Find the SVD of $$A= \begin{bmatrix} 1 & 2\\ 2 & 2 \\ 2 & 1 \end{bmatrix}$$ which has the form A= $U \Sigma V$
1.
$A^TA=\begin{bmatrix} 9 & 8\\ 8 & 9 \\ \end{bmatrix}$
- The eigenvalues are $\lambda_1=17, \lambda_2=1$ The corresponding eigenvectors are
$ \vec{v_1}=\begin{bmatrix} 1 \\ 1 \end{bmatrix}$
and $\vec{v_2}=\begin{bmatrix}1 \\ -1 \end{bmatrix}$
After normalization, $V= \begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{bmatrix}$
3.$\sigma_1=\sqrt{17}$, $\sigma_2=1$ so $\Sigma= \begin{bmatrix} \sqrt{17} & 0 \\ 0 & 1 \\ 0 & 0 \end{bmatrix}$
- The first two columns of U can be represented as
$u_1=\frac{1}{\sqrt{17}}Av_1=\frac{1}{\sqrt{17}} \frac{1}{\sqrt{2}} \begin{bmatrix} 1 & 2\\ 2 & 2 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ \end{bmatrix}=\frac{1}{34} \begin{bmatrix} 3 \\ 4 \\ 3 \end{bmatrix} .$
$u_2 = \frac{1}{1} Av_2= \frac{1}{\sqrt{2}} \begin{bmatrix} 1 & 2 \\ 2 & 2 \\ 2& 1 \end{bmatrix} \begin{bmatrix} 1 \\ -1 \\ \end{bmatrix} = \frac{1}{\sqrt{2}} \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix} $
$U= \begin{bmatrix} \frac{3}{\sqrt{34}} & \frac{-1}{\sqrt{2}} & u_3(1) \\ \frac{4}{\sqrt{34}} & 0 & u_3(2) \\ \frac{3}{\sqrt{34}} & \frac{1}{\sqrt{2}} & u_3(3) \end{bmatrix}$
In order to determine $u_3(i)$, $i \in \{1,2,3\}$ need to satisfy, $u_j^*u_3 = \delta_{j3}, j=1,2,3$
Such a column vector $u_3$ is $\frac{1}{\sqrt{17}} \begin{bmatrix} 2\\ -3\\ 2 \end{bmatrix}$
How was $u_3$ calculated? What exactly is $\delta_{j3}$?
References
Fass, 2006 p. 30 http://www.math.iit.edu/~fass/477577_Chapter_2.pdf
The thing is that the U matrix is not always unqiue. In your example there are two possible $U$ matrices, because the condition $u^{*}_j u_3 = \delta_{j3}$ can be satisfied with 2 vectors. You can find a vector which is orthogonal to the plane formed by $u_1$ and $u_2$, after that you can normalize that vector in this is the first solution $x$. The second solution would be $-x$. In fact if you tried to decompose a matrix $B \in \mathbb{R}^{3 \times 1}$, you'd have an infinite amount of $U$ matrices, because you'd need to find two vectors $u_2$, $u_3$ orthogonal to $u_1$, each other and having unit norm. That's a very weak condition. That's why people usually compute a compact SVD of low rank matrices, they kind of discard unnecessary parts of $U$ (which aren't uniquely identifiable and not don't help to restore the original matrix).