高低阵

Posted 2021-02-01 codedog123

矩阵秩公式：$A in {C^{m imes n}}$

[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

[egin{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} ight|^2} = 0 Rightarrow AX = 0
end{array}]

证明：$rank(A)=rank(A^HX)$

集合${ m{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：${ m{A}} in {C^{m imes n}}$ 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：${ m{A}} in {C^{m imes n}}$ 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，${ m{A}} in {C^{n imes n}}$，存在酉阵Q，使得

[{Q^H}AQ = {Q^{ - 1}}AQ = D = left[ {egin{array}{*{20}{c}}
{{lambda _1}}& otimes & otimes & otimes \
0&{{lambda _2}}& otimes & otimes \
0&0&{...}& otimes \
0&0&0&{{lambda _n}}
end{array}} ight]]

Hermite阵

定义：若A为Hermite阵，则${ m{A}} in {C^{n imes n}}$且$A^{H}=A$

定义：若A为斜Hermite阵，则${ m{A}} in {C^{n imes n}}$且$A^{H}=-A$

Hermite分解定理：若${ m{A = }}{A_{n imes n}}$为Hermite($A^H=A$),则存在酉阵Q，使得

[left[ {egin{array}{*{20}{c}}
{{lambda _1}}&0&0&0\
0&{{lambda _2}}&0&0\
0&0&{...}&0\
0&0&0&{{lambda _n}}
end{array}} ight]]

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[ {egin{array}{*{20}{c}}
{{lambda _1}}& otimes & otimes & otimes \
0&{{lambda _2}}& otimes & otimes \
0&0&{...}& otimes \
0&0&0&{{lambda _n}}
end{array}} ight]]

然后证明$lambda_{1},lambda_{2},...,lambda_{n}$都为实数

[left[ {egin{array}{*{20}{c}}
{{lambda _1}}& otimes & otimes & otimes \
0&{{lambda _2}}& otimes & otimes \
0&0&{...}& otimes \
0&0&0&{{lambda _n}}
end{array}} ight] = left[ {egin{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}} ight]]

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

[egin{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${lambda _k} = frac{{{X^H}AX}}{{{{left| X ight|}^2}}}$(X为特征向量)

[{ m{AX}} = {lambda _k}X Rightarrow {X^H}AX = {lambda _k}{X^H}X Rightarrow {lambda _k} = frac{{{X^H}AX}}{{{{left| X ight|}^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[ {egin{array}{*{20}{c}}
{{lambda _1}}&0&0&0\
0&{{lambda _2}}&0&0\
0&0&{...}&0\
0&0&0&{{lambda _n}}
end{array}} ight]]

令

[sqrt D = left[ {egin{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}} ight] ge 0]

[{(sqrt D )^2} = left[ {egin{array}{*{20}{c}}
{{lambda _1}}&0&0&0\
0&{{lambda _2}}&0&0\
0&0&{...}&0\
0&0&0&{{lambda _n}}
end{array}} ight] = D]

令

[{B^H} = {(Qsqrt D {Q^H})^H} = Q{(sqrt D )^H}{Q^H} = Q(sqrt D ){Q^H} = B Rightarrow B = Q(sqrt D ){Q^H}{ m{ = }}Q(sqrt D ){Q^{{ m{ - }}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 = (Qsqrt D {Q^{ - 1}})(Qsqrt D {Q^{ - 1}}) = Q{(sqrt D )^2}{Q^{ - 1}} = QD{Q^{ - 1}} = A]

(4)任意$A = {A_{m imes n}}$,$A^HA,AA^H$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[ {egin{array}{*{20}{c}}
{{lambda _1}}&0&0&0\
0&{{lambda _2}}&0&0\
0&0&{...}&0\
0&0&0&{{lambda _n}}
end{array}} ight]]

$Q = left[ {egin{array}{*{20}{c}}
{{varepsilon _1}}&{{varepsilon _1}}&{...}&{{varepsilon _n}}
end{array}} ight]$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${varepsilon _1} ot {varepsilon _2} ot ... ot {varepsilon _n}$

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${x_1} ot {x_2} ot ... ot { x_n}$,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[ {egin{array}{*{20}{c}}
{frac{{{X_1}}}{{left| {{X_1}} ight|}}}&{frac{{{X_2}}}{{left| {{X_2}} ight|}}}&{...}&{frac{{{X_n}}}{{left| {{X_n}} ight|}}}
end{array}} ight]]

则Q为酉阵，且$Q^{-1}AQ=Q^HAQ=D$为对角形

 

 

 

 

半正定矩阵：$A=A^H$,${ m{A}} = {A_{n imes n}}$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：$A=A^H$,${ m{A}} = {A_{n imes n}}$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 ight|}^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[ {egin{array}{*{20}{c}}
{{lambda _1}}&0&0&0\
0&{{lambda _2}}&0&0\
0&0&{...}&0\
0&0&0&{{lambda _n}}
end{array}} ight]]

[egin{array}{l}
{Y^H}{Q^H}AQY = {Y^H}left[ {egin{array}{*{20}{c}}
{{lambda _1}}&0&0&0\
0&{{lambda _2}}&0&0\
0&0&{...}&0\
0&0&0&{{lambda _n}}
end{array}} ight]Y = {lambda _1}{left| {{y_1}} ight|^2} + ... + {lambda _n}{left| {{y_n}} ight|^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：$A^H=-A in C^{n imes n}$，则存在酉阵Q，使得

[{Q^H}AQ = {Q^{ - 1}}AQ = left[ {egin{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}} ight]]

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[ {egin{array}{*{20}{c}}
{{lambda _1}}&0&0&0\
0&{{lambda _2}}&0&0\
0&0&{...}&0\
0&0&0&{{lambda _n}}
end{array}} ight]]

[{Q^H}AQ = {Q^{ - 1}}AQ = Di = = left[ {egin{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}} ight]]