Correctness of proof of uniqueness of Fitting decomposition

Question

I would like for someone to check my proof of the uniqueness of the Fitting decomposition of an endomorphism $f$ over a finite space $V$.

Suppose $U,W$ are two subspaces of $V$ s.t.

1. $U \oplus W = V$
2. $f\left(U\right) \subseteq U, f\left(V\right) \subseteq V$: this property is called $f$-invariance
3. $f_U: U \to U$ is nilpotent
4. $f_W: W \to W$ is an isomorphism

Then $U = F_0, W = F_1$ where $F_0,F_1$ are the Fitting modules of $f$

Proof:

1. Because $f_U$ is nilpotent, $F_0\left(f_U\right) = U \implies U \subseteq F_0\left(f\right)$
2. Suppose there exists $v \in W \cap F_0\left(f\right), v \neq 0$. Because $f_W$ is injective and $W$ is $f$-invariant, $f\left(v\right) \in W - \left\{0\right\} \implies \forall n \in \mathbb{N}, f^n\left(v\right) \neq 0$, and so $v \not\in F_0\left(f\right)$, contradiction. This implies that $W \cap F_0\left(f\right) = \{0\}$
3. Because $W \oplus U = V$ and $U \subseteq F_0$, this means that $V \subseteq W \oplus F_0$, but because $W,F_0 \subseteq V$, $W \oplus F_0 \subseteq V$. Applying the Grassmann formula, we get that $\dim U = \dim F_0$, which implies that $U = F_0$
4. Let's set $n$ to be an integer s.t. $\ker f^n = F_0 = U$, and suppose $\exists w \in W, w \not\in F_1$. Now, because $f_W$ is surjective, there exists $w^\star \in W$ s.t. $f^n\left(w^\star\right) = w$ and because $F_0 \oplus F_1 = V$ there are $u \in F_0, v \in F_1$ s.t. $u + v = w^\star$. In particular, this means that $f^n\left(u + v\right) = f^n\left(w^\star\right) = w$, which is equal to $f^n\left(v\right)$ because $u \in F_0$. Recall that $F_1$ is $f$-invariant, which means that $w \in F_1$, contradicting our choice of $w$. This means that $W \subseteq F_1$
5. With the same reasoning of point 3, we can conclude that $W = F_1$ Q.E.D.

Answer 1

I think your reasoning is correct, but you can be a bit more concise:

Given the decomposition $V = W \oplus U$, since $f_{|U}$ is nilpotent, there is a $k>0$ (in fact for all sufficiently large $k$) such that $f^k(U)=0$, and, since $f(W)=W$, it follows that $$ f^k(V)=f^k(U\oplus W) = f^k(U)+f^k(W) = 0+f^k(W)=f^k(W)=W. $$ Thus $U\subseteq \text{ker}(f^k)$ and $W= f^k(V)$ hence $$ \dim(U)+\dim(W) \leq \dim(\text{ker}(f^k))+\dim(f^k(V)) = \dim(V), $$ hence as $V = U \oplus W$, $U = \text{ker}(f^k)$ and $W = f^k(V)$. It follows $U$ and $W$ are unique if they exist, and to see that one just has to show that $f^k(V) \cap \text{ker}(f^k)=\{0\}$, but this is clear since if $u\in \text{ker}(f^k)\cap f^k(V)$ then $f^k(u)=0$ since $u\in \text{ker}(f^k)$ but as $u \in f^k(V)$, we have $f^k(u)=0$ if and only if $u=0$.