标签：amp mit lambda nbsp ilo sch over 分解 解方程
矩阵秩公式：$A \in {C^{m \times n}}$
\[rank(A) = rank({A^H}) = rank(A{A^H}) = rank({A^H}A)\]
引理：$AX=0$和$A^HAX=0$有相同的解
\[\begin{array}{l}
AX = 0 \Rightarrow {A^H}AX = 0\\
{A^H}AX = 0 \Rightarrow {X^H}{A^H}AX = 0 \Rightarrow {(AX)^H}AX = {\left| {AX} \right|^2} = 0 \Rightarrow AX = 0
\end{array}\]
证明：$rank(A)=rank(A^HX)$
集合${\rm{W}} = \{ X|AX = 0\} $和$\widetilde W = \{ X|{A^H}AX = 0\} $
\[\dim W = \dim W \Rightarrow n - rank(A) = n - rank({A^H}A) \Rightarrow rank(A) = rank({A^H}A)\]
令$A=A^H$
\[rank({A^H}) = rank({({A^H})^H}{A^H}) \Rightarrow rank({A^H}) = rank(A{A^H})\]
因为$rank(A^H)=rank(A)$,所以
\[rank(A) = rank({A^H}) = rank(A{A^H}) = rank({A^H}A)\]
定理：${\rm{A}} \in {C^{m \times n}}$ A为列满秩，则A有左侧逆
证明：
$rank({A^H}A) = rank(A) = n$,所以$A^HA$满秩，所以$A^HA$存在逆阵${({A^H}A)^{ - 1}}$
\[{({A^H}A)^{ - 1}}{A^H}A = I\]
所以A的左侧逆为${({A^H}A)^{ - 1}}{A^H}$
定理：${\rm{A}} \in {C^{m \times n}}$ A为行满秩，则A有右侧逆
证明：
$rank(A{A^H}) = rank(A) = m$,所以$A^HA$满秩，所以$A^HA$存在逆阵${(A{A^H})^{ - 1}}$
\[A{A^H}{(A{A^H})^{ - 1}} = I\]
所以A的右侧逆为${A^H}{(A{A^H})^{ - 1}}$
高阵的性质：
(1)若$BCX=0$，B为高阵,则$CX=0$
\[BCX = 0 \Rightarrow {B_L}BCX = 0 \Rightarrow CX = 0\]
(2)若BX=BY，B为高阵，则$X=Y$
\[BX = BY \Rightarrow {B_L}BX = {B_L}BY \Rightarrow X = Y\]
Schur（舒尔分解）:任意方阵A，${\rm{A}} \in {C^{n \times n}}$，存在酉阵Q，使得
\[{Q^H}AQ = {Q^{ - 1}}AQ = D = \left[ {\begin{array}{*{20}{c}}
{{\lambda _1}}& \otimes & \otimes & \otimes \\
0&{{\lambda _2}}& \otimes & \otimes \\
0&0&{...}& \otimes \\
0&0&0&{{\lambda _n}}
\end{array}} \right]\]
Hermite阵
定义：若A为Hermite阵，则${\rm{A}} \in {C^{n \times n}}$且$A^{H}=A$
定义：若A为斜Hermite阵，则${\rm{A}} \in {C^{n \times n}}$且$A^{H}=-A$
Hermite分解定理：若${\rm{A = }}{A_{n \times n}}$为Hermite($A^H=A$),则存在酉阵Q，使得
\[\left[ {\begin{array}{*{20}{c}}
{{\lambda _1}}&0&0&0\\
0&{{\lambda _2}}&0&0\\
0&0&{...}&0\\
0&0&0&{{\lambda _n}}
\end{array}} \right]\]
并且这里的$\lambda_{1},\lambda_{2},...,\lambda_{n}$都为实数根
证明：
根据Schur定理，有酉阵Q，使得
\[{Q^H}AQ = {Q^{ - 1}}AQ = D = \left[ {\begin{array}{*{20}{c}}
{{\lambda _1}}& \otimes & \otimes & \otimes \\
0&{{\lambda _2}}& \otimes & \otimes \\
0&0&{...}& \otimes \\
0&0&0&{{\lambda _n}}
\end{array}} \right]\]
然后证明$\lambda_{1},\lambda_{2},...,\lambda_{n}$都为实数
\[\left[ {\begin{array}{*{20}{c}}
{{\lambda _1}}& \otimes & \otimes & \otimes \\
0&{{\lambda _2}}& \otimes & \otimes \\
0&0&{...}& \otimes \\
