正奇异值：设$A=A_{m \times n}, rank(A)=p>0$，则$\lambda ({A^H}A)$与$\lambda (A{A^H})$恰有p个正特征根，${\lambda _1} > 0,{\lambda _2} > 0,...,{\lambda _p} > 0$
$\lambda ({A^H}A) = \{ {\lambda _1},{\lambda _2},...,{\lambda _p},0,...,0\} $ n-p个0
$\lambda (A{A^H}) = \{ {\lambda _1},{\lambda _2},...,{\lambda _p},0,...,0\} $ m-p个0
称$\sqrt {{\lambda _1}} ,\sqrt {{\lambda _2}} ,,...,\sqrt {{\lambda _p}} $为正奇异值，记为$S^{+}(A) = \{ \sqrt {{\lambda _1}} ,\sqrt {{\lambda _2}} ,,...,\sqrt {{\lambda _p}} \} $
正奇异值分解：任意$A=A_{m \times n}$，秩$rank(A)=r>0$，则有$A = P\Delta {Q^H}$，
\[\Delta {\rm{ = }}\left[ {\begin{array}{*{20}{c}}
{\sqrt {{\lambda _1}} }&0&0&0\\
0&{\sqrt {{\lambda _2}} }&0&0\\
0&0&{...}&0\\
0&0&0&{\sqrt {{\lambda _p}} }
\end{array}} \right]\]
P和Q为半酉阵，$P^HP=Q^HQ=I$
(1)求$\lambda (A{A^H}) = \lambda ({A^H}A) = \{ {\lambda _1},{\lambda _2},...,{\lambda _p}$与特征向量$x_1,x_2,...,x_r$
(2)令这里的$x_1,x_2,...,x_r$得互相垂直
\[P = \left[ {\begin{array}{*{20}{c}}
{\frac{{A{X_1}}}{{\left| {A{X_1}} \right|}}}&{\frac{{A{X_2}}}{{\left| {A{X_2}} \right|}}}&{...}&{\frac{{A{X_r}}}{{\left| {A{X_r}} \right|}}}
\end{array}} \right]\]
\[Q = \left[ {\begin{array}{*{20}{c}}
{\frac{{{X_1}}}{{\left| {{X_1}} \right|}}}&{\frac{{{X_2}}}{{\left| {{X_2}} \right|}}}&{...}&{\frac{{{X_r}}}{{\left| {{X_r}} \right|}}}
\end{array}} \right]\]
$P=P_{m \times r}, Q=Q_{n \times r}$
(3)有$A = P\Delta {Q^H}$
奇异值分解：
(1)根据正奇异值分解得到$A = P\Delta {Q^H}
(2)扩充$W=[Q_{m \times r},Q_{m \times (n-r)}], V=[P_{n \times r},P_{n \times (n-r)}]$使得P和Q为从半酉阵扩展为酉阵
(3)$\Delta_{r \times r}$0扩充为$\Delta_{n \times n}$
\[D {\rm{ = }}\left[ {\begin{array}{*{20}{c}}
\Delta &0\\
0&0
\end{array}} \right]\]
(4)有
\[{\rm{A = W}}\left[ {\begin{array}{*{20}{c}}
\Delta &0\\
0&0
\end{array}} \right]{V^H}\]
证明：
\[{\rm{A = W}}\left[ {\begin{array}{*{20}{c}}
\Delta &0\\
0&0
\end{array}} \right]{V^H} = \left[ {\begin{array}{*{20}{c}}
{{P_{m \times r}}}&{{P_{m \times (m - r)}}}
\end{array}} \right]{\left[ {\begin{array}{*{20}{c}}
\Delta &0\\
0&0
\end{array}} \right]_{m \times n}}{\left[ {\begin{array}{*{20}{c}}
{{Q_{n \times r}}}&{{Q_{n \times (n - r)}}}
\end{array}} \right]^H} = {P_{m \times r}}\Delta {Q_{n \times r}} = A\]
奇异值分解性质：
(1)方阵$A=A_{n \times n}$，有$S_1S_2...S_n=|det(A)|=$|\lambda_1lambda_2...lambda_n|$
\[\lambda (A) = \{ {\lambda _1},{\lambda _1},...,{\lambda _n}\} \]
\[\lambda (A{A^H}) = \lambda ({A^H}A) = \{ {t_1},{t_2},...,{t_n}\} \]
