矩阵论
1. 准备知识——复数域上的矩阵与换位公式)
1. 准备知识——复数域上的内积域正交阵
1. 准备知识——相似对角化与合同&正定阵
2. 矩阵分解—— SVD准备知识——奇异值
2. 矩阵分解——SVD
2. 矩阵分解——QR分解
2. 矩阵分解——乔利斯分解&平方根公式
2. 矩阵分解——正规谱分解——正规阵
2. 矩阵分解——正规分解
2. 矩阵分解——单阵及特征值特征向量一些求法
矩阵论准备知识,很多内容都是线性代数的扩展
设 A、B为n阶方阵,如果存在可逆阵P,使得 P − 1 A P = B P^{-1}AP=B P−1AP=B ,则称A与B相似,记为 A ∼ B A\sim B A∼B
所以,方阵之间的相似关系是一种等价关系
若A与B相似,则有
∣ x I − A ∣ = ∣ x I − B ∣ \begin{aligned} |xI-A|=|xI-B| \end{aligned} ∣xI−A∣=∣xI−B∣
即A与B的特征根公式相同,其中A与B都是n阶方阵
可设 P − 1 A P = B P^{-1}AP = B P−1AP=B ,则有
∣ x I − B ∣ = ∣ x I − P − 1 A P ∣ = ∣ P − 1 ( x I − A ) P ∣ = 行列式计算 ∣ x I − A ∣ \begin{aligned} \mid xI-B\mid &=\mid xI-P^{-1}AP\mid=\mid P^{-1}(xI-A)P\mid \\ &\overset{\text{行列式计算}}{=}\mid xI-A\mid \end{aligned} ∣xI−B∣=∣xI−P−1AP∣=∣P−1(xI−A)P∣=行列式计算∣xI−A∣
由相似,可将A与B矩阵表示为 A ∼ B A\sim B A∼B 或者 A P = P B AP=PB AP=PB ,其中P为可逆矩阵
e g : A = ( 1 1 1 1 ) , λ ( A ) = { 2 , 0 } A = ( 2 1 0 3 ) , λ ( A ) = { 2 , 3 } A = ( 2 1 0 2 ) , λ ( A ) = { 2 , 2 } \begin{aligned} &eg:\\ &A=\left ( \begin{matrix} 1\quad 1\\ 1\quad 1 \end{matrix} \right),\lambda(A)=\{2,0\}\\\\ &A=\left ( \begin{matrix} 2\quad 1\\ 0\quad 3 \end{matrix} \right),\lambda(A)=\{2,3\}\\\\ &A=\left ( \begin{matrix} 2\quad 1\\ 0\quad 2 \end{matrix} \right),\lambda(A)=\{2,2\} \end{aligned} eg:A=(1111),λ(A)={2,0}A=(2103),λ(A)={2,3}A=(2102),λ(A)={2,2}
总结: 相似 ⇔ 特征多项式相同 ⇒ 特征值相等 相似 \Leftrightarrow 特征多项式相同\Rightarrow 特征值相等 相似⇔特征多项式相同⇒特征值相等
特征值相等 ⇒ 实对称矩阵 相似 特征值相等\overset{实对称矩阵}{\Rightarrow} 相似 特征值相等⇒实对称矩阵相似
特征值是相似变化下的不变量
设问:若 A = A n × p , B = B p × n ,且 p ≤ n A=A_{n\times p},B=B_{p\times n},且p\le n A=An×p,B=Bp×n,且p≤n ,则 ⇒ ( A ∙ B ) \Rightarrow (A\bullet B) ⇒(A∙B) 为 n 阶方阵, ( B ∙ A ) (B\bullet A) (B∙A) 为 p 阶方阵,求其特征值
∣ λ I n − A B ∣ = λ n − p ∣ λ I p − B A ∣ \mid \lambda I_n-AB \mid= \lambda^{n-p}\mid \lambda I_p-BA \mid ∣λIn−AB∣=λn−p∣λIp−BA∣
可见AB与BA两个方阵特征值基本相等
若 A = A n × p , B = B p × n ,且 p ≤ n A=A_{n\times p},B=B_{p\times n},且p\le n A=An×p,B=Bp×n,且p≤n ,则 ⇒ ( A ∙ B ) \Rightarrow (A\bullet B) ⇒(A∙B) 为 n 阶方阵, ( B ∙ A ) (B\bullet A) (B∙A) 为 p 阶方阵
若BA的特征根 λ ( B A ) = { λ 1 , λ 2 , . . . , λ p } \lambda(BA)=\{\lambda_1,\lambda_2,...,\lambda_p\} λ(BA)={λ1,λ2,...,λp} ,则 AB 的特征根 λ ( A B ) = { λ 1 , λ 2 . . . , λ p , 0 , . . . , 0 } ( 含 n − p 个零根 ) \lambda(AB)=\{\lambda_1,\lambda_2...,\lambda_p,0,...,0\}(含n-p个零根) λ(AB)={λ1,λ2...,λp,0,...,0}(含n−p个零根) ,可见 A B AB AB 与 B A BA BA 只差 n − p n-p n−p 个零根,其余根相同
即AB与BA必有相同非零根
由于 AB与BA 只相差 n-p 个零根,所以 t r ( A B ) = t r ( B A ) tr(AB) = tr(BA) tr(AB)=tr(BA)
证: t r ( A B ) = λ 1 + λ 2 + ⋯ + λ p + 0 + ⋯ + 0 = t r ( B A ) tr(AB)=\lambda_1+\lambda_2+\cdots+\lambda_p+0+\cdots+0=tr(BA) tr(AB)=λ1+λ2+⋯+λp+0+⋯+0=tr(BA)
∣ I n ± A B ∣ = ∣ I p ± B A ∣ \mid I_n \pm AB\mid = \mid I_p\pm BA\mid ∣In±AB∣=∣Ip±BA∣ ,当 λ = 1 , A 取 − A \lambda = 1,A取-A λ=1,A取−A 时,分别可证
若 n>p,则 ∣ A B ∣ = 0 \mid AB \mid=0 ∣AB∣=0
证: A B 为 n 阶矩阵,由于 r ( A B ) ≤ r ( A ) ≤ p < n ( 矩阵的秩越乘越小 ) 故 ∣ A B ∣ = 0 \begin{aligned} &证:AB为n阶矩阵,由于r(AB)\le r(A)\le p
