Alternative proof for Jordan form statement

Question

I am trying to understand why given a nilpotent matrix $L,$ rank ($L^{k-1}$) - rank ($L^{k}$) is the number of Jordan blocks sized $\geq k \times k$ in the Jordan representation of $L.$ There is a proof in http://www.matrixanalysis.com/SolutionsManual.pdf on page 149, problem 7.7.3, but it seems to use some techniques and refer to results that I haven't learned about. Is there another way to prove this? Or could someone maybe explain the proof in simpler terms by breaking it down?

Answer 1

The key facts that you need here are that a) rank is invariant under similarity transforms, b) the transform of a power of $L$ is the corresponding power of the transform of $L$ and c) the Jordan form of a nilpotent matrix has zeros on the diagonal. Together a) and b) imply that you can look at the Jordan form of $L$ instead of $L$. It's immediate from the form of the Jordan blocks that a block with zero diagonal loses one rank with each power until it becomes zero and has zero rank, and that happens when the power equals its block size.