# Convention on non-negative singular values?

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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For symmetric matrix $A,$ SVD and EVD coincide if eigenvalues of $A$ are positive. – user2468 Jul 14 '12 at 14:22
@J.D. not only positive, nonnegative! (i.e., $\mathbf A$ is symmetric positive semidefinite) – J. M. Jul 14 '12 at 14:49

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.
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? – user506901 Jul 14 '12 at 19:40
Sounds right. ${}$ – Michael Hardy Jul 14 '12 at 22:16