或者考虑 AB 为 n 阶方阵,而只有p个非零特征值,p < n,故必有零特征值, ∣ A B ∣ = ∏ λ i = 0 \mid AB \mid = \prod\lambda_i = 0 ∣AB∣=∏λi=0
例 3 : P = ( I A 0 I ) , 求证 P − 1 = ( I − A 0 I ) \begin{aligned} &例3: P = \left( \begin{matrix} I\quad A \\ 0\quad I \end{matrix} \right),求证P^{-1}=\left( \begin{matrix} I\quad -A\\ 0\quad I \end{matrix} \right) \end{aligned} 例3:P=(IA0I),求证P−1=(I−A0I)
其实也就是二阶矩阵求逆
A = ( B C D E ) , 则 A − 1 = ( E − C − D B ) \begin{aligned} A=\left( \begin{matrix} B\quad C\\ D\quad E \end{matrix} \right),则A^{-1}=\left( \begin{matrix} E\quad -C\\ -D\quad B \end{matrix} \right) \end{aligned} A=(BCDE),则A−1=(E−C−DB)
A = ( a 1 b 1 a 1 b 2 ⋯ a 1 b n a 2 b 1 a 2 b 2 ⋯ a 2 b n ⋮ ⋮ ⋱ ⋮ a n b 1 a n b 2 ⋯ a n b n ) n × n = ( a 1 a 2 ⋮ a n ) ( b 1 b 2 ⋯ b n ) = Δ α β T , 其中 α = ( a 1 a 2 ⋮ a n ) , β = ( b 1 b 2 ⋮ b n ) \begin{aligned} A&=\left( \begin{matrix} a_1b_1\quad &a_1b_2\quad &\cdots\quad &a_1b_n\\ a_2b_1\quad &a_2b_2\quad &\cdots\quad &a_2b_n\\ \vdots\quad &\vdots\quad &\ddots\quad &\vdots\\ a_nb_1\quad &a_nb_2\quad &\cdots\quad &a_nb_n \end{matrix} \right)_{n\times n}\\\\ &=\left( \begin{matrix} a_1\\a_2\\\vdots \\a_n \end{matrix} \right)\left( b_1\quad b_2\quad \cdots \quad b_n \right)\\\\ &\overset{\Delta}{=}\alpha \beta^{T},其中 \alpha=\left( \begin{matrix} a_1\\a_2\\\vdots \\a_n \end{matrix} \right),\beta=\left( \begin{matrix} b_1\\b_2\\\vdots \\b_n \end{matrix} \right) \end{aligned} A=⎝ ⎛a1b1a2b1⋮anb1a1b2a2b2⋮anb2⋯⋯⋱⋯a1bna2bn⋮anbn⎠ ⎞n×n=⎝ ⎛a1a2⋮an⎠ ⎞(b1b2⋯bn)=ΔαβT,其中α=⎝ ⎛a1a2⋮an⎠ ⎞,β=⎝ ⎛b1b2⋮bn⎠ ⎞
∣ λ E − A ∣ = ∣ λ E − α n × 1 β 1 × n T ∣ = λ n − 1 ∣ λ I − β 1 × n T α n × 1 I ∣ = λ n − 1 ( λ I − t r ( A ) ) , 其中 t r ( A ) = ∑ i = 1 n a i b i \begin{aligned} &\mid \lambda E-A\mid = \mid \lambda E-\alpha_{n\times 1}\beta_{1\times n}^T\mid=\lambda^{n-1}\mid\lambda I-\beta_{1\times n}^T\alpha_{n\times 1}I\mid\\ &=\lambda^{n-1}(\lambda I-tr(A)),其中 tr(A)=\sum\limits_{i=1}\limits^{n}a_{i}b_i \end{aligned} ∣λE−A∣=∣λE−αn×1β1×nT∣=λn−1∣λI−β1×nTαn×1I∣=λn−1(λI−tr(A)),其中tr(A)=i=1∑naibi
若n阶方针,秩为1,r(A)=1,则全体特征值为 λ ( A ) = { t r ( A ) , 0 , . . . , 0 } \lambda(A)=\{tr(A),0,...,0\} λ(A)={tr(A),0,...,0} ,其中 t r ( A ) = a 1 b 1 + a 2 b 2 + . . . + a n b n = β T α tr(A)=a_1b_1+a_2b_2+...+a_nb_n=\beta^T\alpha tr(A)=a1b1+a2b2+...+anbn=βTα
由换位公式可知, α n × 1 β 1 × n T \alpha_{n\times 1}\beta_{1\times n}^T αn×1β1×nT 与 β 1 × n T α n × 1 \beta_{1\times n}^T\alpha_{n\times 1} β1×nTαn×1 相差 n-1 个零根,即有一个相等的非零特征根,而 β 1 × n T α n × 1 \beta_{1\times n}^T\alpha_{n\times 1} β1×nTαn×1 为1阶矩阵,所以 λ 1 = β 1 × n T α n × 1 = a 1 b 1 + a 2 b 2 + . . . + a n b n = t r ( A ) \lambda_1=\beta_{1\times n}^T\alpha_{n\times 1}=a_1b_1+a_2b_2+...+a_nb_n=tr(A) λ1=β1×nTαn×1=a1b1+a2b2+...+anbn=tr(A)
A = α β T A=\alpha \beta^T A=αβT 的每个列向量都是 λ 1 = t r ( A ) \lambda_1=tr(A) λ1=tr(A) 的特征向量
证明:
A α = ( α β ) α = λ 1 α \begin{aligned} A\alpha = (\alpha \beta)\alpha=\lambda_1 \alpha \end{aligned} Aα=(αβ)α=λ1α
A + c I A+cI A+cI 称为A的平移矩阵
