Find a decomposition $X=U \Sigma V^{T}$ of the matrix $X=\begin{bmatrix} 2 & 1 & 2\\ -2 & -1 & -2\\ 4 & 2 & 4\\ 2 & 1 & 2 \end{bmatrix}$ where $\Sigma$ is a rectangular diagonal matrix of
size 4x3, $U$ and $V$ are orthogonal matrices, and the upper right element of $V$ is equal to $\frac{1}{\sqrt{2}}$
I found V according to the given constraint as $\begin{bmatrix} \frac{2}{3} & \frac{-\sqrt{2}}{6} & \frac{1}{\sqrt{2}} \\ \frac{1}{3} & \frac{2\sqrt{2}}{3} & 0\\ \frac{2}{3} & \frac{-\sqrt{2}}{6} & \frac{-1}{\sqrt{2}} \end{bmatrix}$ and $\Sigma=\begin{bmatrix} \sqrt{63} & 0 & 0 \\ 0 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix}$
For finding $U$, I am using $U\Sigma=XV$. But I am stuck at this step as to how to proceed further as $\Sigma$ matrix has repeated 0 eigen values.
Can you please let me know how to find $U$ from the above equation?
You already found the eigenvectors of $X^TX$. These are the rows of $V^T$.
You need to find the eigenvectors of $XX^T$. There should be one eigenvector corresponding to the eigenvalue of $\sqrt {63}$ and three corresponding to the zero eigenvalue. Arrange them as column vectors $U=\begin{pmatrix}u_1&u_2&u_3&u_4\end{pmatrix}$.