A Fancy Example of SVD
==Example==
Consider the $4 × 5$ matrix
$$\mathbf{M} = \begin{bmatrix}
1 & 0 & 0 & 0 & 2 \
0 & 0 & 3 & 0 & 0 \
0 & 0 & 0 & 0 & 0 \
0 & 2 & 0 & 0 & 0
\end{bmatrix}
$$
A singular value decomposition of this matrix is given by $UΣV^⁎$
$$\begin{align}
\mathbf{U} &= \begin{bmatrix}
\color{Green}0 & \color{Blue}-1 & \color{Cyan}0 & \color{Emerald}0 \
\color{Green}-1 & \color{Blue}0 & \color{Cyan}0 & \color{Emerald}0 \
\color{Green}0 & \color{Blue}0 & \color{Cyan}0 & \color{Emerald}-1 \
\color{Green}0 & \color{Blue}0 & \color{Cyan}-1 & \color{Emerald}0
\end{bmatrix} \[6pt]
\boldsymbol{\Sigma} &= \begin{bmatrix}
3 & 0 & 0 & 0 & \color{Gray}\mathit{0} \
0 & \sqrt{5} & 0 & 0 & \color{Gray}\mathit{0} \
0 & 0 & 2 & 0 & \color{Gray}\mathit{0} \
0 & 0 & 0 & \color{Red}\mathbf{0} & \color{Gray}\mathit{0}
\end{bmatrix} \[6pt]
\mathbf{V}^* &= \begin{bmatrix}
\color{Violet}0 & \color{Violet}0 & \color{Violet}-1 & \color{Violet}0 &\color{Violet}0 \
\color{Plum}-\sqrt{0.2}& \color{Plum}0 & \color{Plum}0 & \color{Plum}0 &\color{Plum}-\sqrt{0.8} \
\color{Magenta}0 & \color{Magenta}-1 & \color{Magenta}0 & \color{Magenta}0 &\color{Magenta}0 \
\color{Orchid}0 & \color{Orchid}0 & \color{Orchid}0 & \color{Orchid}1 &\color{Orchid}0 \
\color{Purple} - \sqrt{0.8} & \color{Purple}0 & \color{Purple}0 & \color{Purple}0 & \color{Purple}\sqrt{0.2}
\end{bmatrix}
\end{align}$$
The scaling matrix $\mathbf{\Sigma}$ is zero outside of the diagonal (grey italics) and one diagonal element is zero (red bold). Furthermore, because the matrices $U$ and $V^T$ are unitary matrix|unitary, multiplying by their respective conjugate transposes yields identity matrix|identity matrices, as shown below. In this case, because $U$ and $V^T$ are real valued, each is an orthogonal matrix.
$$\begin{align}
\mathbf{U} \mathbf{U}^* &=
\begin{bmatrix}
1 & 0 & 0 & 0 \
0 & 1 & 0 & 0 \
0 & 0 & 1 & 0 \
0 & 0 & 0 & 1
\end{bmatrix} = \mathbf{I}_4 \[6pt]
\mathbf{V} \mathbf{V}^* &=
\begin{bmatrix}
1 & 0 & 0 & 0 & 0 \
0 & 1 & 0 & 0 & 0 \
0 & 0 & 1 & 0 & 0 \
0 & 0 & 0 & 1 & 0 \
0 & 0 & 0 & 0 & 1
\end{bmatrix} = \mathbf{I}_5
\end{align}$$
This particular singular value decomposition is not unique. Choosing $\mathbf V$ such that
$$\mathbf{V}^* = \begin{bmatrix}
\color{Violet}0 & \color{Violet}1 & \color{Violet}0 & \color{Violet}0 & \color{Violet}0 \
\color{Plum}0 & \color{Plum}0 & \color{Plum}1 & \color{Plum}0 & \color{Plum}0 \
\color{Magenta}\sqrt{0.2} & \color{Magenta}0 & \color{Magenta}0 & \color{Magenta}0 & \color{Magenta}\sqrt{0.8} \
\color{Orchid}\sqrt{0.4} & \color{Orchid}0 & \color{Orchid}0 & \color{Orchid}\sqrt{0.5} & \color{Orchid}-\sqrt{0.1} \
\color{Purple}-\sqrt{0.4} & \color{Purple}0 & \color{Purple}0 & \color{Purple}\sqrt{0.5} & \color{Purple}\sqrt{0.1}
\end{bmatrix}$$
is also a valid singular value decomposition.