SVD of $\left[\begin{array}{c}2 \sqrt{2}\\2 \sqrt{2}\end{array}\right]$

Find the SVD of $\left[\begin{array}{c}2 \sqrt{2}\\2 \sqrt{2}\end{array}\right]$.

Find the transpose of the matrix: $\left[\begin{array}{c}2 \sqrt{2}\\2 \sqrt{2}\end{array}\right]^{T} = \left[\begin{array}{cc}2 \sqrt{2} & 2 \sqrt{2}\end{array}\right]$ (for steps, see matrix transpose calculator).

Multiply the matrix with its transpose: $W = \left[\begin{array}{c}2 \sqrt{2}\\2 \sqrt{2}\end{array}\right]\cdot \left[\begin{array}{cc}2 \sqrt{2} & 2 \sqrt{2}\end{array}\right] = \left[\begin{array}{cc}8 & 8\\8 & 8\end{array}\right]$ (for steps, see matrix multiplication calculator).

Now, find the eigenvalues and eigenvectors of $W$ (for steps, see eigenvalues and eigenvectors calculator).

Eigenvalue: $16$, eigenvector: $\left[\begin{array}{c}1\\1\end{array}\right]$.

Eigenvalue: $0$, eigenvector: $\left[\begin{array}{c}-1\\1\end{array}\right]$.

Find the square roots of the nonzero eigenvalues ($\sigma_{i}$):

$\sigma_{1} = 4$

The $\Sigma$ matrix is a zero matrix with $\sigma_{i}$ on its diagonal: $\Sigma = \left[\begin{array}{c}4\\0\end{array}\right]$.

The columns of the matrix $U$ are the normalized (unit) vectors: $U = \left[\begin{array}{cc}\frac{\sqrt{2}}{2} & - \frac{\sqrt{2}}{2}\\\frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2}\end{array}\right]$ (for steps in finding a unit vector, see unit vector calculator).

Now, $v_{i} = \frac{1}{\sigma_{i}}\cdot \left[\begin{array}{c}2 \sqrt{2}\\2 \sqrt{2}\end{array}\right]^{T}\cdot u_{i}$:

$v_{1} = \frac{1}{\sigma_{1}}\cdot \left[\begin{array}{c}2 \sqrt{2}\\2 \sqrt{2}\end{array}\right]^{T}\cdot u_{1} = \frac{1}{4}\cdot \left[\begin{array}{cc}2 \sqrt{2} & 2 \sqrt{2}\end{array}\right]\cdot \left[\begin{array}{c}\frac{\sqrt{2}}{2}\\\frac{\sqrt{2}}{2}\end{array}\right] = \left[\begin{array}{c}1\end{array}\right]$ (for steps, see matrix scalar multiplication calculator and matrix multiplication calculator).

Therefore, $V = \left[\begin{array}{c}1\end{array}\right]$.

The matrices $U$, $\Sigma$, and $V$ are such that the initial matrix $\left[\begin{array}{c}2 \sqrt{2}\\2 \sqrt{2}\end{array}\right] = U \Sigma V^T$.

$U = \left[\begin{array}{cc}\frac{\sqrt{2}}{2} & - \frac{\sqrt{2}}{2}\\\frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2}\end{array}\right]\approx \left[\begin{array}{cc}0.707106781186548 & -0.707106781186548\\0.707106781186548 & 0.707106781186548\end{array}\right]$A

$\Sigma = \left[\begin{array}{c}4\\0\end{array}\right]$A

$V = \left[\begin{array}{c}1\end{array}\right]$A