若 λ ( A ) = { λ 1 + c , λ 2 + c , . . . , λ n + c } \lambda(A)=\{\lambda_1+c,\lambda_2+c,...,\lambda_n+c\} λ(A)={λ1+c,λ2+c,...,λn+c}
A + c I A+cI A+cI 与 A A A 有 相同的特征向量
证明:
令 A 的 n 个特征向量 x 1 , x 2 , . . . , x n ,有 A x 1 = λ 1 x 1 , A x 1 = λ 2 x 2 , . . . A x n = λ 1 x n ⇔ { ( A + c I ) x 1 = λ 1 x 1 + c x 1 = ( λ 1 + c ) x 1 ( A + c I ) x 2 = λ 2 x 2 + c x 2 = ( λ 2 + c ) x 2 ⋯ ( A + c I ) x n = λ n x n + c x n = ( λ n + c ) x n 故 λ ( A ) = { λ 1 , λ 2 , . . . , λ n } \begin{aligned} &令A的n个特征向量x_1,x_2,...,x_n,有\\\\ &Ax_1=\lambda_1 x_1,Ax_1=\lambda_2 x_2,...Ax_n=\lambda_1 x_n\\\\ \Leftrightarrow & \left\{ \begin{aligned} (A+cI)x_1 = \lambda_1x_1+cx_1=(\lambda_1+c)x_1\\ (A+cI)x_2 = \lambda_2x_2+cx_2=(\lambda_2+c)x_2\\ \cdots\\ (A+cI)x_n = \lambda_nx_n+cx_n=(\lambda_n+c)x_n \end{aligned} \right.\\\\ 故&\lambda(A)=\{\lambda_1,\lambda_2,...,\lambda_n\} \end{aligned} ⇔故令A的n个特征向量x1,x2,...,xn,有Ax1=λ1x1,Ax1=λ2x2,...Axn=λ1xn⎩ ⎨ ⎧(A+cI)x1=λ1x1+cx1=(λ1+c)x1(A+cI)x2=λ2x2+cx2=(λ2+c)x2⋯(A+cI)xn=λnxn+cxn=(λn+c)xnλ(A)={λ1,λ2,...,λn}
( 1 ) A − E = ( 0 1 0 0 1 0 0 1 0 ) 由秩 1 公式 , λ ( A − E ) = { t r ( A ) , 0 , 0 } , 由平移公式 λ ( A ) = { t r ( A − E ) + 1 , 1 , 1 } = { 2 , 1 , 1 } 且 ( A − E ) 与 A 的特征向量相等, A − E 的列向量 ( 1 1 1 ) ( 2 ) A − E = ( 3 6 0 − 3 − 6 0 − 3 − 6 0 ) , λ ( A − E ) = { − 3 , 0 , 0 } , 故 λ ( A ) = { − 2 , 1 , 1 } , A 的特征向量为 ( 3 − 3 6 ) \begin{aligned} &(1)A-E=\left( \begin{matrix} 0\quad 1\quad 0\\ 0\quad 1\quad 0\\ 0\quad 1\quad 0 \end{matrix} \right)\\ &由秩1公式,\lambda(A-E)=\{tr(A),0,0\},\\ &由平移公式 \lambda(A)=\{tr(A-E)+1,1,1\}=\{2,1,1\}\\ &且(A-E)与A的特征向量相等,A-E的列向量\left( \begin{matrix} 1\\1\\1 \end{matrix} \right)\\ &(2)A-E=\left( \begin{matrix} 3& 6& 0\\ -3& -6& 0\\ -3& -6& 0\\ \end{matrix} \right),\lambda(A-E)=\{-3,0,0\}, \\&故\lambda(A)=\{-2,1,1\},A的特征向量为\left( \begin{matrix} 3\\-3\\6 \end{matrix} \right) \end{aligned} (1)A−E=⎝ ⎛010010010⎠ ⎞由秩1公式,λ(A−E)={tr(A),0,0},由平移公式λ(A)={tr(A−E)+1,1,1}={2,1,1}且(A−E)与A的特征向量相等,A−E的列向量⎝ ⎛111⎠ ⎞(2)A−E=⎝ ⎛3−3−36−6−6000⎠ ⎞,λ(A−E)={−3,0,0},故λ(A)={−2,1,1},A的特征向量为⎝ ⎛3−36⎠ ⎞
若 λ ( A ) = { λ 1 , . . . , λ n } \lambda(A)=\{\lambda_1,...,\lambda_n\} λ(A)={λ1,...,λn} ,则 λ ( k A ) = { k λ 1 , k λ 2 , . . . , k λ n , } ( k ≠ 0 ) \lambda(kA)=\{k\lambda_1,k\lambda_2,...,k\lambda_n,\}(k\neq 0) λ(kA)={kλ1,kλ2,...,kλn,}(k=0)
k λ ( A ) k\lambda(A) kλ(A) 与 λ ( k A ) \lambda(kA) λ(kA) 有相同的特征向量
A = ( − 1 − 2 6 − 1 0 3 − 1 − 1 4 ) , A − I = ( − 2 − 2 6 − 1 − 1 3 − 1 − 1 3 ) , λ ( A − I ) = { 0 , 0 , 0 } , 则 λ ( A ) = λ ( A − I ) + 1 = { 1 , 1 , 1 } \begin{aligned} &A=\left( \begin{matrix} -1\quad -2\quad 6\\ -1\quad 0\quad 3\\ -1\quad -1\quad 4 \end{matrix} \right),A-I=\left( \begin{matrix} -2\quad -2\quad 6\\ -1\quad -1\quad 3\\ -1\quad -1\quad 3 \end{matrix} \right),\lambda(A-I)=\{0,0,0\}\\ &,则\lambda(A)=\lambda(A-I)+1=\{1,1,1\} \end{aligned} A=⎝ ⎛−1−26−103−1−14⎠ ⎞,A−I=⎝ ⎛−2−26−1−13−1−13⎠ ⎞,λ(A−I)={0,0,0},则λ(A)=λ(A−I)+1={1,1,1}
