How to find the SVD of a matrix with one zero eigenvalue? I am trying to find the SVD of the matrix \begin{pmatrix}1&2\\3&6\end{pmatrix}
$A^TA=\begin{pmatrix}1&3\\2&6\end{pmatrix}\begin{pmatrix}1&2\\3&6\end{pmatrix}=\begin{pmatrix}10&20\\20&40\end{pmatrix}$
$\lambda_1=50$
$\lambda_2=0$
$\lambda_1>\lambda_2>=0$
$\Sigma=\begin{pmatrix}5\sqrt{2}&0\\0&0\end{pmatrix}$
When $\lambda_1=50$:
$v_1=\begin{pmatrix}1\\2\end{pmatrix}$
Normalize:
$u_1=\begin{pmatrix}\frac{1}{\sqrt{5}}\\\frac{2}{\sqrt{5}}\end{pmatrix}$
When $\lambda_1=0$:
$v_2=\begin{pmatrix}-2\\1\end{pmatrix}$
Normalize:
$u_2=\begin{pmatrix}\frac{-2}{\sqrt{5}}\\\frac{1}{\sqrt{5}}\end{pmatrix}$
$V=\begin{pmatrix}\frac{1}{\sqrt{5}}&\frac{-2}{\sqrt{5}}\\\frac{2}{\sqrt{5}}&\frac{1}{\sqrt{5}}\end{pmatrix}$
$V^T=\begin{pmatrix}\frac{1}{\sqrt{5}}&\frac{2}{\sqrt{5}}\\\frac{-2}{\sqrt{5}}&\frac{1}{\sqrt{5}}\end{pmatrix}$
$A=U\Sigma V^T$
$\begin{pmatrix}1&2\\3&6\end{pmatrix}=\begin{pmatrix}u_1&u_2\\u_3&u_4\end{pmatrix}\begin{pmatrix}5\sqrt{2}&0\\0&0\end{pmatrix}\begin{pmatrix}\frac{1}{\sqrt{5}}&\frac{2}{\sqrt{5}}\\\frac{-2}{\sqrt{5}}&\frac{1}{\sqrt{5}}\end{pmatrix}$
$\begin{pmatrix}1&2\\3&6\end{pmatrix}=
\begin{pmatrix}5\sqrt{2}u_1&0\\5\sqrt{2}u_3&0\end{pmatrix}
\begin{pmatrix}\frac{1}{\sqrt{5}}&\frac{2}{\sqrt{5}}\\\frac{-2}{\sqrt{5}}&\frac{1}{\sqrt{5}}\end{pmatrix}$
$\begin{pmatrix}1&2\\3&6\end{pmatrix}=
\begin{pmatrix}\sqrt{10}u_1&2\sqrt{10}u_1\\\sqrt{10}u_3&2\sqrt{10}u_3\end{pmatrix}$
$u_1=\frac{1}{\sqrt{10}}$
$u_3=\frac{3}{\sqrt{10}}$
How can I find $u_2$ and $u_4$?
 A: Note that the given matrix is rank-$1$ and that in $\Bbb R^2$ it is easy to find orthogonal vectors. Hence,
$$\begin{aligned} {\rm A} := \begin{bmatrix} 1 & 2 \\ 3 & 6\end{bmatrix} = \begin{bmatrix} 1 \\ 3\end{bmatrix} \begin{bmatrix} 1 \\ 2\end{bmatrix}^\top &= \begin{bmatrix} 1 & -3 \\ 3 & 1\end{bmatrix} \begin{bmatrix} 1 & 0\\ 0 & 0\end{bmatrix} \begin{bmatrix} 1 & -2\\ 2 & 1\end{bmatrix}^\top \\ &= \left( \frac{1}{\sqrt{10}} \begin{bmatrix} 1 & -3 \\ 3 & 1\end{bmatrix} \right) \begin{bmatrix} \sqrt{50} & 0\\ 0 & 0\end{bmatrix} \left( \frac{1}{\sqrt{5}} \begin{bmatrix} 1 & -2\\ 2 & 1\end{bmatrix} \right)^\top = {\rm U} \Sigma {\rm V}^\top\end{aligned}$$
Lastly, do note that the nonzero singular value is $\sqrt{\mbox{tr} \left( {\rm A}^\top {\rm A} \right) } = \| {\rm A} \|_{\text{F}} = \sqrt{50} = \| {\rm A} \|_2$.

matrices svd singular-values
A: You have correctly noticed that the second column of $U$ doesn't really matter in terms of satisfying $A=U\Sigma V^\top$. In order to be a proper SVD, you just need to find a vector orthonormal to the first column of $U$ to make $U$ an orthogonal matrix (but again, this is not necessary for the equality $A = U \Sigma V^\top$ to hold).
