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Title looks a little bit twisted. What I want to say is the following:

$A\in\mathbb{R}^{n\times n}$, polynomial of matrix $A$: $P(A)=\displaystyle \sum_{k=0}^{n} c_k A^k$. $\lambda(A)$ is the set of eigenvalues of $A$.

So, we want to prove $\lambda(P(A))=P(\lambda(A))$

So far, I can prove it when $A$ is diagonalisable, but when $A$ is not diagonalisable, it seems we can't just simply use Jordan normal form to prove it. (Or this statement won't hold when it's not diagonalisable?)

Can anyone suggest me something and help me out? Thanks!

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3 Answers 3

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The result is true only if you allow complex eigenvalues.

If $\lambda$ is a eigenvalue of $A$ then $Av=\lambda v$ for a nonzero column vector $v\in\mathbb R^n$. It is easy to show that in this case $p(A)v=p(\lambda)v$ for any nonconstant polynomial $p$, which shows that $p\bigl(\lambda(A)\bigr)\subseteq\lambda\bigl(p(A)\bigr)$. For the converse, let $\mu\in\lambda\bigl(p(A)\bigr)$, where $p$ is a nonconstant polynomial. You can factor $p(X)-\mu$ on the complex numbers, obtaining

$$p(X)-\mu=a\prod_{i=1}^m(X-a_i)\,,$$

where $a,a_i\in\mathbb C$. Thus, we have $p(A)-\mu I=a\prod_{i=1}^m(A-a_iI)$. Since the matrix $p(A)-\mu I$ is not invertible, then some matrix $A-a_iI$ is not invertible. This shows that $\mu=p(a_i)$, where $a_i$ is a complex eigenvalue of $A$, that is, for some nonzero column vector $w\in\mathbb C^n$ you have $Aw=a_iw$.

For the necessity of the inclusion of complex eigenvalues, consider the matrix $A=\binom{0\ -1}{1\ \ \ 0}$, which has not real eigenvalues (check it). However $A^2=-I$ do has $-1$ as eigenvalue. What happens in this case is that $i$ and $-i$ are the complex eigenvalues of $A$, so in particular $\lambda(A^2)=\{(\pm i)^2\}=\{-1\}$.

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  • $\begingroup$ This explanation is awesome! $\endgroup$
    – Cancan
    Commented Sep 12, 2013 at 22:56
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    $\begingroup$ Good answer. Why $\mu=p(a_i)$? $\endgroup$
    – Seongqjini
    Commented Sep 28, 2016 at 12:52
  • $\begingroup$ @Planche Actually we have $p(a_j)=0$ for each $j$, because of the factorization of the polynomial $p(X)-\mu$. $\endgroup$ Commented Sep 28, 2016 at 22:12
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    $\begingroup$ @Planche I don't understand Matemáticos Chibchas' answer to your question so I'll give my own. There is an $x$ such that $(p(A)-I\mu)x = (a\Pi (A-I a_i))(x) = 0.$ Notice $a\neq 0$ as $p$ is non-constant. Then $(A-I a_i)(x)=0$ for some $i.$ This gives $Ax = a_i x$ and so $\mu x = p(A)x = p(a_i) x.$ $\endgroup$
    – JKEG
    Commented Mar 12, 2018 at 20:31
  • $\begingroup$ To add to @JKEG 's answer remember $A^2x=a_i^2x$, ie. $A(Ax)=a_i (a_i x)$ $\endgroup$
    – Brofessor
    Commented Jun 7, 2018 at 2:16
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As mentioned by Matemáticos Chibchas, the statement is not true if the underlying field is not algebraically closed. More precisely, the statement is not true if the underlying field is not a splitting field of the characteristic polynomial of $A$.

As mentioned by loup blanc, when the underlying field is a splitting field of the characteristic polynomial of $A$ (so that $A$ has $n$ eigenvalues in the field, counting multiplicity), the statement can be easily proved by triangularising $A$.

Without using triangularisation, the statement can be proved in two steps as follows (see Graham (2018), Matrix Theory and Applications for Scientists and Engineers, theorems 7.6 and 7.7):

Theorem 1. Let $A\in M_n(F)$. If $F$ is a splitting field of the characteristic polynomial of $A$ and the spectrum of $A$ (as a multiset) is $\{\lambda_1,\lambda_2,\ldots,\lambda_n\}$, then for every polynomial $p\in F[x]$, we have $$ \det\left(p(A)\right)=\prod_{i=1}^np(\lambda_i). $$ Proof. By assumption, we have $\det(A-xI)=\prod_{i=1}^n(\lambda_i-x)$. Let $p(x)=c\prod_{k=1}^r(x-c_k)$. Then \begin{aligned} \det(p(A)) &=\det\left(c\prod_{k=1}^r(A-c_kI)\right)\\ &=c^n\prod_{k=1}^r\det(A-c_kI)\\ &=c^n\prod_{k=1}^r\prod_{i=1}^n(c_k-\lambda_i)\\ &=\prod_{i=1}^n\left[c\prod_{k=1}^r(c_k-\lambda_i)\right]\\ &=\prod_{i=1}^np(\lambda_i). \end{aligned}

Theorem 2. With the assumptions in theorem 1, $p(A)$ has $n$ eigenvalues in $F$ and they (counting multiplicity) are $\{p(\lambda_1),p(\lambda_2),\ldots,p(\lambda_n)\}$.

Proof. Let $y$ be an indeterminate independent of $x$. Let $K$ be the algebraic closure of the field of fractions of $F[y]$. Then $f(x)=y-p(x)$ is a polynomial with coefficients in $K$ and hence it splits in $K$. By theorem 1 (with $K$ and $f$ taking the roles of $F$ and $p$), $$ {\det}_{F[y]}(yI-p(A)) ={\det}_{F[y]}(f(A)) ={\det}_{K}(f(A)) \stackrel{\text{thm 1}}{=}\prod_{i=1}^nf(\lambda_i) =\prod_{i=1}^n\left(y-p(\lambda_i)\right). $$ Hence the result follows.

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I am very surprised that @user1551 did not scream when he read this question (4 years ago). Of course, the answer is correct; yet, that is interesting, it is not the set of the eigenvalues but the MULTISET of the eigenvalues -we take into account the multiplicities-. Indeed, when using this result, in 99% of cases, we need the multiset version.

It is well-known that the multiset version proof is a direct consequence of the fact that any complex matrix is triangularizable.

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