Dimension of kernel for nilpotent transformation

Question

Let $T:R^n\rightarrow R^n$ be a nilpotent transformation with index n. (e.g. $T^n=0$). Is it true that for all $n\geq k \geq 0$, $\dim Ker T^k = 0$? How can that be shown?

Answer 1

For $k = 0$ this is true because $T^0 = \mathrm{id}_{\mathbb{R}^m}$. If we are dealing with $\mathbb{R}^m$ for $m = 0$ this is also true.

If $m \geq 1$ and $k \geq 1$ then this is false: Notice that if $\ker T = 0$ then $T$ is injective and thus $T^k$ is injective for all $k \geq 1$, contradicting $T^n = 0$. So $\ker T \neq 0$. For all $k \geq 1$ we have $\ker T^k \supseteq \ker T$, so $\ker T^k \neq 0$ for every $k \geq 1$.

In fact $$ 0 = \ker T^0 \subsetneq \ker T \subsetneq \ker T^2 \subsetneq \dotsb \subsetneq \ker T^{n-1} \subsetneq \ker T^n = \mathbb{R}^m, $$ is stricly increasing, assuming that $n = \min \{k \geq 0 \mid T^k = 0\}$.

Answer 2

We can generate counterexamples with square matrices. For example, take $$ A= \left(\begin{array}{rrrr} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{array}\right) $$ Then \begin{align*} A^2 &= \left(\begin{array}{rrrr} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right) & A^3 &= \left(\begin{array}{rrrr} 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right) & A^4 &= \left(\begin{array}{rrrr} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right) \end{align*} This shows that $A$ is nilpotent with index $4$. What can we say about the nullspaces of $A$, $A^2$, and $A^3$? Do you see how this example generalizes?

Many many other counterexamples can be constructed by choosing an invertible $P$ and taking $B=PAP^{-1}$.