Singular Value Decomposition: Nonsquare matrix, third column of U My textbook says the singular value decomposition of the below matrix:
$A = \left( \begin{matrix} -2&1\\ 1&2 \\ 1 & 2 \end{matrix} \right)$
is given by:
$A = \left( \begin{matrix} 0&1&0 \\ \frac{1}{\sqrt{2}}&0&\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}}&0&-\frac{1}{\sqrt{2}} \end{matrix} \right) \left( \begin{matrix} \sqrt{10}&0\\0&\sqrt{5}\\0&0\end{matrix}\right)\left( \begin{matrix} \frac{1}{5}&\frac{2}{5}\\-\frac{2}{5}&1\end{matrix}\right)=U\Sigma V^T$
However, I don't see how to get the third column of U, not only because A only yields 2 singular values, but also because the third column is multiplied out anyway. How did my textbook come up with this result?

It seems like there are some typos in $V^T$, the result should be:
$$A = \left( \begin{matrix} 0&1&0 \\ \frac{1}{\sqrt{2}}&0&\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}}&0&-\frac{1}{\sqrt{2}} \end{matrix} \right) \left( \begin{matrix} \sqrt{10}&0\\0&\sqrt{5}\\0&0\end{matrix}\right)\left( \begin{matrix} \frac{1}{\sqrt{5}}&\frac{2}{\sqrt{5}}\\-\frac{2}{\sqrt{5}}&\frac{1}{\sqrt{5}}\end{matrix}\right)=U\Sigma V^T$$
If we need to step through the process, please give a yell.