0&0&0&{{\lambda _n}}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{\overline {{\lambda _1}} }&0&0&0\\
\otimes &{\overline {{\lambda _2}} }&0&0\\
\otimes & \otimes &{...}&0\\
\otimes & \otimes & \otimes &{\overline {{\lambda _n}} }
\end{array}} \right]\]
所以$\overline {{\lambda _k}} = {\lambda _k}$且$\otimes = 0$，得证
Hermite阵性质：
(1)若$A=A^{H}, X \in C^{n}$，则$f(X)=X^HAX$为实数
\[\begin{array}{l}
f{(X)^H} = {({X^H}AX)^H} = {X^H}{A^H}X = {X^H}AX = f(X)\\
f{(X)^H} = f(X) \Rightarrow \overline {f(X)} = f(X) \Rightarrow f(X) \in {R^1}
\end{array}\]
(2)若$A=A^{H}, X \in C^{n}$，则${\lambda _k} = \frac{{{X^H}AX}}{{{{\left| X \right|}^2}}}$(X为特征向量)
\[{\rm{AX}} = {\lambda _k}X \Rightarrow {X^H}AX = {\lambda _k}{X^H}X \Rightarrow {\lambda _k} = \frac{{{X^H}AX}}{{{{\left| X \right|}^2}}}\]
(3)若$A \ge 0$,则存在$B \ge 0$，使得$B^2=A$，称B为A的平方根，记为$B = \sqrt A $，可写$A = {(\sqrt A )^2}$
证明：由Hermite分解定理
\[{Q^H}AQ = D = \left[ {\begin{array}{*{20}{c}}
{{\lambda _1}}&0&0&0\\
0&{{\lambda _2}}&0&0\\
0&0&{...}&0\\
0&0&0&{{\lambda _n}}
\end{array}} \right]\]
令
\[\sqrt D = \left[ {\begin{array}{*{20}{c}}
{\sqrt {{\lambda _1}} }&0&0&0\\
0&{\sqrt {{\lambda _2}} }&0&0\\
0&0&{...}&0\\
0&0&0&{\sqrt {{\lambda _n}} }
\end{array}} \right] \ge 0\]
\[{(\sqrt D )^2} = \left[ {\begin{array}{*{20}{c}}
{{\lambda _1}}&0&0&0\\
0&{{\lambda _2}}&0&0\\
0&0&{...}&0\\
0&0&0&{{\lambda _n}}
\end{array}} \right] = D\]
令
\[{B^H} = {(Q\sqrt D {Q^H})^H} = Q{(\sqrt D )^H}{Q^H} = Q(\sqrt D ){Q^H} = B \Rightarrow B = Q(\sqrt D ){Q^H}{\rm{ = }}Q(\sqrt D ){Q^{{\rm{ - }}1}}\]
所以$B \sim \sqrt D \Rightarrow \lambda (B) = \lambda (\sqrt D ) = \{ \sqrt {{\lambda _1}} ,\sqrt {{\lambda _2}} ,...,\sqrt {{\lambda _n}} \} $
\[{B^2} = BB = (Q\sqrt D {Q^{ - 1}})(Q\sqrt D {Q^{ - 1}}) = Q{(\sqrt D )^2}{Q^{ - 1}} = QD{Q^{ - 1}} = A\]
(4)任意$A = {A_{m \times n}}$,$A^HA,AA^H$为Hermite，且$A{A^H},{A^H}A \ge 0$
Hermite分解：
\[A = {A^H} \Rightarrow {Q^H}AQ = {Q^{ - 1}}AQ = \left[ {\begin{array}{*{20}{c}}
{{\lambda _1}}&0&0&0\\
0&{{\lambda _2}}&0&0\\
0&0&{...}&0\\
0&0&0&{{\lambda _n}}
\end{array}} \right]\]
$Q = \left[ {\begin{array}{*{20}{c}}
{{\varepsilon _1}}&{{\varepsilon _1}}&{...}&{{\varepsilon _n}}
\end{array}} \right]$Q为酉阵，Q中列均为特征向量（无关），即为$A{\varepsilon _1} = \lambda {\varepsilon _1},...,A{\varepsilon _n} = \lambda {\varepsilon _n}$,且${\varepsilon _1} \bot {\varepsilon _2} \bot ... \bot {\varepsilon _n}$
Hermite分解定理求解过程：
