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Let $A_n$ be a series of matrices, and let $A$ be another matrix. Let $S(B)$ be an SVD operator that takes a matrix and returns the left singular vectors matrix ordered by largest singular value to smallest singular value. Also, assume all singular values for $A$ are unique.

Is there some matrix norm $\| \cdot \|$ under which if $\|A_n - A\| \to 0$ as $n \to \infty$ then $\|S(A_n) - S(A)\| \to 0$?

Does it happen for the Frobenius norm?

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    Even if all the singular values are unique, the singular vectors are only unique up to factors of $\pm 1$. So if $S(\cdot)$ is some numerical routine that returns a matrix it could be discontinuous. Other than that, it's continuous. And it doesn't matter what norm you use; since matrices are finite dimensional, all norms are equivalent for purposes of continuity.2011-11-29
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    in what sense "other than that" it is continuous? What is the right definition of the SVD operator to make it continuous (even when the singular values are not unique)?2011-11-29
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    I only commented because I wasn't sure. =) As the Robert Israel's answer shows, any matrix-valued function S(A) can't be continuous. However, I believe that the singular spaces are continuous on the grassmanian manifold.2011-11-30

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