Let $V$ be direct sum of its T-variant subspace $V_1,...,V_l$, and $C_T(x)=(x-e_1)^{m_1} \cdot\cdot\cdot (x-e_l)^{m_l}$ be characteristic polynomial of $T$. Assume $w(T-e_iI_n)^{m_i}=0$ for every $w \in V_i$. I want to prove minimal polynomial $m_{T_i}$ of $T_i =T|V_i$ is the form of $(x-e_i)^{d_i}$with $1\leq d_i \leq m_i$.
Many textbook writes this is trivial but I don't see why. so I tired to prove this by setting $T=\begin{pmatrix} T_i&O \\ C&D \end{pmatrix}$. i.e., $$w(T-e_iI_n)^{m_i}=w\begin{pmatrix} a_{m_i}T_i^{m_i}+...+a_1T_i+a_oI_n&O \\ C^*&D^* \end{pmatrix}=0$$ for every $w\in V_i$, but this didn't work for me.
Since $w(T - e_iI_n)^{m_i} = 0$ for all $w \in V_i$, the operator $(T - e_iI_n)^{m_i}$ is zero on $V_i$. Thus $T|_{V_i}$ satisfies the polynomial $(x - e_i)^{m_i}$. It follows that whatever the minimal polynomial of $T|_{V_i}$ is, it must be a divisor of $(x - e_i)^{m_i}$. Hence the minimal polynomial has the form $(x - e_i)^{d_i}$ for some $d_i \le m_i$.