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If $A$ is $n\times n$, the singular values of $A$ are the squares of the eigenvalues of $A^{T}A$. But how does one obtain the eigenvalues of A given the singular values of A (A is unknown)?

EDIT: One way that just popped into my mind would be to use SVD: Multiply the identity matrix (which is an orthonormal basis U) by $\Sigma$ (diagonal matrix with singular values) and then multiply again by the identity(which is an orthonormal basis $V^T$) - basically just ending up with $\Sigma$. Afterwards, simply computing the characteristic polynomial of $\Sigma$, then solving for eigenvalues.

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    @RobertIsrael: Actually, I think what we both said is identical: I just found out that in my example the singular values equal the eigenvalues. Thanks for the help2012-10-16

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The eigenvalues of $A$ are not determined by the singular values of $A$. For example, all orthogonal matrices have $A^T A = I$, so the singular values are all $1$, but the eigenvalues can be anywhere on the unit circle.

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    There are some constraints: the determinant of $A^T A$ is the square of the determinant of $A$, so this tells you what the product of all the eigenvalues is.2012-10-16