Finding minimal polynomial of invariant subspace.

Question

Let us say I have an operator $T:V\rightarrow$$V$, and have been given its minimal polynomial $p(t)$. Let $g(t)$ be a polynomial. We are given (no need to prove) that $W= image(g(T))$ is a T-invariant subspace. How do I find the minimal polynomial of the restriction of $T$ to $W$, i.e. $T_W)$?

Answer 1

Denote $d(t) = \gcd(p(t),g(t))$. By the definition of a minimal polynomial, it suffices to show that $m(t) := p(t)/d(t)$ is such that $q(T|_W) = 0$ if and only if $m(t) \mid q(t)$. The easy part is showing that $m(t) \mid q(t)$ implies that $q(T|_W) = 0$.

For the reverse implication: if $q(T|_W) = 0$, we can say that for every $y \in \operatorname{im}(g(T))$, $q(T)y = 0$. That is, for every $x \in V$ $q(T)[g(T)x] = 0$. In other words, $q(T)g(T) = 0$. Thus, $p(t) \mid q(t) g(t)$.

Now, there exist coefficients $a,b$ such that $d(t) = ap(t) + bg(t)$. With that, $$ p(t) \mid q(t)g(t) \text{ and } p(t) \mid p(t)g(t)\implies\\ p(t) \mid q(t) \cdot (ap(t) + bg(t)) \implies\\ p(t) \mid q(t)d(t). $$ Since $p(t) \mid q(t)d(t)$, it follows that $m(t) = \frac{p(t)}{d(t)} \mid q(t)$. This ends the proof.