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is there any relationship between the eigenvector of sum(AA'+BB') and sum(A'A+B'B) ?

thanks a lot!

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what does the 'sum' notation mean? – Suvrit Dec 20 '12 at 11:59
and what does $A'$ mean? The transpose? The Hermitian transpose? – Dima Pasechnik Dec 20 '12 at 12:15
I guess $A′$ means the transpose. Steven shows the eigenvectors are generally different. However, there is an interesting relation between the eigenvalues, under a mild assumption. See Corollary 2.2. of Lin & Wolkowicz, An eigenvalue majorization inequality for positive semidefinite block matrices, Linear Multilinear Algebra, 60 (2012), 1365-1368. – Betrand Dec 20 '12 at 20:27
SORRY FOR come back later.... A is real matrix. sum(AA′+BB′) means AA′+BB′ and that A′ means the transpose of A. thanks – David liu Dec 21 '12 at 4:44

By way of penance for my earlier "answer":

Take $A=\pmatrix{1&0\cr x&0\cr}$ and $B=\pmatrix{1&y\cr 0&0\cr}$.

Then the eigenvectors of $M=AA'+BB'$ and $N=A'A+B'B$ are in general different. As $x$ goes to 0, the eigenvectors of $M$ go off to zero and infinity while the eigenvectors of $N$ can be anything; as $y$ goes to 0, the eigenvectors of $N$ go off to zero and infinity while the eigenvectors of $M$ can be anything.

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Assuming that $sum(AA'+BB')$ means $AA'+BB'$ and that $A'$ means the transpose of $A$:

Let $x$ and $y$ be arbitrary complex numbers. Then the matrices $X=\pmatrix{0&1\cr 1&x\cr}$ and $Y=\pmatrix{0&1\cr1&y\cr}$ have arbitrary eigenvectors. But in general, the equaions $$\matrix{AA'+BB'=X&A'A+B'B==Y}$$ are solvable for $A,B$. So the answer to your question is no.

Edit: As Terry Tao points out in comments, this system of equations is clearly not solvable (just take traces). So this is not an answer to your question.

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Are you sure about the solvability of this system? It seems to me that $AA'+BB'$ and $A'A+B'B$ necessarily have the same trace, for instance (and are necessarily positive semidefinite, too). – Terry Tao Dec 20 '12 at 16:57
Terry Tao: I did check this before posting, but obviously I got it wrong (as your remark about the trace makes clear). Thanks for the correction. – Steven Landsburg Dec 20 '12 at 17:51

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