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Steven Landsburg
Steven Landsburg
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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.

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.

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.

Source Link
Steven Landsburg
Steven Landsburg
  • 23k
  • 5
  • 95
  • 153

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.