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Some reference say that if rank($A$)=rank($A^2$),then the geometric and algebraic multiplicities of the eigenvalues $\lambda=0$ are equal;that is,all the Jordan blocks correspondint to $\lambda=0$ (if any) in the Jordan form of A are 1*1.

I don't know how to check out this conclusion.Could you help me ? Thank you!

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closed as off topic by Andrés E. Caicedo, Igor Rivin, quid, Federico Poloni, Qiaochu Yuan Oct 21 '12 at 17:29

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read the faq for which questions are appropriate for this site. Ask such questions e.g. in – Julian Kuelshammer Oct 21 '12 at 14:41
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Take one Jordan block with zero eigenvalue. Suppose it is of size mxm. It has zeros everywhere except on the diagonal next to the main, total m-1 ones, so the rank is m-1. Now square this matrix (squaring the whole matrix is equivalent to squaring all its Jordan blocks. The result is the matrix which has zero everywhere, except the ones on the diagonal, SECOND from the main. (That is $a_{i,j}=1$ if and only if $j=i+2$. So there are $m-2$ non-zero rows. Thus the rank of $A^2$ is less than that of $A$.

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Thanks,but you say"squaring the whole matrix is equivalent to squaring all its Jordan blocks. " the words make sense only if A is inverse.But in this case,A is not inverse. So you can not use the conclusion? – Hinn Oct 21 '12 at 15:32
I am sorry, but I do not understand your expression "A is inverse". What does this mean? If A consists of Jordan blocks J_1, J_2,... then $A^2$ consists of the Jordan blocks $J_1^2,J_2^2,...$. Is not this evident? – Alexandre Eremenko Oct 21 '12 at 18:47

The assumption implies $\mathrm{ran}(A)=\mathrm{ran}(A^k)$, hence also $\mathrm{ker}(A)=\mathrm{ker}(A^k)$, for all $k\in\mathbb{N}$. The algebraic multiplicity of an eigenvalue $\lambda$ of the matrix $A$ is $\max _ {k\in\mathbb{N}} \mathrm{dim} \ker (A-\lambda)^k $ (think e.g. to the Jordan form of $A$). So in your assumption the algebraic and geometric multiplicity of the eigenvalue $0$ coincide (and conversely).

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