A = ( 7 4 − 1 4 7 − 1 − 4 − 4 4 ) , A − 3 I = ( 4 4 − 1 4 4 − 1 − 4 − 4 1 ) , λ ( A − 3 I ) = { 9 , 0 , 0 } λ ( A ) = λ ( A − 3 I ) + 3 = { 12 , 3 , 3 } \begin{aligned} &A=\left( \begin{matrix} 7\quad 4\quad -1\\ 4\quad 7\quad -1\\ -4\quad -4\quad 4 \end{matrix} \right),A-3I=\left( \begin{matrix} 4\quad 4\quad -1\\ 4\quad 4\quad -1\\ -4\quad -4\quad 1 \end{matrix} \right),\lambda(A-3I)=\{9,0,0\}\\ \\&\lambda(A)=\lambda(A-3I)+3=\{12,3,3\} \end{aligned} A=⎝ ⎛74−147−1−4−44⎠ ⎞,A−3I=⎝ ⎛44−144−1−4−41⎠ ⎞,λ(A−3I)={9,0,0}λ(A)=λ(A−3I)+3={12,3,3}
C : { z = a + b i ∣ a , b ∈ R } C:\{z=a+bi\mid a,b\in R\} C:{z=a+bi∣a,b∈R} ,其中 i 2 = − 1 , − 1 = i i^2=-1,\sqrt{-1}=i i2=−1,−1=i
若 z = a + b i , 则 z ‾ = a + b i ‾ = a − b i 若z=a+bi,则\overline{z}=\overline{a+bi}=a-bi 若z=a+bi,则z=a+bi=a−bi
n 维实列向量 X = ( x 1 x 2 ⋮ x n ) , x i ∈ R , n 维实复列向量 X = ( x 1 x 2 ⋮ x n ) , x i ∈ C 列向量可表示为转置形式 X = ( x 1 , x 2 , . . . , x n ) T m × n 实矩阵 R m × n = { A = ( a i j ) ∣ a i , j ∈ R , 1 ≤ i ≤ m , 1 ≤ j ≤ n } m × n 复矩阵 C m × n = { A = ( a i j ) ∣ a i j ∈ C , 1 ≤ i ≤ m , 1 ≤ j ≤ n } 且 R m × n ∈ C m × n \begin{aligned} &n维实列向量X=\left( \begin{matrix} x_1\\ x_2 \\ \vdots \\x_n \end{matrix} \right),x_i\in R, n维实复列向量X=\left( \begin{matrix} x_1\\ x_2 \\ \vdots \\x_n \end{matrix} \right),x_i\in C\\\\ &列向量可表示为转置形式 X=\left(x_1,x_2,...,x_n\right)^T\\\\ &m\times n 实矩阵R_{m\times n} = \{A=(a_{ij})\mid a_{i,j}\in R,1\le i\le m,1\le j\le n\}\\ &m\times n复矩阵C^{m\times n}=\{A=(a_{ij})\mid a_{ij}\in C,1\le i\le m,1\le j\le n\}\\ &且 R_{m\times n}\in C^{m\times n} \end{aligned} n维实列向量X=⎝ ⎛x1x2⋮xn⎠ ⎞,xi∈R,n维实复列向量X=⎝ ⎛x1x2⋮xn⎠ ⎞,xi∈C列向量可表示为转置形式X=(x1,x2,...,xn)Tm×n实矩阵Rm×n={A=(aij)∣ai,j∈R,1≤i≤m,1≤j≤n}m×n复矩阵Cm×n={A=(aij)∣aij∈C,1≤i≤m,1≤j≤n}且Rm×n∈Cm×n
A m × n = ( a 11 a 12 ⋯ a 1 n a 21 a 22 ⋯ a 2 n ⋮ ⋮ ⋱ ⋮ a m 1 a m 2 ⋯ a m n ) ∈ C m × n 可表示为 A = ( α 1 , α 2 , ⋯ , α n ) , 且 α i ∈ C m \begin{aligned} &A_{m\times n}=\left( \begin{matrix} a_{11}\quad &a_{12}&\cdots\quad &a_{1n}\\ a_{21}\quad &a_{22}&\cdots\quad &a_{2n}\\ \vdots\quad &\vdots &\ddots\quad &\vdots\\ a_{m1}\quad &a_{m2}&\cdots \quad &a_{mn} \end{matrix} \right)\in C^{m\times n}\\\\ &可表示为A=(\alpha_1,\alpha_2,\cdots,\alpha_n),且\alpha_i\in C^{m} \end{aligned} Am×n=⎝ ⎛a11a21⋮am1a12a22⋮am2⋯⋯⋱⋯a1na2n⋮amn⎠ ⎞∈Cm×n可表示为A=(α1,α2,⋯,αn),且αi∈Cm
由 z z ‾ = ( a + b i ) ( a − b i ) = a 2 + b 2 z\overline{z}=(a+bi)(a-bi)=a^2+b^2 zz=(a+bi)(a−bi)=a2+b2
模公式 z z ‾ = z ‾ z = ∣ z ∣ 2 = a 2 + b 2 z\overline z=\overline zz=\mid z\mid^2=a^2+b^2 zz=zz=∣z∣2=a2+b2
A = ( a i j ) = ( a 11 ⋯ a 1 n ⋮ ⋱ ⋮ a n 1 ⋯ a n n ) , A 的共轭矩阵 A ‾ = ( a i j ‾ ) = ( a 11 ‾ ⋯ a 1 n ‾ ⋮ ⋱ ⋮ a n 1 ‾ ⋯ a n n ‾ ) \begin{aligned} &A=(a_{ij})=\left( \begin{matrix} a_{11}&\cdots &a_{1n}\\ \vdots &\ddots &\vdots\\ a_{n1}&\cdots &a_{nn}\\ \end{matrix} \right),\\\\&A的共轭矩阵 \overline{A} = (\overline{a_{ij}})=\left( \begin{matrix} \overline{a_{11}}&\cdots &\overline{a_{1n}}\\ \vdots &\ddots &\vdots\\ \overline{a_{n1}}&\cdots &\overline{a_{nn}}\\ \end{matrix} \right) \end{aligned} A=(aij)=⎝ ⎛a11⋮an1⋯⋱⋯a1n⋮ann⎠ ⎞,A的共轭矩阵A=(aij)=⎝ ⎛a11⋮an1⋯⋱⋯a1n⋮ann⎠ ⎞
