Singular Value Decomposition

\begin{aligned} X &= \begin{bmatrix} 1 & 1 & \cdots & 1 \\ x_1 & x_2 & \cdots & x_m \\ \vdots & \vdots & \vdots & \vdots \\ 1 & x_1 & \cdots & x_m \end{bmatrix} = U \Sigma V^T \\ &= \begin{bmatrix} 1 & 1 & \cdots & 1 \\ u_{1} & u_{2} & \cdots & u_n \\ \vdots & \vdots & \vdots & \vdots \\ 1 & 1 & \cdots & 1 \end{bmatrix} \begin{bmatrix} \sigma_1 & \cdots & \cdots & \cdots \\ \vdots & \sigma_2 & \cdots & \cdots \\ \vdots & \cdots & \ddots & \cdots \\ \vdots & \cdots & \cdots & \sigma_m \\ 0 & 0 & 0 & 0 \\ \end{bmatrix} {\begin{bmatrix} \vdots & \vdots & \vdots & \vdots \\ v_{1} & v_{2} & \cdots & v_n \\ \vdots & \vdots & \vdots & \vdots \end{bmatrix}}^T \\ \\ x_k &\in \mathbb{R}^n \\ \\ \text{U, V } &: \text{unitary matrices} \\ \Sigma &: \text{diagonal matrix} \end{aligned}

where $\begin{bmatrix} 1 \\ u_{1} \\ \vdots \\ 1 \end{bmatrix}$ are “eigen-faces”

$U$ is orthonormal, meaning:

\begin{aligned} U U^T &= U^T U = \mathbb{I}_{n \times n} \\ V V^T &= V^T V = \mathbb{I}_{m \times m} \\ \\ \Sigma &: \text{diagonal} \quad \sigma_1 \geq \sigma_2 \geq \cdots \geq \sigma_m \geq 0 \end{aligned}

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