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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 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?

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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?2011-11-13
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    @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.2011-11-13
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    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.2011-11-13
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    @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.2011-11-13
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    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."2011-11-13
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    Actually I just did an example where I lost two 1's in the J matrix after one exponentiation, so I am trying to see the pattern you described now. But I still struggle to understand why not every single Jordan block would lose a 1 with exponentiation, since they all have 0 a diagonal.2011-11-13
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    @spzdot: I must be misunderstanding you somewhere. Further up, you wrote "But I can see that the J matrix loses a 1 with every power, not every single Jordan block", and now you write "But I still struggle to understand why not every single Jordan block would lose a 1", which sounds like the opposite to me. Indeed every single Jordan block loses a 1 with each power. The block size enters into it because it only has so many 1s to lose before it's zero.2011-11-13
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    Sorry, I wasn't clear. At first I thought that $J$ matrix would always lose only one 1 with an exponentiation, but then I did an example where I lost two 1's at once. I am currently doing some examples to confirm that for myself, because I did some before (most likely incorrectly) where not every Jordan block would lose a 1.2011-11-13
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    @spzdot: I see. From the way you describe what you're doing, I get the impression that perhaps you're not aware that the blocks are all independent? That is, the powers of the Jordan normal form have block structure, and the blocks are the powers of the Jordan blocks. So you don't really need to calculate powers of the whole matrix; you can just focus on one block and check that it acquires a zero row and column with each power.2011-11-13
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    Yes, block multiplication makes sense now too... Thanks a lot for your time, I got it, I think.2011-11-13