## Return to Answer

2 added 18 characters in body

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.

1

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 infinity while the eigenvectors of $N$ can be anything; as $y$ goes to 0, the eigenvectors of $N$ go off to infinity while the eigenvectors of $M$ can be anything.
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