Dimension of kernel for nilpotent transformation powers

Question

Let $T:\Bbb R^n \to \Bbb R^n$ be a nilpotent transformation with index $n$. (i.e. $T^n=0$). Is it true that for all $n≥k≥0$, $\dim \ker T^k=k$? How can that be shown?

The context is a linear algebra text book and it is the chapter about Jordan canonical forms so I guess it is related. I tried playing around with that - but no luck.

Answer 1

The answer to your question is yes. To see that this is the case, make the following observations:

• $\dim \ker T = 1$
• $\ker T^k = (\ker T \cap \operatorname{image}(T^{k-1}))+ \ker T^{k-1}$
• $\operatorname{image}(T^{k}) \subset \operatorname{image}(T^{k-1})$

See if you can put these facts together