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Singular Value Decoposition

Theory

Calculations

如何算奇異值分解？

Let \(A\in\text{M}_{m\times n}(\mathbb{R})\).

  • Diagnoalize \(AA^T\), obtaining eigenvalues \(\lambda_1\ge\lambda_2\ge\dots\ge\lambda_m\) and its corresponding eigenvectors \(u_i\in\mathbb{R}^m,i=1,2,\dots,m\).

  • Find \(\Sigma\in\text{M}_{m\times n}(\mathbb{R})\), where

\[ \Sigma_{ij}=\begin{cases}\sqrt{\lambda_i},&\text{if }i=j\\0,&\text{if }i\neq j\end{cases}. \]

  • Solve the equation \(\det(A^TA-\lambda_iI_n)=0\) to obtain the eigenvectors \(v_i\in\mathbb{R}^n,i=1,2,\dots,n\) of \(A^TA\) corresponding to \(\lambda_i\).

  • We now get two matrices \(U\) and \(V\)

\[ U=\begin{pmatrix}\dfrac{1}{\sigma_1}u_1&\cdots&\dfrac{1}{\sigma_m}u_m\end{pmatrix}\in\text{M}_{m\times m}(\mathbb{R}) \]
\[ V=\begin{pmatrix}\dfrac{1}{\sigma_1}v_1&\cdots&\dfrac{1}{\sigma_m}v_m\end{pmatrix}\in\text{M}_{n\times n}(\mathbb{R}) \]
  • The singular value decomposition of \(A\) is
\[ A=U\Sigma V^T \]