矩阵论
- 准备知识(上)
- 准备知识(下)
1.7 内积
X=(x1x2⋮xn),Y=(y1y2⋮yn)∈Cn\begin{aligned} &X=\left( \begin{matrix} x_1\\x_2\\\vdots\\x_n \end{matrix} \right),Y=\left( \begin{matrix} y_1\\y_2\\\vdots\\y_n \end{matrix} \right)\in C^n \end{aligned} X=⎝⎛x1x2⋮xn⎠⎞,Y=⎝⎛y1y2⋮yn⎠⎞∈Cn
1.7.1 复向量内积
(X,Y)=ΔYHX=x1y1‾+x2y2‾+⋯+xnyn‾=∑i=1nxiyi‾=tr(YHX)(Y,X)=XHY=x1‾y1+x2‾y2+⋯+xn‾y1=∑i=1nxi‾yi=tr(XHY)\begin{aligned} (X,Y)&\overset{\Delta}{=}Y^HX=x_1\overline{y_1}+x_2\overline{y_2}+\cdots+x_n\overline{y_n}=\sum\limits_{i=1}\limits^{n}x_i\overline{y_i}\\ &=tr(Y^HX)\\ (Y,X)&=X^HY=\overline{x_1}y_1+\overline{x_2}y_2+\cdots+\overline{x_n}y_1=\sum\limits_{i=1}\limits^{n}\overline{x_i}y_i\\ &=tr(X^HY) \end{aligned} (X,Y)(Y,X)=ΔYHX=x1y1+x2y2+⋯+xnyn=i=1∑nxiyi=tr(YHX)=XHY=x1y1+x2y2+⋯+xny1=i=1∑nxiyi=tr(XHY)
若取Y=X,则其内积
(X,X)=XHX=x1x1‾+x2x2‾+⋯+xnxn‾=∑i=1nxixi‾=∣x1∣2+∣x2∣2+⋯+∣xn∣2=∣X∣2=tr(XHX)=tr(XXH)\begin{aligned} (X,X)&=X^HX=x_1\overline{x_1}+x_2\overline{x_2}+\cdots+x_n\overline{x_n}=\sum\limits_{i=1}\limits^{n}x_i\overline{x_i}\\ &=\vert x_1 \vert^2+\vert x_2 \vert^2+\cdots+\vert x_n \vert^2 = \vert X \vert^2\\ &=tr(X^HX)=tr(XX^H) \end{aligned} (X,X)=XHX=x1x1+x2x2+⋯+xnxn=i=1∑nxixi=∣x1∣2+∣x2∣2+⋯+∣xn∣2=∣X∣2=tr(XHX)=tr(XXH)
向量内积性质
- (X,X)≥0(X,X)\ge 0(X,X)≥0 ;若 X≠0,(X,X)≥0X\neq 0 ,(X,X)\ge 0X=0,(X,X)≥0
- (X,Y)=(Y,X)‾(X,Y)=\overline{(Y,X)}(X,Y)=(Y,X)
- (kX,Y)=k(X,Y),(X,kY)=k‾(X,Y)(kX,Y)=k(X,Y),(X,kY)=\overline{k}(X,Y)(kX,Y)=k(X,Y),(X,kY)=k(X,Y)
- (X+Y,W)=(X,Y)+(X,W)
- ∣(X,Y)∣2≤∣X∣⋅∣Y∣\vert (X,Y) \vert^2\le \vert X \vert\cdot\vert Y \vert∣(X,Y)∣2≤∣X∣⋅∣Y∣
1.7.2 复矩阵内积
定义
(A,B)=Δtr(ABH)=tr(AHB)=∑aijbij‾,A,B∈Cm,n(A,A)=tr(AAH)=tr(AHA)=∑aijaij‾=∑∣aij∣2\begin{aligned} &(A,B)\overset{\Delta}{=}tr(AB^H)=tr(A^HB)=\sum a_{ij}\overline{b_{ij}},A,B\in C^{m,n}\\ &(A,A)=tr(AA^H)=tr(A^HA)=\sum a_{ij}\overline{a_{ij}}=\sum \vert a_{ij} \vert^2 \end{aligned} (A,B)=Δtr(ABH)=tr(AHB)=∑aijbij,A,B∈Cm,n(A,A)=tr(AAH)=tr(AHA)=∑aijaij=∑∣aij∣2
矩阵A的模长:∣∣A∣∣=(A,A)=tr(AAH)=∑∣aij∣2\vert \vert A \vert \vert=\sqrt{(A,A)}=\sqrt{tr(AA^H)}=\sqrt{\sum\vert a_{ij} \vert^2}∣∣A∣∣=(A,A)=tr(AAH)=∑∣aij∣2
