TITLE: Proof of a direct sum decomposition
QUESTION [6 upvotes]: I was trying to prove this statement:

If $N: V \to V$ is a nilpotent operator on a complex vector space, $N^k=0$ and $U\subset V$ is a subspace with $U \cap \ker(N^{k-1})= \{0\}$ then there exists a subspace $W \subset V$ with $NW \subset W$ and $V=W\oplus (U + NU + N^2U + \dots +N^{k-1}U)$.

Can you tell me please if my proof is ok? 
Proof by induction on $\dim(V)$. Base case: If $\dim(V) = 1$ then the statement is true because every subspace is either $\{0\}$ or $V$. Assume the statement is true if $\dim(V) = n-1$.
If $\dim(V) = n$ then because $V$ is a complex vector space $N$ has an eigen vector: $Nw = \lambda w$. Two cases: Either $w \in U$ or $w \notin U$.
1) If $w \notin U$: Then $\tilde{V} = V \setminus span(w)$ is a space satisfying the induction hypothesis (for $\tilde{N} = N\mid_{\tilde{V}}$). Therefore there exists  $\tilde{W}\subset\tilde{V}$ with $\tilde{V}=\tilde{W}\oplus(U + NU +\dots +N^{k-1}U)$ and $N\tilde{W} \subset \tilde{W}$. Then (for $W=\operatorname{span}(w)+\tilde{W}$) $V= W\oplus (U+NU+\dots+N^{k-1}U)$ and $NW \subset W$.
2) If $w \in U$: Then $\tilde{V} = V \setminus \operatorname{span}(w)$ is a space satisfying the induction hypothesis with $\tilde{U}=U\setminus \operatorname{span}(w)$. Therefore there exists  $\tilde{W}\subset\tilde{V}$ with $\tilde{V}=\tilde{W}\oplus(\tilde{U} + N\tilde{U} +...N^{k-1}\tilde{U})$ and $N\tilde{W} \subset \tilde{W}$. Then $V= \tilde{W}\oplus (U+NU+....+N^{k-1}U)$.

REPLY [5 votes]: Since $N$ is nilpotent, then $\lambda=0$ and $w\in \ker N\subset \ker(N^{k-1})$, so $w\not\in U$ and the case $2)$ is superfluous.
Further, $\tilde{V} = V \setminus\operatorname{span}(w)$ is not a space.
Addition. Below a proof is for $k=2$. I am sure that a similar proof is true for any $k$.
Let $N^2=0$. Denote $U+NU=A$. The direct sum $U\oplus  \ker N$ is a subspace in $V$. There is a direct complement: $V=U\oplus  \ker N\oplus B$. Denote $B+NB=C$ and prove that $A\cap C=0$. Indeed, let  $x\in A\cap C$. Then $x=Nu=Nb$ for some $u\in U,b\in B$ (since $U\cap C=A\cap B=0$). Hence $u-b\in  \ker N$ what is impossible. 
Now let $D$ is a direct complement: $\ker N=NU\oplus  NB\oplus D$. Then set $W=C\oplus D$.