TITLE: Sign of det(UV) in SVD
QUESTION [5 upvotes]: Let $A\in \mathbb{R}^{n\times n}$. Its Singular Value Decomposition (SVD) is
$$A=U\Sigma V^T$$
We know $U$ and $V$ are orthogonal matrices. Sometimes $\det UV=1$ and sometimes $\det UV=-1$. My question is: what kind of matrices give $\det UV=1$? Can we say something about the sign of $\det UV$ based on some properties of $A$ before we do SVD? Thanks.

Edit: A trivial example is: if $A$ is a rotation matrix, i.e., $AA^T=I$ and $\det A=1$, then of course any SVD of $A$ gives $\det UV=1$.
Now I am focusing on $A=R+ab^T$ with $R$ as a rotation matrix and $a,b\in \mathbb{R}^n$. Can we say $det UV=1$ for $A$ all the time?

REPLY [7 votes]: I figured it out. $\det A=\det U\Sigma V^T=\det UV \det \Sigma$. Since $\det \Sigma >0$

$\det UV=1$ when $\det A>0$
$\det UV=-1$ when $\det A<0$

But what happens when $\det A=0$?
Edit: Thanks to Darij Grinberg. When $\det A=0$, the SVD of $A$ is like
$$A=U\Sigma V^T=(U_1, U_2)
\left(\begin{array}{cc}
\Sigma_1 & 0 \\
0 & 0 \\
\end{array}\right)
\left(\begin{array}{cc}
V_1^T \\
V_2^T \\
\end{array}\right)
$$
We can arbitrarily flip the sign of each column of $U_2$ and $V_2$ without change the SVD. Therefore, for $\det A=0$, $U$ and $V$ can be rotation or reflection matrices.
The main reason I'm concern about this problem is: in SVD $U$ and $V$ are orthogonal. But in some applications, we require $U$ or $V$ to be rotation matrices.