Let $V=F^n$ be vector space over a field $F$, and $T$ a linear map represented by matrix $A$. The following conditions are given
(i) $C_A=(x-e_1)^{m_1}\ldots(x-e_l)^{m_l}$ (so $m_1 +\cdots+m_l =n$)
(ii) $I_n=E_1+\cdots+E_l$
(iii) $E_i E_j=0$ if $i\neq j$ and $E_i^2=0$
(iv) $AE_i=E_iA$ and $(A-e_iI_n)^{m_i}E_i=0$
(v) $V_i=\{\,vE_i\mid v\in V\,\}$ for $i=1,\ldots,l$
(vi) $V$ is direct sum of $V_1,\ldots,V_l$
where $C_A$ is characteristic polynomial of $A$.
I know that $V_i$ is $T$-invariant. I need to prove the minimal polynomial of $T_i=T|_{V_i}$ is the form of $(x-e_i)^{d_i}$ for some $1\leq d_i \leq m_i$. my textbook(Abstarct Linear Algebra p.101) says this is true by (iv). but I don't see why.
I searched another and it writes since $w(A-e_iI_n)^{m_i}=0$ for every $w\in V_i$, above sentence is true. I see $w(A-e_iI_n)^{m_i}=0$ but I don't catch why $w(A-e_iI_n)^{m_i}=0$ is related to minimal polynomial of $T|_{V_i}$ and this implies such result.
(in this book, vector $v\in F^n$ is row vector, so I wrote $vA$ not $Av$)
By condition (iv) one has $E_i(A-e_iI_n)^{m_i}=0$ which says that every $v\in V_i$ has $v_i(T-e_iI_n)^{m_i}=0$. Then $(x-e_iI_n)^{m_i}$ is an annihilating polynomial of $T|_{V_i}$. Since the minimal polynomial of $T|_{V_i}$ divides any annihilating polynomial, it must be of the form $(x-e_iI_n)^d$ for some $0\leq d\leq m_i$. But $d=0$ is not possible since $V_i$ contains the eigenspace for the eigenvalue$~e_i$, so it has nonzero dimension.