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I had been working on this problem here below, but seem to not know a precise and clean way to show the proof to the question below. I had about a few ways of doing it, but the statements/operations were pretty loosely used. The problem is as follows:

Suppose $\mathbf{A}$ is an $n \times n$ matrix with (not necessarily distinct) eigenvalues $\lambda_{1}, \lambda_{2}, \ldots, \lambda_{n}$. Can it be shown that the polynomial matrix

$p( \mathbf{A} ) = k_{m} \mathbf{A}^{m}+k_{m-1} \mathbf{A}^{m-1}+\ldots+k_{1} \mathbf{A} +k_{0} \mathbf{I} $

has the eigenvalues

$p(\lambda_{j}) = k_{m}{\lambda_{j}}^{m}+k_{m-1}{\lambda_{j}}^{m-1}+\ldots+k_{1}\lambda_{j}+k_{0}$

where $j = 1,2,\ldots,n$ and the same eigenvectors as $\mathbf{A}$.


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Looks like homework due to the imperative tone of the question, so here's a hint: Cayley-Hamilton. – user8285 Mar 15 '11 at 9:48
@Jerry: 1) I would prefer if you did not give short hints in answers. Either write something substantive or leave a comment. 2) You don't need Cayley-Hamilton for this. – Qiaochu Yuan Mar 15 '11 at 10:16

Here's another hint: If $X$ is an eigenvector of $A$, say $AX=\lambda X$, then you can use that to simplify $p(A)X$ into $(\text{some scalar value})X$, and that thing in front of $X$ is then an eigenvalue of $p(A)$, corresponding to the eigenvector $X$.

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This is very interesting question. I don't have the capabilites of answering this, but could you please elaborate on this a bit further. I would surely like to see how this question can be answered. – user7789 Mar 17 '11 at 19:45
@Christopher: OK. For example, a term like $A^2 X$ can be simplifed as follows: $A^2 X = AAX = A\lambda X=\lambda AX = \lambda \lambda X = \lambda^2 X$. Here I used the relation $AX=\lambda X$ twice, and I also used that the order of multiplication doesn't matter for a scalar times a matrix (so that I can change $A\lambda$ into $\lambda A$). Similiarly, $A^3 X=\lambda^3 X$, and so on, and this will have the consequence that $p(A)X = p(\lambda)X$ for any polynomial $p$, whenever $X$ is an eigenvector of $A$ with eigenvalue $\lambda$. – Hans Lundmark Mar 17 '11 at 20:47

Hint: write $A$ in upper triangular form. Locate the eigenvalues of $A$. Show that $p(A)$ is upper triangular as well and compute its diagonal. Conclude.

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You can start by writing $\mathbf{A} = P \mathbf{\Lambda} P^{-1}$, where $P$ is the matrix of eigenvectors and $\mathbf{\Lambda}$ is the diagonal matrix of eigenvalues. Therefore, you can write $\mathbf{A}^k$ as $P\mathbf{\Lambda}^k P^{-1}$, where $\mathbf{\Lambda}^k$ is simply the diagonal matrix made up of the $n^{th}$ powers of the eigenvalues of $\mathbf{A}$.

Using this fact, you can write

\begin{equation} p( \mathbf{A} ) = P(k_{m}\mathbf{\Lambda}^{m}+k_{m-1} \mathbf{\Lambda}^{m-1}+\ldots+k_{1} \mathbf{\Lambda} +k_{0} \mathbf{I})P^{-1} \ \end{equation}

This shows that $P(\mathbf{A})$ has eigenvalues given by $p(\lambda_i)$ for $i = 1,2,\ldots,n$ and the same eigenvectors as $\mathbf{A}$

EDIT: My proof works only for diagonalizable matrices as Theo Buehler has commented below.

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First: $A$ was not assumed diagonalizable. Second: why should there be an orthonormal basis of eigenvectors? In the complex case this is equivalent to the fact that the matrix is normal. Just replace $P^T$ by $P^{-1}$ throughout and then the diagonalizable case is fine. – t.b. Mar 18 '11 at 2:22
@Theo: I see your comment to replace everything in the statements to $P^{-1}$ from $P^{T}$. By doing this, would one of the statements be redundant where you have in the third line: $P^{-1} = P^{T}$. By changing this, we would have $P^{-1} = P^{-1}$. I don't know, but that would just look funny mathematically, unless thats what was trying to be proved. But I understand your stand point for the matrix to be diagonalizable, but maybe not every $P^{T}$ needs to be replaced by $P^{-1}$. Please correct me if I am having the wrong idea. – user7789 Mar 18 '11 at 3:24
@Theo - You're right. I messed up my answer a bit and I've edited it. – svenkatr Mar 18 '11 at 3:26
@Christopher: sure, this should simply be suppressed as svenkatr just did. I reiterate: this argument only works if $A$ is diagonalizable, which need not be the case. Actually, if $N \neq 0$ is nilpotent then $N^{n} = 0$, so $N^{n}$ has more eigenvectors than $N$. – t.b. Mar 18 '11 at 3:31
@Theo: Hmm, this is very interesting that is the case. So I'm wondering Theo, how could this argument be implemented into a working case for any general matrix or square matrix without being strictly for diagonal ones? Thanks – user7789 Mar 18 '11 at 3:47

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