The minimal polynomial of restricted T divides the minimal polynomial of T

Question

Given the linear operator $$T:V\to V,$$ $m(x)$ its characteristic polynomial and $$W⊆V$$ invariant subspace, I want to prove that $$m_{T|W}\:\left(x\right)|m_T\:\left(x\right),$$ etc: the minimal polynomial of restricted T divides the minimal polynomial of T.

How can I show that ?

Answer 1

This seems to be a question about minimal polynomials. But you say $m_T$ is the characteristic polynomial. Since that's inconsistent with various things, including standard notation, I'm going to assume you meant to say it's the minimal polynomial.

Basic Fact: Suppose $T$ is a linear operator with minimal polynomial $m$ and $p$ is a polynomial. Then $p(T)=0$ if and only if $m|p$.

Proof (this must be in the book, near the place where minimal polynomials are introduced): If $p=mq$ then $p(T)=m(T)q(T)=0q(T)=0$. Suppose on the other hand that $p(T)=0$. The division algorithm gives $$p=mq+r,$$where $q$ and $r$ are polynomials and $\deg(r)<\deg(m)$. Now $m(T)=0$ and $p(T)=0$ imply $r(T)=0$; hence the minimality of $m$ implies $r=0$, so $m|p$.

So it's enough to show that $$m_T(T|_W)=0,$$which is obvious: $$m_T(T|_W)=m_T(T)|_W=0|_W=0.$$