Let $T$ an operator over a $\mathbb{F}$ vector space $\mathbb{V}$, with $\dim(\mathbb{V})<\infty$. Let $p=p_{1}^{r_{1}}\cdots p_{k}^{r_{k}}$ the minimal polynomial of $T$, and $\mathbb{V}=W_{1}\oplus\cdots\oplus W_{k}$ the primary decomposition of $\mathbb{V}$ for $T$.
Let $\mathbb{W}$ a subspace T-invariant of $\mathbb{V}$. I need to show that
$$\mathbb{W}=(\mathbb{W}\cap W_{1})\oplus\cdots\oplus (\mathbb{W}\cap W_{k}) $$
Here's my attempt:
Let $T_w$ be a linear operator obtained by restricting $T$ to $W$.
Then we know that the minimal polynomial of $T_w$ divides the minimal polynomial for $T$.
So the minimal polynomial of $T_w$ = $p_1^{e_1}\cdots p_k^{e_k}$ where $0 \le e_i \le r_i$
Now we know that by the primary decomposition theorem
$$\mathbb{W} = \mathbb{W_1} \oplus \cdots \oplus \mathbb{W_k} $$
where $\mathbb{W_i} = N(p_i(T)^{e_i})$
We see that $\mathbb{W_i} \subset W_i $ [As by the primary decomposition theorem $W_i = N(p_i(T)^{r_i})]$
Also $\mathbb{W_i} \subset \mathbb{W}$ then we see that $\mathbb{W_i} \subset \mathbb{W} \cap W_i$
Let $w \in \mathbb{W} \cap W_i$ then $p_i(T)^{r_i}(w) = 0$ then from here can we conclude that $p_i(T)^{e_i}(w) = 0?$ This is where I am stuck. Can someone help me out with this?
An answer to your immediate question was already provided in the comments, but it's worth noting that the result follows from the fact that the projection operators $\pi_i:\mathbb{V}\to\mathbb{V}$ onto the generalized eigenspaces $W_i$ of $T$ are polynomials in $T$. Since $\mathbb{W}$ is invariant under $T$, $\mathbb{W}$ is also invariant under each $\pi_i$, and for every $w\in\mathbb{W}$ we have $w=\sum_i\pi_iw$ so $$\mathbb{W}=(\mathbb{W}\cap W_1)+\cdots+(\mathbb{W}\cap W_k)$$ This sum is direct since the sum of the $W_i$'s is direct.
The fact that this holds for any $T$-invariant subspace $\mathbb{W}$ is actually equivalent to the $W_i$'s being sums of generalized eigenspaces of $T$.