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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
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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
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2 Answers 2

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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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For complex matrices, one can use the fact that diagonalisable matrices are dense. Since the property you want is obviously true for diagonalisable ones, just argue that you can take limits of sequences of diagonalisable ones and get it for all matrices.

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Dear genneth: Could you detail your argument? –  Pierre-Yves Gaillard Sep 29 '11 at 6:43
    
@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
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