The minimal polynomial of $T|_W$ divides the minimal polynomial for $T$ I'm stuck on proving the following theorem.


Let $W$ be an invariant subspace for $T$. The minimal polynomial for $T|_W$ divides the minimal polynomial for $T$.


We have \begin{equation} A = \begin{pmatrix}
 B& C\\ 
 0& D
\end{pmatrix}
\end{equation}
where $A$ is the matrix representing $T$ on the ordered basis for $V$, and $B$ is the matrix representing $T|_W$ on the ordered basis for $W$. And it is obvious that the $k$-th power of $A$ is \begin{equation} A^k = \begin{pmatrix}
 B^k& C_k\\ 
 0& D^k
\end{pmatrix}
\end{equation}
Therefore, any polynomial which annihilates $A$ also annihilates $B$ (and $D$ too). So, the minimal polynomial for $B$ divides the minimal polynomial for $A$.
I got the above logic for proving the theorem, but i don't understand the last bold sentence. Explain me the last sentence more detailed.
 A: If $m(\lambda) = \lambda^{k}+a_{k-1}\lambda^{k-1}+\cdots a_1\lambda +a_0$ is the minimal polynomial for $T$, then
$$
            0=m(T) = T^{k}+a_{k-1}T^{k-1}+\cdots + a_1 T + a_{0} I.
$$
If $p$ is any other non-zero polynomial for which $p(T)=0$, then $m$ divides $p$.
The restriction $T_W$ of $T$ to the invariant subspace $W$ also satisfies
$$
            0 = T_{W}^{k}+a_{k-1}T_{W}^{k-1}+\cdots+a_1 T_{W}+a_{0}I_W
$$
Therefore, the minimal polynomial $m_W$ for $T_{W}$ must divide $m$.
A: The set of polynomials in $\mathbb{F}[T]$ annihilating $V$ are closed under multiplication by $\mathbb{F}[T]$ and addition, and so form an ideal of the polynomial ring $\mathbb{F}[T]$. Since $\mathbb{F}$ is a field, $\mathbb{F}[T]$ is a principal ideal domain, and hence the ideal of polynomials in $T$ annihilating $V$ is generated by $T$'s minimal polynomial $\mu_V$. Every element of the ideal $\left\langle \mu_V \right\rangle$ also annihilates the subspace $W$, and so must be contained in the ideal of polynomials in $T$ annihilating $W$, generated by the minimal polynomial of $T_W$, $\left\langle \mu_W \right\rangle$.
Since $\left \langle \mu_W \right\rangle \supset \left \langle \mu_V \right \rangle$ and both are principal ideals, we can conclude that the generator $\mu_W$ divides $\mu_V$.
A: To complement TrialAndError's answer, notice that
$$m_T(T) = \begin{pmatrix}
\sum_{i=0}^{k-1} a_i (T|_W)^i + T|_W^{k} & * \\
0 & * \\
\end{pmatrix} = 0$$
Which implies that $m_T(T|_W) = 0$, hence $m_T$ belongs to the ideal of polynomials that annihilate $T|_W$, meaning that the unique monic generator of the ideal ($m_{T|_W}$) must divide it:
$$m_{T|_W} | m_T$$
