Finding dim null T for nilpotent operators I was working through some problems when I came across this one that I was having some trouble with.
Assume that $\dim V = n$. Assume that $n > 20130520$. Let $T$ be an operator on $V$.
(1) If $T^{n-1} \neq  0$, $T^n = 0$. Find $\dim \operatorname{null} T$.
(2) If $T^{n-2} \neq  0$, $T^{n-1} = 0$. Find $\dim \operatorname{null} T$.
I wasn't too sure how I should go about this. One thing I was thinking of was that for (1), the assumption implies that $\operatorname{null} T^{n-1} \neq \operatorname{null} T^n$ which would imply that the left side would be a subset of the right side and you could continue until you get to $\operatorname{null} T$.
Any tips or advice would be appreciated. Thanks in advance.
 A: I am not sure why $n$ is given such a large lower bound. Anyway, consider the chain of subspaces:
$$
T^n(V) \subseteq T^{n-1}(V) \subseteq \ldots \subseteq T(V) \subseteq V.
$$
If we have $T^{N+1}(V) = T^N(V)$ for some $i$, then $T^n(V) = T^N(V)$ for all $n \ge N$. That means if $T^n(V) = 0$ and $T^{n-1}(V) \ne 0$, we must have
$$
0 = \dim T^n(V) < \dim T^{n-1}(V) < \ldots < \dim T(V) < \dim(V) = n.
$$
By simple counting, we must have $\dim T^i(V) = n - i$, so $\dim \ker T = 1$.
For the second case, this argument is not good enough, as it only tells us that $\dim \ker T \le 2$.
By the counting argument, we know that somewhere in the chain, we must have
$$\dim\ker T|_{T^I(V)} = \dim T^I(V) - \dim T^{I+1}(V) = 2$$
with $I \ge 1$. But since $T^I(V) \subseteq V$, we know
$$
\dim\ker T|_{T^I(V)} \le \dim\ker T
$$
and so
$$
2 \le \dim \ker T.
$$
(The general statement is that $\dim \ker T^{i+1}|_{T^i(V)}$ is a non-increasing sequence, and $\sum_{i=0}^{n-1} \dim\ker T^{i+1}|_{T^i(V)} = \dim V = n$ if $T$ is nilpotent.)
