# Eigenvalues of operator $p(T)$ in terms of the eigenvalues of $T$, where $p$ is a polynomial

Let $T$ be a linear operator on a finite dimensional vector space over an algebraically closed field $F.$ Let $f$ be a polynomial over $F.$ Prove that $c$ is a characteristic value of $f(T)$ iff $c=f(t)$, where $t$ is a characteristic value of $T.$

I have proved that $c$ is a characteristic value of $f(T)$, where $c=f(t)$ and $t$ is a characteristic value of $T$, but I cannot show rigorously that all characteristic values of $f(T)$ are of the form $f(t)$ where $t$ is the characteristic value of $T$.By intuition I know it is so $K$ is an algebraically closed field and because if the space $V$ has dimension $n$,then $T$ has $n$ eigenvalues (counted with multiplicity) and $f(T)$ also has also $n$ eigenvalues (counted with multiplicity). So, since for all $n$ eigenvalues $t$ of $T$, I have $n$ eigenvalues $f(t)$ of $f(T)$ then I exhaust the eigenvalues of $f(T)$ but I feel that this way of proving is not rigourous.

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Hint: use Jordan canonical form of $T$.

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Actually you can get away with just upper-triangularizing $T$. –  Qiaochu Yuan Aug 22 '12 at 4:34

By choosing a basis of the vector space, we can regard $T$ as a square matrix. There exists an invertible matrix $S$ such that $STS^{-1}$ is an upper triangular matrix. Since $Sf(T)S^{-1} = f(STS^{-1})$, $Sf(T)S^{-1}$ is also an upper triangular matrix. Every element of the diagonal of $f(STS^{-1})$ is of the form $f(t)$, where $t$ is an element of the diagonal of $STS^{-1}$. Hence we are done.

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By regular matrix you mean invertible matrix? Or what the definition of regular matrix? –  Karim Jonson Aug 22 '12 at 4:58
Invertible matrix –  Makoto Kato Aug 22 '12 at 5:46
@user38432 Please feel free to ask me about my answer. –  Makoto Kato Aug 22 '12 at 9:35
Thank you. I did the computations and they seem ok. –  Karim Jonson Aug 22 '12 at 13:38

Alternatively:

It's easy to see that if there's a nonzero vector $\alpha$ such that $T\alpha = t\alpha$ for some scalar $t$, then $$f(T)\alpha = f(t)\alpha,$$ for any $f(x)\in F[x]$.

For the other implication, first notice that the statement holds trivially if $f$ is a constant polynomial, so let's assume that $f$ is not constant.

Suppose that for the scalar $c$, there's a nonzero vector $\alpha$ such that $f(T)\alpha = c\alpha$. Consider the $T$-annhihilator of $\alpha$ $$S_T(\alpha;\{0\})=\{h(x)\in F[x] : h(T)\alpha = 0\}.$$

Let $p$ be the minimal polynomial of $T$ and $g$ the monic generator of $S_T(\alpha;\{0\})$. Since $F$ is algebraically closed \begin{gather*} p(x) = (x-a_1)^{d_1}\cdots (x-a_k)^{d_k}\\ \text{with $a_j$s characteristic values of $T$ and $d_j$s nonnegative integers.} \end{gather*}

Given that $p$ annihilates $T$, we know that $g(x)\mid p(x)$, so \begin{gather*} g(x) = (x-a_1)^{e_1}\cdots (x-a_k)^{e_k}\\ \text{with $0\leq e_j\leq d_j$ for each $j$.} \end{gather*}

The monic polynomial $f(x) - c$ is in $S_T(\alpha;\{0\})$ so $g(x)\mid (f(x)-c)$ and then there's a $j$ such that $(x-a_j)\mid (f(x)-c)$. This says that $a_j$ is a root of $f(x)-c$, that is $$f(a_j) = c$$ with $a_j$ a characteristic value of $T$.

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