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




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**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](https://math.stackexchange.com/a/3163368/104576): 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.



Can we choose smoothly the singular vectors of a matrix?
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).


