In SVD, $A=U\Sigma V^{T}=(UV^{T})(V \Sigma V^{T})=QS$
Here is an example.
$A= \left(\begin{array}{cc}
1 & - 2\\
3 & - 1
\end{array}\right) = \left(\begin{array}{cc}
0 & - 1\\
1 & 0
\end{array}\right) \left(\begin{array}{cc}
3 & - 1\\
- 1 & 2
\end{array}\right) =QS$
I can't draw this answer.
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.
Can you help me to get that answer?