Degree of T-conductor larger than zero but less than the full vector space I am from Physics background. I have been trying to figure out this proof for sometime now. Can anyone please solve it?
Let $T$ be a linear operator on a finite dimensional vector space $V$ and let $W$ be an invariant subspace of $T$. For any $\alpha\in V$ show that,
$0<max_{\alpha}\text{deg }s(\alpha;W)\leq\text{dim }V$,
where $s(\alpha;W)$ is the $T$-conductor of $\alpha$ into $W$.
For clarification of the term $T$-conductor: The set of all polynomial functions of $T$, that take a vector $\alpha$ into an invariant space $W$ forms an ideal. The minimal polynomial (i.e. least common factor) of that ideal is called $T$-conductor of $\alpha$ into $W$ and denoted by $s(\alpha;W)$. It can be found in the book of Linear Algebra by Hoffman and Kunze.
Thanks in advance. 

Hint: Show that, as a consequence the Cayley-Hamilton theorem, the characteristic polynomial $f$ satisfies $\operatorname{deg}(f) = \dim V$ and $f \in s(\alpha;W)$ for all $\alpha \in V$.  
