# Linear operator problem

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 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.