性质
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(A,A)=tr(AAH)=∑∣aij∣2≥0(A,A)=tr(AA^H)=\sum\vert a_{ij} \vert^2 \ge 0(A,A)=tr(AAH)=∑∣aij∣2≥0 ;若 A≠0A\neq 0A=0 ,则 (A,A)>0(A,A)>0(A,A)>0
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(A,B)=(B,A)‾(A,B)=\overline{(B,A)}(A,B)=(B,A)
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(kA,B)=k(A,B),(A,kB)=k‾(A,B)(kA,B)=k(A,B),(A,kB)=\overline{k}(A,B)(kA,B)=k(A,B),(A,kB)=k(A,B)
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(A+B,D)=(A,D)+(B,D),(D,A+B)=(D,A)+(D,B)
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∣(A,B)∣2≤∣A∣⋅∣B∣\vert (A,B) \vert^2 \le \vert A\vert\cdot\vert B \vert∣(A,B)∣2≤∣A∣⋅∣B∣
矩阵的内积形式
列分块
A=(a11⋯a1p⋮⋱⋮an1⋯anp)∈Cn×p=(α1,⋯,αp),其中αi为n维列向量(n×1阶矩阵)AH=(a11‾⋯an1‾⋮⋱⋮a1p‾⋯anp‾)∈Cp×n=(α1‾T⋮αp‾T),其中α1‾T是n维行向量(1×n阶矩阵)\begin{aligned} A&=\left( \begin{matrix} &a_{11}\quad&\cdots&a_{1p}\\ &\vdots\quad&\ddots&\vdots\\ &a_{n1}\quad&\cdots&a_{np} \end{matrix} \right)\in C^{n\times p}\\ &=(\alpha_1,\cdots,\alpha_p),其中\alpha_i为n维列向量(n\times 1阶矩阵)\\\\ A^H&=\left( \begin{matrix} &\overline{a_{11}}\quad&\cdots\quad &\overline{a_{n1}}\\ &\vdots\quad &\ddots&\vdots\\ &\overline{a_{1p}}\quad&\cdots\quad &\overline{a_{np}}\\ \end{matrix} \right)\in C^{p\times n}\\ &=\left( \begin{matrix} \overline{\alpha_1}^T\\ \vdots\\ \overline{\alpha_p}^T \end{matrix} \right),其中\overline{\alpha_1}^T是n维行向量(1\times n阶矩阵)\\\\ \end{aligned} AAH=⎝⎛a11⋮an1⋯⋱⋯a1p⋮anp⎠⎞∈Cn×p=(α1,⋯,αp),其中αi为n维列向量(n×1阶矩阵)=⎝⎛a11⋮a1p⋯⋱⋯an1⋮anp⎠⎞∈Cp×n=⎝⎛α1T⋮αpT⎠⎞,其中α1T是n维行向量(1×n阶矩阵)
AHA=(α1‾T⋮αp‾T)(α1,⋯,αp)=(α1‾Tα1α1‾Tα2⋯α1‾Tαpα2‾Tα1α2‾Tα2⋯α2‾Tαp⋮⋮⋱⋮αp‾Tα1αp‾Tα2⋯αp‾Tαp)=((α1,α1)(α2,α1)⋯(αp,α1)(α1,α2)(α2,α2)⋯(αp,α2)⋮⋮⋱⋮(α1,αp)(α2,αp)⋯(αp,αp))=((α1,α1)‾(α1,α2)‾⋯(α1,αp)‾(α2,α1)‾(α2,α2)‾⋯(α2,αp)‾⋮⋮⋱⋮(αp,α1)‾(αp,α2)‾⋯(αp,αp)‾)\begin{aligned} A^HA&=\left( \begin{matrix} \overline{\alpha_1}^T\\ \vdots\\ \overline{\alpha_p}^T \end{matrix} \right)(\alpha_1,\cdots,\alpha_p)\\\\ &=\left( \begin{matrix} &\overline{\alpha_1}^T\alpha_1\quad&\overline{\alpha_1}^T\alpha_2\quad &\cdots&\overline{\alpha_1}^T\alpha_p\\ &\overline{\alpha_2}^T\alpha_1\quad&\overline{\alpha_2}^T\alpha_2\quad &\cdots&\overline{\alpha_2}^T\alpha_p\\ &\vdots\quad&\vdots\quad &\ddots\quad &\vdots\\ &\overline{\alpha_p}^T\alpha_1\quad&\overline{\alpha_p}^T\alpha_2\quad &\cdots&\overline{\alpha_p}^T\alpha_p \end{matrix} \right)\\\\ &=\left( \begin{matrix} &(\alpha_1,\alpha_1)\quad &(\alpha_2,\alpha_1)\quad&\cdots\quad &(\alpha_p,\alpha_1)\\ &(\alpha_1,\alpha_2)\quad &(\alpha_2,\alpha_2)\quad&\cdots\quad &(\alpha_p,\alpha_2)\\ &\vdots\quad&\vdots\quad &\ddots\quad &\vdots\\ &(\alpha_1,\alpha_p)\quad &(\alpha_2,\alpha_p)\quad&\cdots\quad &(\alpha_p,\alpha_p) \end{matrix} \right)\\\\ &=\left( \begin{matrix} &\overline{(\alpha_1,\alpha_1)}\quad &\overline{(\alpha_1,\alpha_2)}\quad&\cdots\quad &\overline{(\alpha_1,\alpha_p)}\\ &\overline{(\alpha_2,\alpha_1)}\quad &\overline{(\alpha_2,\alpha_2)}\quad&\cdots\quad &\overline{(\alpha_2,\alpha_p)}\\ &\vdots\quad&\vdots\quad &\ddots\quad &\vdots\\ &\overline{(\alpha_p,\alpha_1)}\quad &\overline{(\alpha_p,\alpha_2)}\quad&\cdots\quad &\overline{(\alpha_p,\alpha_p)} \end{matrix} \right)\\ \end{aligned} AHA=⎝⎛α1T⋮αpT⎠⎞(α1,⋯,αp)=⎝⎛α1Tα1α2Tα1⋮αpTα1α1Tα2α2Tα2⋮αpTα2⋯⋯⋱⋯α1Tαpα2Tαp⋮αpTαp⎠⎞=⎝⎛(α1,α1)(α1,α2)⋮(α1,αp)(α2,α1)(α2,α2)⋮(α2,αp)⋯⋯⋱⋯(αp,α1)(αp,α2)⋮(αp,αp)⎠⎞=⎝⎛(α1,α1)(α2,α1)⋮(αp,α1)(α1,α2)(α2,α2)⋮(αp,α2)⋯⋯⋱⋯(α1,αp)(α2,αp)⋮(αp,αp)⎠⎞
行分块
A=(a11⋯a1p⋮⋱⋮an1⋯anp)∈Cn×p=(α1α2⋮αn),其中αi为p维行向量(1×p阶矩阵)AH=(α1‾Tα2‾T⋯αn‾T),其中αi‾T为p维列向量(p×1阶矩阵)\begin{aligned} A&=\left( \begin{matrix} &a_{11}\quad&\cdots&a_{1p}\\ &\vdots\quad&\ddots&\vdots\\ &a_{n1}\quad&\cdots&a_{np} \end{matrix} \right)\in C^{n\times p}\\ &=\left( \begin{matrix} \alpha_1\\ \alpha_2\\ \vdots\\ \alpha_n \end{matrix} \right),其中\alpha_i为p维行向量(1\times p阶矩阵)\\\\ A^H&=\left( \begin{matrix} \overline{\alpha_1}^T\quad \overline{\alpha_2}^T\quad \cdots\quad \overline{\alpha_n}^T \end{matrix} \right),其中\overline{\alpha_i}^T 为p维列向量(p\times 1阶矩阵) \end{aligned} AAH=⎝⎛a11⋮an1⋯⋱⋯a1p⋮anp⎠⎞∈Cn×p=⎝⎛α1α2⋮αn⎠⎞,其中αi为p维行向量(1×p阶矩阵)=(α1Tα2T⋯αnT),其中αiT为p维列向量(p×1阶矩阵)