乘后共轭=共轭后乘 : A ∙ B ‾ = A ‾ ∙ B ‾ \overline{A\bullet B}=\overline{A}\bullet \overline{B} A∙B=A∙B
共轭转置记为 Hemite 变换,即 A H = A ‾ T = A T ‾ A^H = \overline{A}^T=\overline{A^T} AH=AT=AT ,
A = ( a i j ) = ( a 11 ⋯ a 1 n ⋯ ⋱ ⋯ a m 1 ⋯ a m n ) ∈ C , A ‾ = ( a 11 ‾ ⋯ a 1 n ‾ ⋮ ⋱ ⋮ a m 1 ‾ ⋯ a m n ‾ ) ∈ C X = ( x 1 x 2 ⋮ x n ) ∈ C n , 则 X H = ( x 1 ‾ , x 2 ‾ , . . . , x n ‾ ) \begin{aligned} &A=(a_{ij})=\left( \begin{matrix} a_{11}&\cdots &a_{1n}\\ \cdots &\ddots &\cdots\\ a_{m1}&\cdots &a_{mn} \end{matrix} \right)\in C,\\ &\overline{A}=\left( \begin{matrix} &\overline{a_{11}}&\cdots &\overline{a_{1n}}\\ &\vdots &\ddots &\vdots\\ &\overline{a_{m1}}&\cdots &\overline{a_{mn}} \end{matrix} \right)\in C\\ &X=\left( \begin{matrix} x_1\\ x_2\\ \vdots\\ x_n \end{matrix} \right)\in C_n,则X^H=(\overline{x_1},\overline{x_2},...,\overline{x_n}) \end{aligned} A=(aij)=⎝ ⎛a11⋯am1⋯⋱⋯a1n⋯amn⎠ ⎞∈C,A=⎝ ⎛a11⋮am1⋯⋱⋯a1n⋮amn⎠ ⎞∈CX=⎝ ⎛x1x2⋮xn⎠ ⎞∈Cn,则XH=(x1,x2,...,xn)
共轭不会使矩阵变型,转置使矩阵变型 A ∈ C n × p ⇒ A H ∈ C p × n A\in C^{n\times p}\Rightarrow A^H\in C^{p\times n} A∈Cn×p⇒AH∈Cp×n
eg:
A = ( 1 i 1 i 1 i ) ∈ C 3 × 2 , 则 A H = ( 1 − i 1 − i 1 − i ) T = ( 1 1 1 − i − i − i ) ∈ C 2 × 3 \begin{aligned} A=\left( \begin{matrix} 1\quad i\\ 1\quad i\\ 1\quad i \end{matrix} \right)\in C^{3\times2},则A^H=\left( \begin{matrix} 1\quad -i\\ 1\quad -i\\ 1\quad -i \end{matrix} \right)^T =\left( \begin{matrix} 1\quad 1\quad 1\\ -i\quad -i\quad -i \end{matrix} \right) \in C^{2\times 3} \end{aligned} A=⎝ ⎛1i1i1i⎠ ⎞∈C3×2,则AH=⎝ ⎛1−i1−i1−i⎠ ⎞T=(111−i−i−i)∈C2×3
Hermite变换 | 转置 |
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( A H ) H = A (A^H)^H=A (AH)H=A | ( A T ) T = A (A^T)^T=A (AT)T=A |
( k A ) H = k ‾ A H (kA)^H=\overline{k}A^H (kA)H=kAH | ( k A ) T = k A T (kA)^T=kA^T (kA)T=kAT |
( A + B ) H = A H + B H (A+B)^H=A^H+B^H (A+B)H=AH+BH | ( A + B ) T = A T + B T (A+B)^T=A^T+B^T (A+B)T=AT+BT |
( A B ) H = B H A H , ( A B C ) H = C H B H A H (AB)^H=B^HA^H,(ABC)^H=C^HB^HA^H (AB)H=BHAH,(ABC)H=CHBHAH | ( A B ) T = B T A T (AB)^T=B^TA^T (AB)T=BTAT |
实数阵的 Hermite 变换仍是其本身
定理: a ∈ C 是实数 ⟺ a ‾ = a ⟺ a H = a a\in C 是实数 \iff \overline{a}=a\iff a^H=a a∈C是实数⟺a=a⟺aH=a
Hermite变换 | 转置 |
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A H = A A^H=A AH=A Hermite矩阵 | A T = A A^T=A AT=A ,对称阵 |
A H = − A A^H=-A AH=−A 斜Hermite矩阵 | A T = − A A^T=-A AT=−A ,反对称阵 |
Hermite矩阵性质:如果A是Hermite矩阵,则其对角线上元素全是实数
A = ( a 11 ∗ a 22 ⋱ ∗ a n n ) , 而 A H = ( a 11 ‾ ∗ a 22 ‾ ⋱ ∗ a n n ‾ ) , H e r m i t e 矩阵 A H = A a 11 = a 11 ‾ , a 22 = a 22 ‾ , . . . , a n n = a n n ‾ , 可见 H e r m i t e 矩阵对角线元素是实数 \begin{aligned} &A=\left( \begin{matrix} &a_{11}&\quad&\quad &*\\ &\quad&a_{22}&\quad&\quad \\ &\quad &\quad&\ddots&\quad\\ &*&\quad&\quad&a_{nn} \end{matrix} \right),\\ &而A^H=\left( \begin{matrix} &\overline{a_{11}}&\quad&\quad &*\\ &\quad&\overline{a_{22}}&\quad&\quad \\ &\quad &\quad&\ddots&\quad\\ &*&\quad&\quad&\overline{a_{nn}} \end{matrix} \right),Hermite矩阵A^H=A\\ &a_{11}=\overline{a_{11}},a_{22}=\overline{a_{22}},...,a_{nn}=\overline{a_{nn}},可见Hermite矩阵对角线元素是实数 \end{aligned} A=⎝ ⎛a11∗a22⋱∗ann⎠ ⎞,而AH=⎝ ⎛a11∗a22⋱∗ann⎠ ⎞,Hermite矩阵AH=Aa11=a11,a22=a22,...,ann=ann,可见Hermite矩阵对角线元素是实数
