# How to find the SVD (Singular Value Decomposition )

How to find the SVD (Singular Value Decomposition ) of $\ A=\begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} \$

$A^T A=\begin{pmatrix} 1 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} \ =1+1=2$

Then how to find the eign value , orthogonal basis in order to find the SVD.

Help me out

Note that $$A= \begin{bmatrix} 1 \\ 0 \\ 1\end{bmatrix}= \left( \frac{1}{\sqrt2}\begin{bmatrix} 1 \\ 0 \\ 1\end{bmatrix}\right)\left( \sqrt{2}\right)(1)$$
If you want to express $A$ in terms of $UDV^T$ where $U$ and $V$ are orthogonal matrix and $D$ is a $3$ by $1$ vector. Try to find two additional vectors that form an orthogonal basis with $(1,0,1)$ by inspection.
Remark: From your approach $$2=(1)(\sqrt{2})^2(1)$$
• the two other orthogonal vectors with $(1,0, ) \$ will be $(0,0,1) \ \ and \ \ (0,1,0) \$ . Then what would be $UDV^T$ ? – M. A. SARKAR Nov 21 '17 at 7:38
• $(1,0,1)$ and $(0,0,1)$ are not orthogonal. Try again? – Siong Thye Goh Nov 21 '17 at 7:41
• Great! Can you make them an orthonomal basis of $\mathbb{R}^3$, once you can do that, you would have gotten your $U$. – Siong Thye Goh Nov 21 '17 at 7:47
• Guide: $U$ is a $3$ by $3$ matrix. $D$ is a $3$ by $1$ matrix. $V$ is a $1$ by $1$ matrix. $V=[1]$. and I believe your earlier work would enable you to construct $U$. $D$ is a diagonal rectangular matrix. Only the first entry is non-zero. – Siong Thye Goh Nov 21 '17 at 8:03
• There was a typesetting error in your $U$, hence I can't tell for sure. Anyway, after you write it out, we should be able to verify it right? check that $U^TU=I$, $V^TV=1$ , and $UDV^T=A$. – Siong Thye Goh Nov 21 '17 at 8:09