AAH=(α1α2⋮αn)(α1‾Tα2‾T⋯αn‾T)=(α1α1‾Tα1α2‾T⋯α1αn‾Tα2α1‾Tα2α2‾T⋯α2αn‾T⋮⋮⋱⋮αnα1‾Tαnα2‾T⋯αnαn‾T)=((α1,α1)(α1,α2)⋯(α1,αn)(α2,α1)(α2,α2)⋯(α2,αn)⋮⋮⋱⋮(αn,α1)(αn,α2)⋯(αn,αn))\begin{aligned} AA^H&=\left( \begin{matrix} \alpha_1\\ \alpha_2\\ \vdots\\ \alpha_n \end{matrix} \right)\left( \begin{matrix} \overline{\alpha_1}^T\quad \overline{\alpha_2}^T\quad \cdots\quad \overline{\alpha_n}^T \end{matrix} \right)\\\\ &=\left( \begin{matrix} &\alpha_1\overline{\alpha_1}^T\quad &\alpha_1\overline{\alpha_2}^T\quad&\cdots\quad &\alpha_1\overline{\alpha_n}^T\\ &\alpha_2\overline{\alpha_1}^T\quad &\alpha_2\overline{\alpha_2}^T\quad&\cdots\quad &\alpha_2\overline{\alpha_n}^T\\ &\vdots\quad &\vdots\quad &\ddots\quad &\vdots\\ &\alpha_n\overline{\alpha_1}^T &\quad\alpha_n\overline{\alpha_2}^T\quad&\cdots\quad &\alpha_n\overline{\alpha_n}^T \end{matrix} \right)\\\\ &=\left( \begin{matrix} &(\alpha_1,\alpha_1)\quad &(\alpha_1,\alpha_2)\quad &\cdots\quad &(\alpha_1,\alpha_n)\\ &(\alpha_2,\alpha_1)\quad &(\alpha_2,\alpha_2)\quad &\cdots\quad &(\alpha_2,\alpha_n)\\ &\vdots\quad &\vdots\quad &\ddots\quad &\vdots\\ &(\alpha_n,\alpha_1)\quad &(\alpha_n,\alpha_2)\quad &\cdots\quad &(\alpha_n,\alpha_n) \end{matrix} \right) \end{aligned} AAH=⎝⎛α1α2⋮αn⎠⎞(α1Tα2T⋯αnT)=⎝⎛α1α1Tα2α1T⋮αnα1Tα1α2Tα2α2T⋮αnα2T⋯⋯⋱⋯α1αnTα2αnT⋮αnαnT⎠⎞=⎝⎛(α1,α1)(α2,α1)⋮(αn,α1)(α1,α2)(α2,α2)⋮(αn,α2)⋯⋯⋱⋯(α1,αn)(α2,αn)⋮(αn,αn)⎠⎞
1.8 正交
1.8.1 向量正交
X=(x1x2⋮xn),Y=(y1y2⋮yn)∈Cn\begin{aligned} X&=\left( \begin{matrix} x_1\\ x_2\\ \vdots\\ x_n \end{matrix} \right),Y=\left( \begin{matrix} y_1\\ y_2\\ \vdots\\ y_n \end{matrix} \right)\in C^{n} \end{aligned} X=⎝⎛x1x2⋮xn⎠⎞,Y=⎝⎛y1y2⋮yn⎠⎞∈Cn
X⊥Y⟺(X,Y)=0=x1y1‾+x2y2‾+⋯+xnyn‾=x1‾y1+x2‾y2+⋯+xn‾yn‾=(Y,X)\begin{aligned} X\bot Y\iff (X,Y)=0&=x_1\overline{y_1}+x_2\overline{y_2}+\cdots+x_n\overline{y_n}\\ &=\overline{\overline{x_1}y_1+\overline{x_2}y_2+\cdots+\overline{x_n}y_n}\\=(Y,X) \end{aligned} X⊥Y⟺(X,Y)=0=(Y,X)=x1y1+x2y2+⋯+xnyn=x1y1+x2y2+⋯+xnyn
