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?
Best Answer
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.