Let $T:V→V$ be linear operator whose minimal polynomial factors into monic, irreducible, pairwise coprime polynomials: $m_T (t)=p_1 (t)\cdots p_k (t)$. Then:
$V$ decomposes as a direct sum of the $W_i$, $V=W_1 \oplus \cdots \oplus W_k$
The $W_i$ are $T$-invariant
$p_i$ is the minimal polynomial of $T_i$, where $T_i$ is $T$ restricted to the image $W_i$
The proof for part 3 of the theorem is below. I would be grateful for clarification on misunderstandings or efficiencies. Many thanks.
Proof of part 3
To be the minimal polynomial of $T_i$, $p_i$ must be satisfied by $T_i$. From part 2 of the theorem, $W_i$=Ker $p_i (T)$, therefore $p_i (T)w_i=0$. As $p_i$ is restricted to $W_i$, Ker $p_i (T)$=Ker $p_i (T_i)$, therefore $p_i (T_i)w_i=0$. $p_i (T_i)$ is an annihilating polynomial and $p_i (T_i)=0$. Therefore $p_i$ is satisfied by $T_i$.
Now we take ${f_j=\frac{\prod\limits_{j=1}^k p_j}{p_i}}$, therefore $m_T=p_if_j$ and $m_T (T)=p_i (T_i) f_j (T_j)$ where $j=1,...,k$ and $j\neq i$.
As $m_T(T)=0$ and $p_i (T_i)=0$, $f_j (T_j)=0$ as $p_i$ can be any polynomial from $1$ to $k$ and a $p_i$ can be a polynomial of differing finite degree, therefore $f_j (T_j)=0$.
We take another polynomial $g$ where $g(T_i)=0$. $i$ can equal any of $1,...,k$ and $p_i$, $f_j$ can be polynomials of any finite degree, therefore we take $g(T_i)f_j(T_j)=0$ to potentially equal $m_T(T)$. As $g(T_i)f_j(T_j)=0$, $m_T(T)\space|\space g(T_i)f_j(T_j)$:
$\implies m_T(T)=p_i(T_i)f_j(T_j)\space|\space g(T_i)f_j(T_j)$
$\implies m_T(T)=p_i(T_i)\space|\space g(T_i)$
$\implies m_T(T)=p_i(T_i)$
$\implies m_i(T_i)=p_i(T_i)$
Therefore $p_i$ is the minimal polynomial of $T_i$.