正交性质
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X⊥Y⇒aX⊥bYX\bot Y\Rightarrow aX\bot bYX⊥Y⇒aX⊥bY
证:(aX,bY)=b‾YHaX=ab‾YHX=ab‾(X,Y)=0(aX,bY)=\overline{b}Y^HaX=a\overline{b}Y^HX=a\overline{b}(X,Y)=0(aX,bY)=bYHaX=abYHX=ab(X,Y)=0
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勾股定理:X1⊥X2⊥⋯±Xn⇒∣c1X1±c2X2±⋯±cnXn∣2=∣c1X1∣2+∣c2X2∣2+⋯+∣cnXn∣2X_1\bot X_2\bot \cdots \pm X_n\Rightarrow \vert c_1X_1\pm c_2X_2\pm \cdots \pm c_nX_n\vert^2=\vert c_1X_1\vert^2+\vert c_2X_2\vert^2+\cdots+\vert c_nX_n\vert^2X1⊥X2⊥⋯±Xn⇒∣c1X1±c2X2±⋯±cnXn∣2=∣c1X1∣2+∣c2X2∣2+⋯+∣cnXn∣2
此时,X1,X2,⋯,xnX_1,X_2,\cdots,x_nX1,X2,⋯,xn 称为一个正交组
1.8.2 单位向量
若 X≠0⃗X\neq \vec{0}X=0 ,X∣X∣\frac{X}{\vert X \vert}∣X∣X 是一个单位向量(∣X∣X∣∣=1\vert \frac{X}{\vert X \vert} \vert=1∣∣X∣X∣=1)
1.8.3 U阵(正交阵)
预:非单位列向量
半:p个n维列向量(p<n)
预-半U阵(预-半正交阵)
α1,α2,⋯,αp是n维列向量,且p≤n,且α1⊥α2⊥⋯⊥αp则称A=(α1,α2,⋯,αp)为预半U阵\alpha_1,\alpha_2,\cdots,\alpha_p 是n维列向量,且p\le n,且\alpha_1\bot\alpha_2\bot \cdots\bot\alpha_p\\ 则称A=(\alpha_1,\alpha_2,\cdots,\alpha_p) 为预半U阵 α1,α2,⋯,αp是n维列向量,且p≤n,且α1⊥α2⊥⋯⊥αp则称A=(α1,α2,⋯,αp)为预半U阵
判定
A=(α1,α2,⋯,αp)是预半U阵⟺AHA=((α1,α1)⋯0⋮⋱00⋯(αp,αp))是对角阵其中,α1,α2,⋯,αp是n维列向量\begin{aligned} A&=(\alpha_1,\alpha_2,\cdots,\alpha_p) 是预半U阵\\ &\iff A^HA=\left( \begin{matrix} &(\alpha_1,\alpha_1)&\cdots&0\\ &\vdots&\ddots&0\\ &0&\cdots&(\alpha_p,\alpha_p) \end{matrix} \right)是对角阵\\ &其中,\alpha_1,\alpha_2,\cdots,\alpha_p是n维列向量 \end{aligned} A=(α1,α2,⋯,αp)是预半U阵⟺AHA=⎝⎛(α1,α1)⋮0⋯⋱⋯00(αp,αp)⎠⎞是对角阵其中,α1,α2,⋯,αp是n维列向量
证明:
区分 : AHAA^HAAHA 是 p×pp \times pp×p 阶满秩方阵,而 AAHAA^HAAH 是 n×nn\times nn×n 不满秩方阵
半U阵(半正交阵)
A=(α1,α2,⋯,αp)是预半U阵,其中αi是n维列向量,若满足∣α1∣=∣α2∣=⋯=∣αp∣=1,则A为半U阵\begin{aligned} &A=(\alpha_1,\alpha_2,\cdots,\alpha_p)是预半U阵,其中\alpha_i是n维列向量,若满足\\ &\vert \alpha_1 \vert=\vert \alpha_2 \vert=\cdots=\vert \alpha_p \vert = 1,则A为半U阵 \end{aligned} A=(α1,α2,⋯,αp)是预半U阵,其中αi是n维列向量,若满足∣α1∣=∣α2∣=⋯=∣αp∣=1,则A为半U阵
判定
A=(α1,⋯,αp)是半U阵⟺α1⊥⋯⊥αp,且∣α1∣=⋯=∣αp∣=1⟺AHA=Ip\begin{aligned} A=(\alpha_1,\cdots,\alpha_p)是半U阵&\iff \alpha_1\bot\cdots\bot\alpha_p,且\vert \alpha_1 \vert=\cdots=\vert \alpha_p \vert=1\\ &\iff A^HA=I_{p} \end{aligned} A=(α1,⋯,αp)是半U阵⟺α1⊥⋯⊥αp,且∣α1∣=⋯=∣αp∣=1⟺AHA=Ip
