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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 $\mathbf{A}$.

Thanks.

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    Looks like homework due to the imperative tone of the question, so here's a hint: Cayley-Hamilton.2011-03-15
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    @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.2011-03-15

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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.2011-03-17
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    @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 $\lambda$.2011-03-17
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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.2011-03-18
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    @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.2011-03-18
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    @Theo - You're right. I messed up my answer a bit and I've edited it.2011-03-18
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    @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 $N$.2011-03-18
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    @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? Thanks2011-03-18
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    @Christopher: That's the worst that can happen, as far as I can see (but I haven't checked carefully). I'd use the Jordan normal form: http://en.wikipedia.org/wiki/Jordan_normal_form. The blocks to zero eigenvalues can be annihilated, the others can't, I think.2011-03-18
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    @Christopher: Another way to save it is to notice that the diagonalizable matrices are dense in the set of all matrices. Thus, if you have a continuous function $f: M_n(\mathbb{R})\rightarrow\mathbb{R}$ which is conjugation invariant and you show some property holds on the diagonal matrices, it follows that it holds everywhere. Of course, you'd need to prove that the characteristic polynomial (say, you think of $a_nx^n+ ...+a_0$ as the point $(a_n,...,a_0)$ in $\mathbb{R}^n$) as a function of the matrix is continuous first, but this method is often useful nonetheless.2011-03-19