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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 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?

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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.

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thanks but what do you mean by L instead of L? And when you talk of transforms, what exactly does transform mean? – spzdot Nov 13 '11 at 18:01
@spzdot: It parses as "(the Jordan form of $L$) instead of $L$" :-) About the transforms: Sorry, I wrote out "similarity transform" once and then wrote "transform" for short -- they're all similarity transforms. Google and Wikipedia tell me that "similarity transformation" is the more common term. – joriki Nov 13 '11 at 18:01
Thanks that makes sense. Yes this is the way I attempted my proof, I realized that the rank will decrease by one with every power, and will hit 0 at some point, and that can be seen from the way the Jordan form is structured. But I can see that the J matrix loses a 1 with every power, not every single Jordan block (is that what you meant? Because they all have 0 diagonal), which is why I struggle to tie this with the block sizes. – spzdot Nov 13 '11 at 18:09
@spzdot: What do you mean by "J matrix"? I did mean the individual blocks. Why is it an argument against that that they all have zeros on the diagonal? If you form the first few powers of a block with zero diagonal, you should see that it acquires one more zero row and column with every power. – joriki Nov 13 '11 at 18:19
By $J$ matrix I mean the Jordan matrix in the decomposition of $L = PJP^{-1}.$ I have done some examples, and whenever I exponentiate this $L$ I lose a 1 in the $J$ matrix. But there are usually several Jordan blocks in that matrix, all with 0's on the diagonal. Maybe I am misunderstanding what you mean by "a block with zero diagonal loses one rank with each power until it becomes zero and has zero rank." – spzdot Nov 13 '11 at 18:22

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