X = ( x 1 x 2 ⋮ x n ) ∈ C , 则其模长 ∣ X ∣ = ∣ x 1 ∣ 2 + ∣ x 2 ∣ 2 + ⋯ ∣ x n ∣ 2 \begin{aligned} &X=\left( \begin{matrix} x_1\\x_2\\\vdots\\x_n \end{matrix} \right)\in C,则其模长 \mid X \mid = \sqrt{ \mid x_1 \mid^2+ \mid x_2 \mid^2+ \cdots \mid x_n \mid^2} \end{aligned} X=⎝ ⎛x1x2⋮xn⎠ ⎞∈C,则其模长∣X∣=∣x1∣2+∣x2∣2+⋯∣xn∣2
区分 :复数的模平方和复数平方的模
模长性质
∣ k ∙ X ∣ = ∣ k ∣ ∙ ∣ X ∣ , k ∈ C \mid k\bullet X \mid = \mid k \mid \bullet \mid X \mid ,k\in C ∣k∙X∣=∣k∣∙∣X∣,k∈C
∣ X k ∣ = ∣ X ∣ ∣ k ∣ , ( k ≠ 0 ⃗ ) \vert \frac{X}{k} \vert=\frac{\vert X\vert}{\vert k \vert},(k\neq \vec{0}) ∣kX∣=∣k∣∣X∣,(k=0)
∣ X ± Y ∣ ≤ ∣ X ∣ + ∣ Y ∣ \vert X\pm Y\vert \le \vert X\vert + \vert Y \vert ∣X±Y∣≤∣X∣+∣Y∣
令 X = ( x 1 x 2 ⋮ x n ) ∈ C , 则 { ① X H X = ∣ X ∣ 2 ② t r ( X H X ) = t r ( X X H ) = ∣ X ∣ 2 , 其中 ∣ X ∣ 2 = ∣ x 1 ∣ 2 + ∣ x 2 ∣ 2 + ⋯ + ∣ x n ∣ 2 = ∑ i = 1 n ∣ x i ∣ 2 \begin{aligned} &令X=\left( \begin{matrix} x_1\\ x_2\\ \vdots\\ x_n \end{matrix} \right)\in C,则 \left\{ \begin{aligned} &①\quad X^HX=\mid X \mid^2\\\\ &②\quad tr(X^HX)=tr(XX^H)=\mid X \mid^2 \end{aligned} \right.,其中\\ &\mid X \mid^2 = \mid x_1 \mid ^2 + \mid x_2 \mid^2+\cdots+\mid x_n \mid^2=\sum\limits_{i=1}\limits^{n}\mid x_i \mid^2 \end{aligned} 令X=⎝ ⎛x1x2⋮xn⎠ ⎞∈C,则⎩ ⎨ ⎧①XHX=∣X∣2②tr(XHX)=tr(XXH)=∣X∣2,其中∣X∣2=∣x1∣2+∣x2∣2+⋯+∣xn∣2=i=1∑n∣xi∣2
eg
X = ( 1 i 1 ) ∈ C 3 , X X H = ( 1 i 1 ) ( 1 − i 1 ) = ( 1 ∗ 1 ∗ 1 ) ∴ t r ( X H X ) = t r ( X X H ) = ∣ 1 ∣ 2 + ∣ − i 2 ∣ 2 + ∣ 1 ∣ 2 = 3 \begin{aligned} &X=\left( \begin{matrix} 1\\i\\1 \end{matrix} \right)\in C^3,XX^H=\left( \begin{matrix} 1\\i\\1 \end{matrix} \right)(1\quad -i\quad 1)=\left( \begin{matrix} &1\quad &\quad &*\\ &\quad &1&\quad\\ &*&\quad&1 \end{matrix} \right)\\ &\therefore tr(X^HX)=tr(XX^H)=\vert 1 \vert^2+\vert -i^2 \vert^2+\vert 1 \vert^2 = 3 \end{aligned} X=⎝ ⎛1i1⎠ ⎞∈C3,XXH=⎝ ⎛1i1⎠ ⎞(1−i1)=⎝ ⎛1∗1∗1⎠ ⎞∴tr(XHX)=tr(XXH)=∣1∣2+∣−i2∣2+∣1∣2=3
令 A = ( a i j ) ∈ C , 则模长 ( 矩阵的 F 范数 ) ∣ ∣ A ∣ ∣ = ∑ i = 1 n ∣ a i j ∣ 2 \begin{aligned} &令A=(a_{ij})\in C,则模长(矩阵的F范数)\mid \mid A \mid\mid=\sqrt{\sum\limits_{i=1}\limits^{n}\mid a_{ij} \mid^2} \end{aligned} 令A=(aij)∈C,则模长(矩阵的F范数)∣∣A∣∣=i=1∑n∣aij∣2
设 A = ( a i j ) n × p , 则 t r ( A H A ) = t r ( A A H ) = ∣ ∣ A ∣ ∣ = ∑ i = 1 n ∣ a i j ∣ 2 \begin{aligned} &设A=(a_{ij})_{n\times p},则 \quad tr(A^HA)=tr(AA^H)=\mid\mid A\mid\mid = \sum\limits_{i=1}\limits^{n}\mid a_{ij} \mid^2 \end{aligned} 设A=(aij)n×p,则tr(AHA)=tr(AAH)=∣∣A∣∣=i=1∑n∣aij∣2
eg
