Find the SVD of a matrix with one zero eigenvalue

linear algebramatricessvd

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 $u_4$?

Best Answer

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 $\sqrt{\mbox{tr} \left( {\rm A}^\top {\rm A} \right) } = \| {\rm A} \|_{\text{F}} = \sqrt{50} = \| {\rm A} \|_2$.

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