singular value decomposition of $\left[\begin{smallmatrix}0& 1\\0& 0\\0&1\end{smallmatrix}\right]$ 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. 
 A: $\left(
\begin{array}{ccc}
 \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 0 \\
 0 & 0 & 1 \\
 \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\
\end{array}
\right)$
$\left(
\begin{array}{cc}
 \sqrt{2} & 0 \\
 0 & 0 \\
 0 & 0 \\
\end{array}
\right)$
$\left(
\begin{array}{cc}
 0 & 1 \\
 1 & 0 \\
\end{array}
\right)$
A: A procedure of finding a SVD is 


*

*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 $V$). 

*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 $U$).

*Then $A V = U S$ where $S$ is a diagonal matrix with diagonal entries $\sqrt{\lambda_1}, \sqrt{\lambda_2}, \ldots$.

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