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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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And how exactly do you compute the characteristic polynomial of $A$ if $A$ is unknown? – EuYu Oct 16 '12 at 22:22
There will be more than one option for A, but any one that fits the crieria will be fine. – Wuschelbeutel Kartoffelhuhn Oct 16 '12 at 22:23
So let me get this straight. You have a matrix $S$ which is positive (semi)definite. You want to take some random Gram factorization $A^\mathrm{T}A$ of $S$ and find the eigenvalues of $A$? – EuYu Oct 16 '12 at 22:26
Are you asking for inverse eigenvalue problem? – Mhenni Benghorbal Oct 16 '12 at 22:28
@Mhenni Benghorbal: Never heard that term before but it sounds about right. – Wuschelbeutel Kartoffelhuhn Oct 16 '12 at 22:36
up vote 0 down vote accepted

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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Arent the eigenvalues at least constrained by the singular values of A? Any arbitrary spectrum is acceptable. – Wuschelbeutel Kartoffelhuhn Oct 16 '12 at 22:39
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. – Robert Israel Oct 16 '12 at 22:42

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