Hoffman Kunze problem 12 section6.4

Question

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

Solution: I am using this lemma:

Lemma: Suppose that $T\alpha=c\alpha$. If $f$ is any polynomial , then $f(T)\alpha=f(c)\alpha$.

$\Rightarrow$) If $c$ is characteristic value of $f(T)$ then there exist non zero vector $\alpha$ in $V$ such that $f(T)\alpha=c\alpha$ . If $f$ is constant polynomial say $f(t)$ then $f(t)=c$.

$\Leftarrow$) If $c=f(t)$ ,where $t$ is eigen value of $T$ then there exist non zero vector $\alpha$ in $V$ such that $T\alpha=t\alpha$. By Lemma $f(T)\alpha=f(t)\alpha$.

Is this solution correct?I don't need other solution. I have an alternative solution. Any help or suggestions regarding this solution.

Thanks in advance.

Answer 1

Let $c$ be an eigenvalue of $f(T)$. Then $W=\text{ker}(f(T)-cI)$ is nontrivial and also a $T$ invariant subspace.

Let $g(x)=f(x)-c,$ then $g(x)$ annhilates $T|_W$. So, minimal polynomial of $T|_W$, $m_W(x)$ divides $g(x)$. But $m_W(x)$ divides the minimal polynomial, $m(x)$ of $T$ also. Then take $t$ any root of $m_W(x)$ which is common root of $m(x)$ and $g(x)$.