$\newcommand{\GLm}{\text{GL}_n^-}$Let $A$ be a real $n \times n$ matrix with non-positive determinant. Suppose that the smallest singular value of $A$ is strictly smaller than all the others (it has multiplicity $1$).

Question: Do there exist an open neighbourhood $O$ of $A$, and smooth maps $U:O \to \operatorname{SO}_n$, $V:O \to \operatorname{O}_n^-$ such that $$ X=U(X)\Sigma(X)V(X)^T$$ holds for every $X \in O$, where $\Sigma(X) = \operatorname{diag}\left( \sigma_1(X),\dots\sigma_n(X) \right)$, and $\sigma_1(X)$ is the smallest singular value of $X$?

Note that I don't care about the ordering of $\sigma_2,\dotsc,\sigma_n$, but I specifically want the minimal singular value to be in a fixed position.

Comment 1: If we replace the requirement that $\sigma_1 $ has multiplicity $1$ by the requirement that all the singular values are distinct, then the answer is positive: In that case the map \begin{align*} \mu: \operatorname{SO}_n\times \mathcal D\times \operatorname{O}_n^-\to Y \\ (U,\Sigma,V)\mapsto U\Sigma V^T \end{align*} is a local diffeomorphism, so is locally invertible (here $\mathcal D$ is the space of $n\times n$ diagonal matrices with strictly increasing positive entries, and $Y=\{A \,|\, \text{$\det A<0$ and all the singular values of $A$ are distinct}\}$). This works when $\det A <0$. When $\det A=0$ (and all its singular values are distinct) a slight adaptation of this argument works.

Comment 2: If such maps $U$, $V$ exist, then the map $A \to \Sigma(A)$ is also smooth. In general the ordered singular values cannot be chosen smoothly when they "cross", but here I don't require to keep the ordering of $\sigma_2,\dotsc,\sigma_n$ fixed, so I think there is no obstruction, but I may be wrong.

  • $\begingroup$ There is some literature on the "analytic SVD", which seems like it would suit you; I suggest you to start a search with this term. $\endgroup$ – Federico Poloni Oct 8 '19 at 16:38
  • $\begingroup$ Doesn’t even continuity fail at diag(-2,1,1), even if you restrict the maps to diagonal matrices with $-2$ in the top left corner? $\endgroup$ – Anthony Quas Oct 9 '19 at 15:21
  • $\begingroup$ @AnthonyQuas Note that I assumed that the smallest singular value is strictly smaller than all the others. (and I don't require a fixed ordering on the other singular values). $\endgroup$ – Asaf Shachar Oct 10 '19 at 11:28
  • $\begingroup$ I think continuity already fails for the circle $O(2)\setminus SO(2)\subset GL^-(2)$. If you follow an eigenvector $v$ once around the circle, you end up at $-v$, so you cannot possibly get continuity. Or am I missing something? --- EDIT: it seems your edit has fixed this problem - sorry. $\endgroup$ – Sebastian Goette Oct 10 '19 at 13:48
  • $\begingroup$ @AnthonyQuas Singular values are, by definition, nonnegative, so the smallest singular value of diag(-2,1,1) is 1. $\endgroup$ – Federico Poloni Oct 10 '19 at 17:13

Consider the set of matrices $$ \begin{pmatrix} -1&0&0\\ 0&2-a&b\\ 0&b&2+a \end{pmatrix} $$

For $a=0$ and $b$ small and positive, the singular vectors are $(1,0,0)$, $(0,1,1)/\sqrt 2$, $(0,1,-1)/\sqrt 2$.

For $a>0$ small and $b=0$, the singular vectors are the coordinate directions.

Hence the singular vectors fail to even be continuous in a neighbourhood of diag(-1,2,2).

  • $\begingroup$ But in that example the singular vector corresponding to the smallest singular value 1 is constant. Of course, if two singular values are equal, then it's impossible in general to choose continuously the corresponding singular vectors for a small perturbation of the matrix. But the question is about separated singular values. $\endgroup$ – Oleg Eroshkin Oct 11 '19 at 3:53
  • $\begingroup$ @OlegEroshkin: I think you missed comment 1 in the OP. In the case where the singular values are distinct, this is true (and known to be so). $\endgroup$ – Anthony Quas Oct 11 '19 at 4:14
  • $\begingroup$ I thought that OP want to choose continuously only the singular vector corresponding to the smallest singular value. Otherwise the condinion that the smallest value is distinct is totally useless - just take the direct sum of a family of matrix without continous singular vectors and a matrix with a very small singular value. But that is exactly your example. So I just misread the question. $\endgroup$ – Oleg Eroshkin Oct 11 '19 at 4:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.