高低阵
矩阵秩公式:A∈Cm×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]
posted on 2019-01-07 01:48 codeDog123 阅读(669) 评论(0) 收藏 举报
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