# Can we choose smoothly the singular vectors of a matrix?

$$\newcommand{\GLm}{\text{GL}_n^-}$$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.

• 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. – Federico Poloni Oct 8 '19 at 16:38
• 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? – Anthony Quas Oct 9 '19 at 15:21
• @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). – Asaf Shachar Oct 10 '19 at 11:28
• 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. – Sebastian Goette Oct 10 '19 at 13:48
• @AnthonyQuas Singular values are, by definition, nonnegative, so the smallest singular value of diag(-2,1,1) is 1. – 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.