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In the literature I have on disposal it is stated that singular values are non-negative values, and that, for a symmetric matrix $A$, the SVD and EVD coincide. This would mean that singular values of $A$ are the eigenvalues of $A$, but the eigenvalues of $A$ can be negative, regardless of $A$ being symmetric.

So, I wonder if the choice of singular values being exclusively positive is some kind of convention? If so, how degenerate that is given the above observation the equivalence of SVD and EVD for symmetric matrices?

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  • $\begingroup$ For symmetric matrix $A,$ SVD and EVD coincide if eigenvalues of $A$ are positive. $\endgroup$
    – user2468
    Jul 14, 2012 at 14:22
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    $\begingroup$ @J.D. not only positive, nonnegative! (i.e., $\mathbf A$ is symmetric positive semidefinite) $\endgroup$ Jul 14, 2012 at 14:49

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You can factor a (not necessarily square) matrix as orthogonal times diagonal times orthogonal, and the diagonal entries need not all be non-negative. But multiplying a row or a column of an orthogonal matrix by $-1$ still gives an orthogonal matrix, and you can do that and change a minus to a plus in the diagonal matrix. In that way, the two orthogonal matrices can be chosen so that the diagonal entries in that matrix are all non-negative. Those are what are taken to be the singular values.

It's a convention to define it that way. But I suspect there are theorems that say that's the only way to define it that makes it have specified nice properties, and those theorems would not be mere conventions.

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    $\begingroup$ OK. A practical way (I know of) to find singular values of some square matrix $A$ is to find the eigenvalues of $A^TA$ or $AA^T$, which must be non-negative. However, by plainly making EVD on $A$, one might get negative eigenvalues. What would that imply about some other SVD method that does not take a detour via EVD on $A^TA, AA^T$? One has to make sure that the left singular vectors and right singular vectors are of sings that imply positive singular values? $\endgroup$
    – user506901
    Jul 14, 2012 at 19:40
  • $\begingroup$ Sounds right. ${}$ $\endgroup$ Jul 14, 2012 at 22:16
  • $\begingroup$ I just fixed what I realized must have been a confusing typo in the first sentence. I meant "need not all be nonnegative". $\endgroup$ Jul 17, 2012 at 21:46
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    $\begingroup$ @user506901, forming the cross-product matrix is in fact a bad idea (in inexact arithmetic), since the singular values are not as accurately determined in this case. $\endgroup$ Jul 31, 2012 at 13:14

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