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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?

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  • $\begingroup$ If $A$ is orthogonally diagonalizable, i.e., $A= U\Lambda U^T$, then we can get $A= U\Sigma V^T$ by changing the sign of the eigenvectors corresponding to negative eigenvalues (but only on the right matrix, for example). $\endgroup$ – yasmar May 3 '11 at 5:55
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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.

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    $\begingroup$ Then you can find a pair $\left(U,V\right)$ with $\det \left(UV\right)=1$ and a pair $\left(U,V\right)$ with $\det \left(UV\right)=-1$. In fact, at least one diagonal entry of $\Sigma$ is zero, so you can flip the sign of the corresponding row (or column?) of $V$ without changing anything. $\endgroup$ – darij grinberg May 3 '11 at 9:22

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