X = ( 1 i 1 ) ∈ C 3 × 1 , 则 ∣ X ∣ 2 = X H X = ( 1 , − i , 1 ) ( 1 i 1 ) = 1 + − i 2 + 1 = 3 \begin{aligned} X=\left( \begin{matrix} 1\\i\\1 \end{matrix} \right)\in C^{3\times 1},则\mid X \mid^2 = X^HX=(1,-i,1)\left( \begin{matrix} 1\\i\\1 \end{matrix} \right)=1+-i^2+1=3 \end{aligned} X=⎝ ⎛1i1⎠ ⎞∈C3×1,则∣X∣2=XHX=(1,−i,1)⎝ ⎛1i1⎠ ⎞=1+−i2+1=3
对于方阵 A = ( a i j ) n × n A=(a_{ij})_{n\times n} A=(aij)n×n ,有
{ t r ( A ) = Δ a 11 + a 22 + ⋯ + a n n = ∑ i = 1 n λ i d e t ( A ) = ∣ A ∣ = ∏ i = 1 n λ i \begin{aligned} \left\{ \begin{aligned} &tr(A)\overset{\Delta}{=} a_{11}+a_{22}+\cdots+a_{nn}=\sum\limits_{i=1}\limits^{n}\lambda_i\\ &det(A) = \mid A \mid = \prod\limits_{i=1}\limits^{n}\lambda_i \end{aligned} \right. \end{aligned} ⎩ ⎨ ⎧tr(A)=Δa11+a22+⋯+ann=i=1∑nλidet(A)=∣A∣=i=1∏nλi
任一矩阵 A n × p , A H A 与 A A H A_{n\times p},A^HA与AA^H An×p,AHA与AAH 都是Hermite矩阵
( A H A ) H = A H A , ( A A H ) H = A \begin{aligned} &(A^HA)^H=A^HA,(AA^H)^H=A \end{aligned} (AHA)H=AHA,(AAH)H=A
A n × p = ( a 11 ⋯ a 1 p ⋮ ⋱ ⋮ a n 1 ⋯ a n p ) ∈ C n × p , A p × n H = ( a 11 ‾ ⋯ a n 1 ‾ ⋮ ⋱ ⋮ a 1 p ‾ ⋯ a n p ‾ ) ∈ C p × n A A H = ( a 11 ⋯ a 1 p ⋮ ⋱ ⋮ a n 1 ⋯ a n p ) ( a 11 ‾ ⋯ a n 1 ‾ ⋮ ⋱ ⋮ a 1 p ‾ ⋯ a n p ‾ ) = ( a 11 a 11 ‾ + a 12 a 12 ‾ + ⋯ + a 1 p a 1 p ‾ ∗ a 21 a 21 ‾ + a 22 a 22 ‾ + ⋯ + a 2 p a 2 p ‾ ⋱ ∗ a n 1 a n 1 ‾ + a n 2 a n 2 ‾ + ⋯ + a n p a n p ‾ ) \begin{aligned} &A_{n\times p}=\left( \begin{matrix} &a_{11}&\cdots&a_{1p}\\ &\vdots&\ddots&\vdots\\ &a_{n1}&\cdots&a_{np} \end{matrix} \right)\in C^{n\times p},\\ &A^H_{p\times n}=\left( \begin{matrix} &\overline{a_{11}}&\cdots&\overline{a_{n1}}\\ &\vdots&\ddots&\vdots\\ &\overline{a_{1p}}&\cdots&\overline{a_{np}} \end{matrix} \right)\in C^{p\times n}\\ &AA^H=\left( \begin{matrix} &a_{11}&\cdots&a_{1p}\\ &\vdots&\ddots&\vdots\\ &a_{n1}&\cdots&a_{np} \end{matrix} \right)\left( \begin{matrix} &\overline{a_{11}}&\cdots&\overline{a_{n1}}\\ &\vdots&\ddots&\vdots\\ &\overline{a_{1p}}&\cdots&\overline{a_{np}} \end{matrix} \right)\\ &=\left( \begin{matrix} &a_{11}\overline{a_{11}}+a_{12}\overline{a_{12}}+\cdots+a_{1p}\overline{a_{1p}}&\quad &\quad&\ast\\ &\quad&a_{21}\overline{a_{21}}+a_{22}\overline{a_{22}}+\cdots+a_{2p}\overline{a_{2p}}&\quad&\quad\\ &\quad &\quad &\ddots&\quad\\ &\ast &\quad&\quad&a_{n1}\overline{a_{n1}}+a_{n2}\overline{a_{n2}}+\cdots+a_{np}\overline{a_{np}} \end{matrix} \right) \end{aligned} An×p=⎝ ⎛a11⋮an1⋯⋱⋯a1p⋮anp⎠ ⎞∈Cn×p,Ap×nH=⎝ ⎛a11⋮a1p⋯⋱⋯an1⋮anp⎠ ⎞∈Cp×nAAH=⎝ ⎛a11⋮an1⋯⋱⋯a1p⋮anp⎠ ⎞⎝ ⎛a11⋮a1p⋯⋱⋯an1⋮anp⎠ ⎞=⎝ ⎛a11a11+a12a12+⋯+a1pa1p∗a21a21+a22a22+⋯+a2pa2p⋱∗an1an1+an2an2+⋯+anpanp⎠ ⎞
t r ( A H A ) = t r ( A A H ) = ( ∣ a 11 ∣ 2 + ∣ a 12 ∣ 2 + . . . + ∣ a 1 p ∣ 2 ) + ( ∣ a 21 ∣ 2 + ∣ a 22 ∣ 2 + . . . + ∣ a 2 p ∣ 2 ) + . . . + ( ∣ a n 1 ∣ 2 + ∣ a n 2 ∣ 2 + . . . + ∣ a n p ∣ 2 ) = ∑ i = 1 , j = 1 i = n , j ∣ a i j ∣ 2 \begin{aligned} &tr(A^HA)=tr(AA^H)=\\ &(\mid a_{11} \mid^2+\mid a_{12} \mid^2+...+\mid a_{1p} \mid^2)+(\mid a_{21} \mid^2+\mid a_{22} \mid^2+...+\mid a_{2p} \mid^2)+\\ &...+(\mid a_{n1} \mid^2+\mid a_{n2} \mid^2+...+\mid a_{np} \mid^2) =\sum\limits_{i=1,j=1}\limits^{i=n,j}\mid a_{ij} \mid^2 \end{aligned} tr(AHA)=tr(AAH)=(∣a11∣2+∣a12∣2+...+∣a1p∣2)+(∣a21∣2+∣a22∣2+...+∣a2p∣2)+...+(∣an1∣2+∣an2∣2+...+∣anp∣2)=i=1,j=1∑i=n,j∣aij∣2
