sufficiently continuous, and taking eigenvalues should be too. –  genneth Sep 29 '11 at 15:12