singular value decomposition of $\left[\begin{smallmatrix}0& 1\\0& 0\\0&1\end{smallmatrix}\right]$

Question

I'm currently looking for the singular value decomposition of $$\begin{bmatrix} 0 & 1 \\ 0 & 0 \\ 0 & 1 \end{bmatrix}$$ but I am having a struggle, because $A^TA$ has a 0 singular value, so I cant really use our formula. I have the $V$ and $\sum$ (that is I guess the incorrect way to write the sigma matrix in LaTeX) marices. I have for $U$ only one vector: $$u_1=\frac{1}{\sqrt{2}}(1,0,1)$$ How do I get one more?

Thanks in advance

UPDATE:

I thought that since $$(0,1,0)A=\sigma_2 A$$ where $\sigma_2=0$, $u_2=(0,1,0)$

and also

$$u_3=u_1\times u_2=(-\frac{1}{\sqrt{2}},0,\frac{1}{\sqrt{2}})$$

I cheated a little bit, I checked it on Wolfram Alpha, but I'm still unsure whether my conclusions are correct.

Answer 1

$\left( \begin{array}{ccc} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 0 \\ 0 & 0 & 1 \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\ \end{array} \right)$

$\left( \begin{array}{cc} \sqrt{2} & 0 \\ 0 & 0 \\ 0 & 0 \\ \end{array} \right)$

$\left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \\ \end{array} \right)$

Answer 2

A procedure of finding a SVD is

1. Find all eigenvalues of $A^T A$ in decreasing order and unit eigenvectors so that the eigenvectors form an orthonormal basis -> Let's call them $\lambda_1, \lambda_2, \ldots$ and $v_1, v_2, \ldots$ (right singular vectors $V$).
2. Normalize all nonzero vectors from $Av_1, Av_2, \ldots$ and put additional unit vectors if necessary so that the resulting vectors form an orthonormal basis (left singular vectors $U$).
3. Then $A V = U S$ where $S$ is a diagonal matrix with diagonal entries $\sqrt{\lambda_1}, \sqrt{\lambda_2}, \ldots$.
4. As a result, $A = U S V^T$.

In this question, $\lambda_1=2, \lambda_2=0$ and $v_1=(0, 1), v_2=(1, 0)$.

Now $Av_1 = (1, 0, 1)$ and $Av_2 = (0, 0, 0)$, so one of left singular vectors should be $(1/\sqrt{2}, 0, 1/\sqrt{2})$ and you can choose any additional vectors as long as they form an orthonormal basis.

Let $V = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} $ and $U = \begin{pmatrix} 1/\sqrt{2} & 0 \\ 0 & 1 \\ 1/\sqrt{2} & 0 \end{pmatrix}$.

Since $S = \begin{pmatrix} \sqrt{2} & 0 \\ 0 & 0 \end{pmatrix}$, $\quad A = U S V^T$