(1)$A=A^H$恰好有n个正交特征向量$x_1,x_2,...,x_n$，可求解方程$AX_i=\lambda_iX_i$找出${x_1} \bot {x_2} \bot ... \bot { x_n}$,令$P=[x_1,x_2,...,x_n]$，P为预备半酉阵，有$P_{-1}AP=D$为对角形。
令
\[Q = \left[ {\begin{array}{*{20}{c}}
{\frac{{{X_1}}}{{\left| {{X_1}} \right|}}}&{\frac{{{X_2}}}{{\left| {{X_2}} \right|}}}&{...}&{\frac{{{X_n}}}{{\left| {{X_n}} \right|}}}
\end{array}} \right]\]
则Q为酉阵，且$Q^{-1}AQ=Q^HAQ=D$为对角形
半正定矩阵：$A=A^H$,${\rm{A}} = {A_{n \times n}}$且$f(X) = {X^H}AX \ge 0$，则称A为半正定阵，记为"$A \ge 0$"
正定矩阵：$A=A^H$,${\rm{A}} = {A_{n \times n}}$且$f(X) = {X^H}AX > 0$，则称A为正定阵，记为"$A > 0$"
性质：
(1) $A \ge 0 \Leftrightarrow {\lambda _1} \ge 0,{\lambda _2} \ge 0,...,{\lambda _n} \ge 0，A^H=A$
(2)$A > 0 \Leftrightarrow {\lambda _1} > 0,{\lambda _2} > 0,...,{\lambda _n} > 0，A^H=A$
证明：
必要性：
\[f(X) = {X^H}AX > 0 \Rightarrow \frac{{{X^H}AX}}{{{{\left| X \right|}^2}}} = {\lambda _k} > 0\]
充分性($\lambda _k>0 A^H=A$,证任意X，有$X^HAX>0$)：
因为A为Hermite矩阵，根据Hermite分解定理
\[{Q^H}AQ = D = \left[ {\begin{array}{*{20}{c}}
{{\lambda _1}}&0&0&0\\
0&{{\lambda _2}}&0&0\\
0&0&{...}&0\\
0&0&0&{{\lambda _n}}
\end{array}} \right]\]
\[\begin{array}{l}
{Y^H}{Q^H}AQY = {Y^H}\left[ {\begin{array}{*{20}{c}}
{{\lambda _1}}&0&0&0\\
0&{{\lambda _2}}&0&0\\
0&0&{...}&0\\
0&0&0&{{\lambda _n}}
\end{array}} \right]Y = {\lambda _1}{\left| {{y_1}} \right|^2} + ... + {\lambda _n}{\left| {{y_n}} \right|^2} > 0\\
{Y^H}{Q^H}AQY > 0 \Rightarrow {(QY)^H}AQY > 0
\end{array}\]
令$X=QY$
\[{(QY)^H}AQY > 0 \Rightarrow f(X) = {X^H}AX > 0\]
斜Hermite分解定理：$A^H=-A \in C^{n \times n}$，则存在酉阵Q，使得
\[{Q^H}AQ = {Q^{ - 1}}AQ = \left[ {\begin{array}{*{20}{c}}
{{\lambda _1}i}&0&0&0\\
0&{{\lambda _2}i}&0&0\\
0&0&{...}&0\\
0&0&0&{{\lambda _n}i}
\end{array}} \right]\]
其中$\lambda (A)={\lambda_{1}i, \lambda_{2}i, ..., \lambda_{n}i }$特征根为纯虚数或0，$\lambda (\frac{A}{i}) = \{ {\lambda _1},{\lambda _2},...,{\lambda _n}\} $
证明：
\[{A^H} = - A \Rightarrow {(\frac{A}{i})^H} = \frac{{{A^H}}}{{ - 1}} = \frac{{ - A}}{{ - 1}} = A \Rightarrow \frac{A}{i} = {(\frac{A}{i})^H}\]
\[{Q^H}(\frac{A}{i})Q = {Q^{ - 1}}(\frac{A}{i})Q = D = \left[ {\begin{array}{*{20}{c}}
{{\lambda _1}}&0&0&0\\
0&{{\lambda _2}}&0&0\\
0&0&{...}&0\\
0&0&0&{{\lambda _n}}
\end{array}} \right]\]
\[{Q^H}AQ = {Q^{ - 1}}AQ = Di = = \left[ {\begin{array}{*{20}{c}}
{{\lambda _1}i}&0&0&0\\
0&{{\lambda _2}i}&0&0\\
0&0&{...}&0\\
0&0&0&{{\lambda _n}i}
\end{array}} \right]\]
标签：amp mit lambda nbsp ilo sch over 分解 解方程
原文地址：https://www.cnblogs.com/codeDog123/p/10205267.html