性质
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保模长 A为半U阵,则 ∣Ax∣2=∣x∣2\vert Ax \vert^2=\vert x \vert^2∣Ax∣2=∣x∣2
∣Ax∣2=(Ax)H(Ax)=xHAHAx=∣X∣2\vert Ax \vert^2=(Ax)^H(Ax)=x^HA^HAx=\vert X\vert^2∣Ax∣2=(Ax)H(Ax)=xHAHAx=∣X∣2
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保正交 A为半U阵,x⊥yx\bot yx⊥y ,则Ax⊥AyAx\bot AyAx⊥Ay
预-U阵(预-单位正交阵)
α1,α2,⋯,αn是n维列向量,且α1⊥α2⊥⋯⊥αn,则A=(α1,α2,⋯,αn)是预U阵\begin{aligned} &\alpha_1,\alpha_2,\cdots,\alpha_n是n维列向量,且\alpha_1\bot\alpha_2\bot\cdots\bot\alpha_n ,\\ &则A=(\alpha_1,\alpha_2,\cdots,\alpha_n)是预U阵 \end{aligned} α1,α2,⋯,αn是n维列向量,且α1⊥α2⊥⋯⊥αn,则A=(α1,α2,⋯,αn)是预U阵
eg
X1=(1ii),X2=(2i11),X3=(01−1)(X1,X2)=0,(X2,X3)=0,(X1,X3)=0,∴X1⊥X2⊥X3,A=(X1,X2,X3)是预−U阵\begin{aligned} &X_1=\left( \begin{matrix} 1\\i\\i \end{matrix} \right),X_2=\left( \begin{matrix} 2i\\1\\1 \end{matrix} \right),X_3=\left( \begin{matrix} 0\\1\\-1 \end{matrix} \right)\\\\ &(X_1,X_2)=0,(X_2,X_3)=0,(X_1,X_3)=0,\\\\ &\therefore X_1\bot X_2\bot X_3,A=(X_1,X_2,X_3)是预-U阵 \end{aligned} X1=⎝⎛1ii⎠⎞,X2=⎝⎛2i11⎠⎞,X3=⎝⎛01−1⎠⎞(X1,X2)=0,(X2,X3)=0,(X1,X3)=0,∴X1⊥X2⊥X3,A=(X1,X2,X3)是预−U阵
判定
A=(α1,α2,⋯,αn)⟺AHA=((α1,α1)⋯0⋮⋱00⋯(αn,αn))是对角阵其中,α1,α2,⋯,αp是n维列向量\begin{aligned} A&=(\alpha_1,\alpha_2,\cdots,\alpha_n)\\ &\iff A^HA=\left( \begin{matrix} &(\alpha_1,\alpha_1)&\cdots&0\\ &\vdots&\ddots&0\\ &0&\cdots&(\alpha_n,\alpha_n) \end{matrix} \right)是对角阵\\ &其中,\alpha_1,\alpha_2,\cdots,\alpha_p是n维列向量 \end{aligned} A=(α1,α2,⋯,αn)⟺AHA=⎝⎛(α1,α1)⋮0⋯⋱⋯00(αn,αn)⎠⎞是对角阵其中,α1,α2,⋯,αp是n维列向量
等价判定
- A=An×nA=A_{n\times n}A=An×n 为U阵(AHA=IA^HA=IAHA=I),即A的列向量 α1,α2,⋯,αn\alpha_1,\alpha_2,\cdots,\alpha_nα1,α2,⋯,αn 互正交,且长度为1
- A−1=AHA^{-1}=A^HA−1=AH
- AHA=I,且AAH=IA^HA=I,且AA^H=IAHA=I,且AAH=I
U阵(A正交阵)
A是预U阵且∣α1∣=⋯=∣αn∣=1,则A是一个U阵(正交阵)\begin{aligned} A是预U阵且 \vert \alpha_1 \vert=\cdots=\vert \alpha_n \vert=1,则A是一个U阵(正交阵) \end{aligned} A是预U阵且∣α1∣=⋯=∣αn∣=1,则A是一个U阵(正交阵)
判定
