Let $T\in \text{End}(V)$, and let $Y\subset V$ be a T-invariant subspace with $Y\neq 0$ and $Y\neq V$.
Show the minimal polynomial of the induced map $T_{V/Y}:V/Y\rightarrow V/Y$ divides the minimal polynomial of $T$.
This is what I have done so far:
Let $m(x)$ be the minimal polynomial of $T$, and $m_{V/Y}(x)$ be the minimal polynomial of the induced mapping $T_{V/Y}$.
I have first shown that for any polynomial $f(x)\in \mathbb{F}[x]$, $f(T_{V/Y})[v]=[f(T)v]$, where $[v]$ is the corresponding equivalence class in $V/Y$. This thus implies that $m(T_{V/Y})[v]=[0]$ for any $v\in V$.
Then by the division algorithm, $m(x)=m_{V/Y}(x)q(x)+r(x)$, for some $q(x),r(x)\in \mathbb{F}[x]$, and $0\leq deg \ r(x)<deg \ m_{V/Y}(x)$
$$\implies r(T_{V/Y})[v]=m(T_{V/Y})[v]-m_{V/Y}(T_{V/Y})q(T_{V/Y})[v]=0$$
Suppose by contradiction, that $0<deg \ r(x)<deg \ m_{V/Y}(x)$. Let $g(x)=\frac {r(x)}{a}$, where $a\neq 0$ and $a$ is the leading coefficient of $r(x)$. Clearly $g(x)$ is monic and $deg \ r(x)=deg \ g(x)$.
I note that $g(T_{V/Y})[v]=\frac {r(T_{V/Y})[v]}{a}=0$
This is where I am stuck. I need to show that $g(T_{V/Y})=0$, which will thus results in a contradiction showing $deg \ g(x)<deg \ m_{V/Y}(x)$, which is impossible. Thus implying $r(x)=0$, proving the claim.
Is this even the right approach, what am I missing? Any help would be much appreciated. Thanks in advance!
Here is a simpler argument.
Since $Y$ is invariant, $m_V(T)=0$ implies $m_V(T_{V/Y})=0$. Therefore, $m_{V/Y}(x)$ divides $m_V(x)$, since $m_{V/Y}(x)$ divides all polynomials that kill $T_{V/Y}$.
Indeed, let $p(x)$ be a polynomial that kills $T_{V/Y}$. Write $p(x)=m_{V/Y}(x) q(x) + r(x)$, where $r(x)=0$ or $\deg r(x) < \deg m_{V/Y}(x)$. Then $r(T_{V/Y})=0$. By minimality of $m_{V/Y}(x)$, we cannot have $\deg r(x) < \deg m_{V/Y}(x)$ and so must have $r(x)=0$.