$A$ unipotent and $A^k(U)=U$ for a subspace $U\subset \mathbb{C}^n$, does $A(U)=U$? Let $A$ be a unipotent $n\times n$ complex matrix.
Let $k\geq 1$ be an integer.
Let $U$ be a subspace of $\mathbb{C}^n$ such that $A^k(U)=U$.
Does this mean that $A(U)=U$?
EDIT: A matrix $M$ is called unipotent if $(M-I)^r=0$ for some integer $r\geq 1$.
 A: EDIT:  I am doubting my answer now. I think I wrongly assumed that $((I+N)^k(I+N)^{j+1}x \in U$ when $x \in U$. This might be fixed by having some $k'$ such that $(1+N)^{k'}N^jx \in U$, which we can have by choosing $k'+j+1$ as a multiple of $k$. I'm not sure.
Since $A$ is unipotent, $A = I + N$, $N^m = 0$ for some $m$.
I claim that for $x \in U$ and $j \geq 0$,
$$(I+N)^k N^j x \in U.$$
The base case $j=0$ clearly holds. If it is true for $j$, then 
$$(I+N)^{k}(I+N)^{j+1}x = (I+N)^{k}(P(N)+N^{j+1})x \in U$$
where $P$ is a polynomial of degree $j$. But using the induction hypothesis, $(I+N)^k P(N)x \in U$, so we conclude $(I+N)^k N^{j+1}x \in U$.
Next I claim that for $x \in U$, then $N^j x \in U$ for $j \leq m$. By the nilpotency of $N$ we can write
$$(I+N)^k = I+P(N)N$$
where $P(N)$ is a polynomial of degree at most $m-2$. Now for $x \in U$, using our previous result,
$$(I+N)^k N^{m-1} x = N^{m-1}x + P(N)N^mx = N^{m-1} x \in U.$$
Now we have
$$(I+N)^kN^{m-2}x=N^{m-2}x + P(N)N^{m-1}x = N^{m-2}x + p_0 N^{m-1}x \in U.$$
Because $N^{m-1}x \in U$, we conclude $N^{m-2}x \in U$. We can continue this method to finally achieve $Nx \in U$. From this we conclude $A(U) \subseteq U$.
Any unipotent matrix is invertible because $0=(I-A)^m=I+P(A)A$ means the inverse is $-P(A)$. So we can conclude that $A(U)=U$ identically.