A=(α1,⋯,αn)⟺AHA=(10⋯001⋯0⋮⋮⋱⋮00⋯1)=I是单位阵\begin{aligned} A&=(\alpha_1,\cdots,\alpha_n)\\ &\iff A^HA=\left( \begin{matrix} &1&0&\cdots&0\\ &0&1&\cdots&0\\ &\vdots&\vdots&\ddots&\vdots\\ &0&0&\cdots&1 \end{matrix} \right)=I是单位阵 \end{aligned} A=(α1,⋯,αn)⟺AHA=⎝⎛10⋮001⋮0⋯⋯⋱⋯00⋮1⎠⎞=I是单位阵
性质
1.A是U阵⟺AHA=I⟺A−1A=I⟺AAH=I⟺A=(α1,α2,⋯,αn),且α1⊥α2,⋯⊥αn,且∣α1∣=⋯=∣αn∣=12.∣Ax∣2=∣x∣2,A是U阵∵∣Ax∣2=(Ax)H(Ax)=xHAHAx=xHIx=(x,x)=∣x∣23.x⊥y⇒Ax⊥Ay,A是U阵∵(Ax,Ay)=(Ay)HAx=yHAHAx=(x,y)=0⟺Ax⊥Ay4.(Ax,Ay)=(x,y),A是U阵\begin{aligned} &1. A是U阵\iff A^HA=I\iff A^{-1}A=I\iff AA^H=I\\ &\iff A=(\alpha_1,\alpha_2,\cdots,\alpha_n) ,且\alpha_1\bot\alpha_2,\cdots\bot\alpha_n,且\vert \alpha_1\vert=\cdots=\vert\alpha_n\vert=1\\\\ &2.\vert Ax\vert^2=\vert x \vert^2,A是U阵\\ &\because \vert Ax\vert^2 = (Ax)^H(Ax)=x^HA^HAx=x^HIx=(x,x)=\vert x \vert^2\\\\ &3.x\bot y \Rightarrow Ax\bot Ay,A是U阵\\ &\because (Ax,Ay)=(Ay)^HAx=y^HA^HAx=(x,y)=0\iff Ax\bot Ay\\\\ &4.(Ax,Ay)=(x,y),A是U阵 \end{aligned} 1.A是U阵⟺AHA=I⟺A−1A=I⟺AAH=I⟺A=(α1,α2,⋯,αn),且α1⊥α2,⋯⊥αn,且∣α1∣=⋯=∣αn∣=12.∣Ax∣2=∣x∣2,A是U阵∵∣Ax∣2=(Ax)H(Ax)=xHAHAx=xHIx=(x,x)=∣x∣23.x⊥y⇒Ax⊥Ay,A是U阵∵(Ax,Ay)=(Ay)HAx=yHAHAx=(x,y)=0⟺Ax⊥Ay4.(Ax,Ay)=(x,y),A是U阵
预U阵与U阵
A=(α1,⋯,αn)是预U阵⇒A=(α1∣α1∣,α2∣α2∣,⋯,αn∣αn∣)是U阵\begin{aligned} A=(\alpha_1,\cdots,\alpha_n) 是预U阵\Rightarrow A=(\frac{\alpha_1}{\vert \alpha_1\vert},\frac{\alpha_2}{\vert \alpha_2\vert},\cdots,\frac{\alpha_n}{\vert \alpha_n \vert})是U阵 \end{aligned} A=(α1,⋯,αn)是预U阵⇒A=(∣α1∣α1,∣α2∣α2,⋯,∣αn∣αn)是U阵
U阵构造新U阵
若A=(α1,α2,⋯,αn)为U阵,则1.k=±1,kA=(kα1,kα2,⋯,kαn)为U阵2.B=(β1,β2,⋯,βn)为U阵,其中β组为α组的重排3.(封闭性)若A、B为同阶U阵,则AB也为U阵\begin{aligned} 若A&=(\alpha_1,\alpha_2,\cdots,\alpha_n) 为U阵,则\\ &1.k=\pm1,kA=(k\alpha_1,k\alpha_2,\cdots,k\alpha_n)为U阵\\ &2.B=(\beta_1,\beta_2,\cdots,\beta_n) 为U阵,其中\beta组为\alpha组的重排\\ &3.(封闭性)若A、B为同阶U阵,则AB也为U阵 \end{aligned} 若A=(α1,α2,⋯,αn)为U阵,则1.k=±1,kA=(kα1,kα2,⋯,kαn)为U阵2.B=(β1,β2,⋯,βn)为U阵,其中β组为α组的重排3.(封闭性)若A、B为同阶U阵,则AB也为U阵
向量构造U阵
