# Polar decomposition

In SVD, $A=U\Sigma V^{T}=(UV^{T})(V \Sigma V^{T})=QS$

Here is an example.
$A= \left(\begin{array}{cc} 1 & - 2\\ 3 & - 1 \end{array}\right) = \left(\begin{array}{cc} 0 & - 1\\ 1 & 0 \end{array}\right) \left(\begin{array}{cc} 3 & - 1\\ - 1 & 2 \end{array}\right) =QS$

Since $U$ contains columns which are eigenvectors of $AA^{T}$, and by calculation I get root value, so I think Q also has root value.
$\begin{pmatrix}0&-1\\ 1&10\end{pmatrix} \begin{pmatrix}3&-1\\ -1&2\end{pmatrix} \neq \begin{pmatrix}1&-2\\ 3&-1\end{pmatrix}$ in the first place, and $\begin{pmatrix}0&-1\\ 1&10\end{pmatrix}$ is not an orthogonal matrix. –  user1551 Dec 7 '12 at 14:55