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

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

2012-12-20T11:57:50Z

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

thanks a lot!

Answer by Steven Landsburg for is there any relationship between the eigenvector of sum(AA'+BB') and sum(A'A+B'B) ?

2012-12-20T15:37:50Z

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.

Answer by Steven Landsburg for is there any relationship between the eigenvector of sum(AA'+BB') and sum(A'A+B'B) ?

2012-12-20T18:59:11Z

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.

Answer by David liu for is there any relationship between the eigenvector of sum(AA'+BB') and sum(A'A+B'B) ?

2012-12-21T04:44:49Z

SORRY FOR come back later....

A is real matrix. sum(AA′+BB′) means AA′+BB′ and that A′ means the transpose of A.

thanks