若α=(a1a2⋮an)∈C,A=In−2ααH∣α∣2是一个U阵1.AH=A且A2=I(A−1=A)2.A为U阵(AHA=I)证明:1.A2=(In−2ααH∣α∣2)(In−2ααH∣α∣2)=In2−4ααH∣α∣2+4(ααH)(ααH)∣α∣4=In−4ααH∣α∣2+4α(αHα)αH∣α∣4=In−4ααH∣α∣2+4α(∣α∣2)αH∣α∣4=In−4ααH∣α∣2+4ααH∣α∣2=In2.AHA=(In−2ααH∣α∣2)H(In−2ααH∣α∣2)=(In−2ααH∣α∣2)(In−2ααH∣α∣2)=A2=I∴A是U阵\begin{aligned} 若\alpha&=\left( \begin{matrix} a_1\\a_2\\\vdots \\ a_n \end{matrix} \right)\in C,A=I_n-\frac{2\alpha\alpha^H}{\vert \alpha \vert^2}是一个U阵\\\\ &1.A^H=A且A^2=I(A^{-1}=A)\\ &2.A为U阵(A^HA=I)\\\\ 证明: 1.A^2&=(I_n-\frac{2\alpha\alpha^H}{\vert \alpha \vert^2})(I_n-\frac{2\alpha\alpha^H}{\vert \alpha \vert^2})\\ &=I_n^2-\frac{4\alpha\alpha^H}{\vert \alpha \vert^2}+\frac{4(\alpha\alpha^H)(\alpha\alpha^H)}{\vert \alpha\vert^4}=I_n-\frac{4\alpha\alpha^H}{\vert \alpha \vert^2}+\frac{4\alpha(\alpha^H\alpha)\alpha^H}{\vert \alpha\vert^4} \\ &=I_n-\frac{4\alpha\alpha^H}{\vert \alpha \vert^2}+\frac{4\alpha(\vert \alpha\vert^2)\alpha^H}{\vert \alpha\vert^4}=I_n-\frac{4\alpha\alpha^H}{\vert \alpha \vert^2}+\frac{4\alpha\alpha^H}{\vert \alpha \vert^2}=I_n\\ 2.A^HA&=(I_n-\frac{2\alpha\alpha^H}{\vert \alpha \vert^2})^H(I_n-\frac{2\alpha\alpha^H}{\vert \alpha \vert^2})\\ &=(I_n-\frac{2\alpha\alpha^H}{\vert \alpha \vert^2})(I_n-\frac{2\alpha\alpha^H}{\vert \alpha \vert^2})=A^2=I\\ &\therefore A是U阵 \end{aligned} 若α证明:1.A22.AHA=⎝⎛a1a2⋮an⎠⎞∈C,A=In−∣α∣22ααH是一个U阵1.AH=A且A2=I(A−1=A)2.A为U阵(AHA=I)=(In−∣α∣22ααH)(In−∣α∣22ααH)=In2−∣α∣24ααH+∣α∣44(ααH)(ααH)=In−∣α∣24ααH+∣α∣44α(αHα)αH=In−∣α∣24ααH+∣α∣44α(∣α∣2)αH=In−∣α∣24ααH+∣α∣24ααH=In=(In−∣α∣22ααH)H(In−∣α∣22ααH)=(In−∣α∣22ααH)(In−∣α∣22ααH)=A2=I∴A是U阵
eg :
α=(111),其U阵为I3−2ααH∣α∣2=(100010001)−23(111111111)=(13−23−23−2313−23−23−2313)\begin{aligned} \alpha=\left( \begin{matrix} 1\\1\\1 \end{matrix} \right),其U阵为 I_3-\frac{2\alpha\alpha^H}{\vert \alpha\vert^2}&=\left( \begin{matrix} &1&0&0\\&0&1&0\\&0&0&1\\ \end{matrix} \right)-\frac{2}{3}\left( \begin{matrix} &1&1&1\\&1&1&1\\&1&1&1\\ \end{matrix} \right)\\ &=\left( \begin{matrix} &\frac{1}{3}&-\frac{2}{3}&-\frac{2}{3}\\ &-\frac{2}{3}&\frac{1}{3}&-\frac{2}{3}\\ &-\frac{2}{3}&-\frac{2}{3}&\frac{1}{3}\\ \end{matrix} \right) \end{aligned} α=⎝⎛111⎠⎞,其U阵为I3−∣α∣22ααH=⎝⎛100010001⎠⎞−32⎝⎛111111111⎠⎞=⎝⎛31−32−32−3231−32−32−3231⎠⎞