\[S(A) = \{ \sqrt {{t_1}} ,\sqrt {{t_2}} ,...,\sqrt {{t_n}} \} \]
有：
\[\sqrt {{t_1}} \sqrt {{t_2}} ...\sqrt {{t_n}} = \left| {\det (A)} \right| = \left| {{\lambda _1}{\lambda _1}...{\lambda _n}} \right|\]
证明：
\[\det ({A^H}A) = \det (A{A^H}) = \det ({A^H})\det (A) = \det (\overline {{A^T}} )\det (A) = \overline {\det (A)} \det (A) = {\left| {\det (A)} \right|^2} \Rightarrow \det ({A^H}A) = {\left| {\det (A)} \right|^2}\]
\[\det ({A^H}A) = {t_1}{t_2}...{t_n} = {({S_1}{S_2}...{S_n})^2}\]
\[{\left| {\det (A)} \right|^2} = {\left| {{\lambda _1}{\lambda _1}...{\lambda _n}} \right|^2}\]
\[{({S_1}{S_2}...{S_n})^2} = {\left| {\det (A)} \right|^2} = {\left| {{\lambda _1}{\lambda _1}...{\lambda _n}} \right|^2} \Rightarrow ({S_1}{S_2}...{S_n}) = \left| {\det (A)} \right| = \left| {{\lambda _1}{\lambda _1}...{\lambda _n}} \right|\]
所以：
\[({S_1}{S_2}...{S_n}) = \left| {\det (A)} \right| = \left| {{\lambda _1}{\lambda _1}...{\lambda _n}} \right|\]
正SVD分解性质
(1)${\rm{A}} = P\Delta {Q^H} \Rightarrow {A^H} = Q{\Delta ^H}{P^H}$
(2)由正SVD：$A=P \delta Q^H$：
\[\Delta {\rm{ = }}\left[ {\begin{array}{*{20}{c}}
{\sqrt {{\lambda _1}} }&0&0&0\\
0&{\sqrt {{\lambda _2}} }&0&0\\
0&0&{...}&0\\
0&0&0&{\sqrt {{\lambda _p}} }
\end{array}} \right]\]
\[P = \left[ {\begin{array}{*{20}{c}}
{{X_1}}&{{X_2}}&{...}&{{X_r}}
\end{array}} \right]({X_1} \bot {X_2} \bot ... \bot {X_r})\]
\[Q = \left[ {\begin{array}{*{20}{c}}
{{Y_1}}&{{Y_2}}&{...}&{{Y_r}}
\end{array}} \right]({Y_1} \bot {Y_2} \bot ... \bot {Y_r})\]
有：
\[A = \left[ {\begin{array}{*{20}{c}}
{{X_1}}&{{X_2}}&{...}&{{X_r}}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
{\sqrt {{\lambda _1}} }&0&0&0\\
0&{\sqrt {{\lambda _2}} }&0&0\\
0&0&{...}&0\\
0&0&0&{\sqrt {{\lambda _r}} }
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
{Y_1^H}\\
{\begin{array}{*{20}{c}}
{Y_2^H}\\
{...}
\end{array}}\\
{Y_r^H}
\end{array}} \right] = \sqrt {{\lambda _1}} {X_1}Y_1^H + ... + \sqrt {{\lambda _r}} {X_r}Y_r^H\]
可以令$\sqrt {{\lambda _1}} \ge \sqrt {{\lambda _2}} \ge ... \ge \sqrt {{\lambda _r}} $
\[A{\rm{ = }}\sqrt {{\lambda _1}} {X_1}Y_1^H + ... + \sqrt {{\lambda _r}} {X_r}Y_r^H \approx \sqrt {{\lambda _1}} {X_1}Y_1^H\]
可以近似表达A
(3)
\[A = {A_{m \times n}},{S^ + }(A) = \{ {S_1},{S_2},...,{S_r}\} \Rightarrow tr({A^H}A) = tr(A{A^H}) = \sum {{{\left| {{a_{ij}}} \right|}^2}} \]
原文地址：https://www.cnblogs.com/codeDog123/p/10206787.html