Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
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

3 Answers 3

up vote 0 down vote accepted

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.

share|improve this answer

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.

share|improve this answer
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

A more elementary proof. (Note that although the text here is certainly my own work, I did not invent the idea of this proof. It came from an old book I once read, but I can't find a reference for it.)

We will begin by showing that the determinant of $f(A)$ is $f(\lambda_1) \dots f(\lambda_n)$.

By the fundamental theorem of algebra, the polynomial $f(x)$ can be factorised into $k$ linear factors over the complex numbers. Hence we can write $f(x) = a_k (x - c_1) \dots (x-c_k)$ for some complex numbers $c_1$, ..., $c_k$. Now a matrix commutes with all its powers, and with the identity, so it is also possible to write $f(A)$ as $f(A) = a_k (A - c_1 I) \dots (A - c_k I)$.

Also, denote the characteristic polynomial of $A$ by $p(\lambda) = | \lambda I - A|$. Since the eigenvalues of $A$ are $\lambda_1, \dots, \lambda_n$, the characteristic polynomial can be factorised as $p(\lambda) = (\lambda - \lambda_1)\dots(\lambda - \lambda_n)$.

Consider the determinant of $f(A)$: $$ \begin{align} |f(A)| &= |a_k (A - c_1 I) \dots (A - c_k I)|\\ &= (a_k)^n |A - c_1 I| \dots |A - c_k I| \\ &= (a_k)^n |-(c_1 I - A)| \dots |-(c_k I - A)|\\ &= (a_k)^n (-1)^n |c_1 I - A| \dots (-1)^n |c_kI - A|\\ &= (a_k)^n (-1)^{nk} |c_1 I - A| \dots |c_k I -A| \end{align} $$ Now $|c_i I - A|$ is $|\lambda I - A|$ with $\lambda$ replaced by $c_i$, that is, it is the characteristic polynomial of $A$ evaluated at $\lambda = c_i$. Thus $|c_i I - A| = p(c_i) = (c_i - \lambda_1)\dots(c_i - \lambda_n)$. $$ \begin{align} \text{So, } |f(A)| &= (a_k)^n (-1)^{nk} p(c_1) \dots p(c_k)\\ &= (a_k)^n (-1)^{nk}\times (c_1 - \lambda_1) \dots (c_1 - \lambda_n)\\ &\phantom{= (a_k)^n (-1)^{nk}}\ \times \dots\\ &\phantom{= (a_k)^n (-1)^{nk}}\ \times (c_k - \lambda_1) \dots (c_k - \lambda_n)\\ &= (a_k)^n \times (\lambda_1 - c_1) \dots (\lambda_n - c_1)\\ &\phantom{= (a_k)^n}\ \times \dots\\ &\phantom{= (a_k)^n}\ \times (\lambda_1 - c_k) \dots (\lambda_n - c_k)\\ &= (a_k)^n \times (\lambda_1 - c_1) \dots (\lambda_1 - c_k)\\ &\phantom{= (a_k)^n}\ \times \dots\\ &\phantom{= (a_k)^n}\ \times (\lambda_n - c_1) \dots (\lambda_n - c_k)\\ &= \phantom{\times}\ a_k (\lambda_1 - c_1) \dots (\lambda_1 - c_k)\\ &\ \phantom{=} \times \dots\\ & \ \phantom{=} \times a_k (\lambda_n - c_1) \dots (\lambda_n - c_k)\\ &= f(\lambda_1) \times \dots \times f(\lambda_n) \end{align} $$

The above argument shows that if $f(x)$ is any polynomial, then $|f(A)| = f(\lambda_1)\dots f(\lambda_n)$.

Now we will show that the eigenvalues of $f(A)$ are $f(\lambda_1)$, \dots, $f(\lambda_n)$.

Let $a$ be any number and consider the polynomial $h(x) = a - f(x)$. Then $h(A) = aI - f(A)$, and the argument above shows that $|h(A)| = h(\lambda_1) \dots h(\lambda_n)$. Substituting the formulas for $h(x)$ and $h(A)$ into this equation gives that $|aI - f(A)| = (a - f(\lambda_1))\dots (a-f(\lambda_n))$.

Since this is true for all possible $a$, it can be concluded that as polynomials, $|\lambda I - f(A)| = (\lambda - f(\lambda_1))\dots(\lambda - f(\lambda_n))$. But $|\lambda I - f(A)|$ is the characteristic polynomial of $f(A)$, which has been fully factorised here, so this implies that the eigenvalues of $f(A)$ are $f(\lambda_1)$, ..., $f(\lambda_n)$.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.