How do I show that all of the chains in T's chain basis have the same length?

Question

Let $T: \mathbb R^n \rightarrow \mathbb R^n$ be a nilpotent operator, $h \in \mathbb N$ s.t. $KerT^h=ImT^h$.

How do I show that all of the chains in T's chain basis have the same length?

Thanks in advance!

Answer 1

We can prove the claim as follows. Note that the span of any chain is an invariant subspace (in particular, a generalized eigenspace) of $T$. Suppose that $U_1, \dots, U_k$ are the subspaces spanned by each of the different chains.

Let $T|_U$ denote the restriction of $T$ to $U$. Note that if $\operatorname{Im}(T^h) = \ker(T^h)$, then we must have $\operatorname{Im}(T^h) \cap U_j = \ker(T^h) \cap U_j$ for each $j$, which is to say that $$ \operatorname{Im}(T|_{U_j}^h) = \ker(T|_{U_j}^h) \quad j = 1,\dots,k $$ However, because each $U_j$ is the span of a chain, we have $\dim \ker T|_{U_j}^h = h$. By the rank-nullity theorem, we have $$ \dim(U_j) - h = h \implies \dim(U_j) = 2h $$ Thus, each $U_j$ has dimension $2h$. That is, the length of each chain must be $2h$. So, each chain has the same length, as desired.