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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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    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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There is no way to make the operator continuous. Consider the positive semidefinite matrices $ A(t) = \pmatrix{\cos^2(t) & \cos(t) \sin(t)\cr \cos(t) \sin(t) & \sin^2(t)\cr}$ which have eigenvalues (and singular values) $1$ and $0$. The normalized left singular vector for singular value $1$ is $\pm [\cos(t), \sin(t)]$. To make this vector a continuous function of $t$, you must take either $[\cos(t), \sin(t)]$ for all $t$ or $[-\cos(t), -\sin(t)]$ for all $t$, resulting in the vector at $t=\pi$ being the negative of the vector at $t=0$. But $A(\pi) = A(0)$, so this can't happen.

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    $A(t)$ is a continuous function of $t \in [0,1]$. But $S(A(t))$ can't be a continuous function of $t$. If $t_0$ is some point where $S(A(t))$ is discontinuous, there will be a sequence $t_n \to t_0$ such that $S(A(t_n))$ does not converge to $S(A(t_0))$ as $n \to \infty$.2011-11-30