$\newcommand{\GLm}{\text{GL}_n^-}$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. $\endgroup$ Oct 8, 2019 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? $\endgroup$ Oct 9, 2019 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). $\endgroup$ Oct 10, 2019 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. $\endgroup$ Oct 10, 2019 at 13:48
  • $\begingroup$ @AnthonyQuas Singular values are, by definition, nonnegative, so the smallest singular value of diag(-2,1,1) is 1. $\endgroup$ Oct 10, 2019 at 17:13

1 Answer 1


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. $\endgroup$ Oct 11, 2019 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). $\endgroup$ Oct 11, 2019 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. $\endgroup$ Oct 11, 2019 at 4:24

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