Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

1 Answer 1

up vote 1 down vote accepted

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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.