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


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. 


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. 

