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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?2012-10-16
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    There will be more than one option for A, but any one that fits the crieria will be fine.2012-10-16
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    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 $A$?2012-10-16
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    Are you asking for inverse eigenvalue problem?2012-10-16
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    @Mhenni Benghorbal: Never heard that term before but it sounds about right.2012-10-16
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    You can't. This is similar to asking what the value of $z$ is given only the value of $|z|$.2012-10-16
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    @EuYu: Not sure if I can conclude the matrix is pos. semidefinite but my gut feeling says yes. Not familiar with Gram factorization. Will research it a bit. Thanks for all feedback so far.2012-10-16
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    @copper.hat: Yes but I can make an educated guess: If I get abs value equal to 7. -7 would be one of the answers thats sufficient for me, i.e. a possible valid/feasible solution.2012-10-16
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    @WuschelbeutelKartoffelhuhn: It is a well known theory.2012-10-16
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    Well, if you're just looking for one possible solution, you could take $A$ to be the positive semidefinite square root of $A^T A$: its nonzero eigenvalues are then the same as the singular values.2012-10-16
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    @RobertIsrael: Would the eigenvalues of a diagonal matrix D (with diagonal entries corresponding to the **given** singular values) also be correct? (my sample calculations show that this is indeed the case)2012-10-16
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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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    Arent the eigenvalues at least constrained by the singular values of A? Any arbitrary spectrum is acceptable.2012-10-16
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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