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If I am correct, for a complex square matrix $A$, $AA^H$ and $A^HA$ are positive semidefinite, and therefore their eigenvalues are all nonnegative. They share the same set of nonzero eigenvalues, and may or may not share zero eigenvalue if any.

For any shared nonzero eigenvalue $\lambda$, there is a bijection between the eigenvectors of $AA^H$ and of $A^HA$ via $ A^H y = \sqrt{\lambda} x$ $ A x = \sqrt{\lambda} y. $ where $y$ is an eigenvector of $AA^H$, and $x$ is an eigenvector of $A^HA$, both for eigenvalue $\lambda$.

For a shared zero eigenvalue, I was wondering if the following similar statement is true: $ A^H y = 0$ $ A x = 0. $ for any eigenvector $y$ of $AA^H$, and any eigenvector $x$ of $A^HA$, both for eigenvalue $0$? If not, is there a truth close to this statement?

Thanks!

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$A$ and $A^H$ being square matrices, $A A^H$ and $A^H A$ have the same eigenvalues, both zero and nonzero. If they were rectangular matrices, one of $A A^H$ and $A^H A$ might not have a zero eigenvalue while the other would have.

If $y$ is an eigenvector of $A A^H$ for eigenvalue $0$, then $0 = y^H A A^H y = (A^H y)^H (A^H y)$ so $A^H y = 0$, i.e. $y$ is an eigenvector of $A^H$ for eigenvalue $0$. The converse is easy. Similarly with $A$ and $A^H$ interchanged.

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    Yes, the proof didn't require $A$ to be a square matrix.2012-11-27