(adsbygoogle = window.adsbygoogle || []).push({}); Linear algebra questions (rank, generalized eigenspaces)

Hi,

This seems to be an easy question on rank, but somehow I can't get it.

Let U be a linear operator on a finite-dimensional vector space V. Prove:

If rank(U^m)=rank(U^m+1) for some posiive integer m, then rank(U^m)=rank(U^k) for any positive integer k>=m.

It's in the section introducing Jordan canonical forms, so I assume the proof involves that. I tried induction and got to 'dim(U^m+p+1)<=rank(U^m) where p>1,' but I'm not sure how useful that is.

Also, I'm having trouble with this problem (not rank question, but i can't edit the title)

Let T be a linear operator on a finite-dimensional vector space V whose characteristic polynomical splits. Suppose B is Jordan basis for T, and let lambda be an eigenvalue of T. Let B'=B union K(lambda). Prove that B' is a basis for K(lambda). (K(lambda) is the generalized eigenspace corresponding to lambda)

I'd definitely appreciate some help with this!

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# Homework Help: Linear algebra rank question

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