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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.

Rob Jordan
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    Considering the Jordan form would probably be the easiest. – EuYu May 27 '13 at 01:48
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    @EuYu And look at the last name of the OP! – Tunococ May 27 '13 at 02:27
  • Looking at the nonin/de/creasing chains of nullspaces and ranges of the iterates $T^j$ is the way, as shown by Tunococ's answer. An interesting alternative for the easy case (1) is to note that any $x$ such that $T^{n-1}x\neq 0$ is cyclic, i.e. ${x,Tx,T^2x,\ldots,T^{n-1}x}$ is a basis of $V$. What is the matrix of $T$ in this basis? These techniques are the essence of the Jordan normal form mentioned by EuYu. Of course, the lower bound on $n$ is a joke. – Julien May 27 '13 at 02:31

1 Answers1

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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.)

Tunococ
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    Yes, +1. It might be worhtwhile to note that simultaneously, we have the non-decreasing chain of nullspaces. And that both chains become stationary for the same index. – Julien May 27 '13 at 02:27
  • @julien Thanks for the addition. It could be simpler to write it that way too. – Tunococ May 27 '13 at 02:34
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    For (1) yes, we don't need the ranges. But for (2), we do. – Julien May 27 '13 at 02:35