推论
t r ( A B H ) = t r ( B H A ) = ∑ a i j b i j ‾ tr(AB^H)=tr(B^HA)=\sum a_{ij}\overline{b_ij} tr(ABH)=tr(BHA)=∑aijbij
将 A B 矩阵按列分块,可验证 t r ( B H A ) tr(B^HA) tr(BHA)
A = ( α 1 , α 2 ) , B = ( β 1 , β 2 ) , B H = ( β 1 ‾ T β 2 ‾ T ) B H A = ( β 1 ‾ T β 2 ‾ T ) ( α 1 , α 2 ) = ( β 1 ‾ T α 1 β 1 ‾ T α 2 β 2 ‾ T α 1 β 2 ‾ T α 2 ) t r ( B H A ) = β 1 ‾ T α 1 + β 2 ‾ T α 2 = ( a 11 b 11 ‾ + a 21 b 21 ‾ + a 31 b 31 ‾ ) + ( a 12 b 12 ‾ + a 22 b 22 ‾ + a 32 b 32 ‾ ) = ∑ a i j b i j ‾ \begin{aligned} &A=(\alpha_1,\alpha_2),B=(\beta_1,\beta_2),B^H=\left( \begin{matrix} \overline{\beta_1}^T\\ \overline{\beta_2}^T\\ \end{matrix} \right)\\ &B^HA=\left( \begin{matrix} \overline{\beta_1}^T\\ \overline{\beta_2}^T\\ \end{matrix} \right)(\alpha_1,\alpha_2)=\left( \begin{matrix} &\overline{\beta_1}^T\alpha_1\quad &\overline{\beta_1}^T\alpha_2\\ &\overline{\beta_2}^T\alpha_1\quad &\overline{\beta_2}^T\alpha_2 \end{matrix} \right)\\ &tr(B^HA)=\overline{\beta_1}^T\alpha_1+\overline{\beta_2}^T\alpha_2=\\ &\quad (a_{11}\overline{b_{11}}+a_{21}\overline{b_{21}}+a_{31}\overline{b_{31}})+ (a_{12}\overline{b_{12}}+a_{22}\overline{b_{22}}+a_{32}\overline{b_{32}})\\ &=\sum a_{ij}\overline{b_{ij}} \end{aligned} A=(α1,α2),B=(β1,β2),BH=(β1Tβ2T)BHA=(β1Tβ2T)(α1,α2)=(β1Tα1β2Tα1β1Tα2β2Tα2)tr(BHA)=β1Tα1+β2Tα2=(a11b11+a21b21+a31b31)+(a12b12+a22b22+a32b32)=∑aijbij
将 A B矩阵按行分块,可验证 t r ( A B H ) tr(AB^H) tr(ABH)
A = ( α 1 α 2 α 3 ) , B = ( β 1 β 2 β 3 ) , B H = ( β 1 ‾ T , β 2 ‾ T , β 3 ‾ T ) A B H = ( α 1 α 2 α 3 ) ( β 1 ‾ T , β 2 ‾ T , β 3 ‾ T ) = ( α 1 β 1 ‾ T α 1 β 2 ‾ T α 1 β 3 ‾ T α 2 β 1 ‾ T α 2 β 2 ‾ T α 3 β 3 ‾ T α 3 β 1 ‾ T α 3 β 2 ‾ T α 3 β 3 ‾ T ) t r ( A B H ) = ( a 11 b 11 ‾ + a 12 b 12 ‾ ) + ( a 21 b 21 ‾ + a 22 b 22 ‾ ) + ( a 31 b 31 ‾ + a 32 b 32 ‾ ) = ∑ a i j b i j ‾ \begin{aligned} &A=\left( \begin{matrix} \alpha_1\\ \alpha_2\\ \alpha_3 \end{matrix} \right), B=\left( \begin{matrix} \beta_1\\ \beta_2\\ \beta_3 \end{matrix} \right),B^H=(\overline{\beta_1}^T,\overline{\beta_2}^T,\overline{\beta_3}^T)\\ &AB^H=\left( \begin{matrix} \alpha_1\\ \alpha_2\\ \alpha_3 \end{matrix} \right)(\overline{\beta_1}^T,\overline{\beta_2}^T,\overline{\beta_3}^T)=\left( \begin{matrix} &\alpha_1\overline{\beta_1}^T\quad&\alpha_1\overline{\beta_2}^T\quad&\alpha_1\overline{\beta_3}^T\\ &\alpha_2\overline{\beta_1}^T\quad&\alpha_2\overline{\beta_2}^T\quad&\alpha_3\overline{\beta_3}^T\\ &\alpha_3\overline{\beta_1}^T\quad&\alpha_3\overline{\beta_2}^T\quad&\alpha_3\overline{\beta_3}^T \end{matrix} \right)\\ &tr(AB^H)=(a_{11}\overline{b_{11}}+a_{12}\overline{b_{12}})+(a_{21}\overline{b_{21}}+a_{22}\overline{b_{22}})+(a_{31}\overline{b_{31}}+a_{32}\overline{b_{32}})\\ &\quad\quad\quad\quad = \sum a_{ij}\overline{b_{ij}} \end{aligned} A=⎝ ⎛α1α2α3⎠ ⎞,B=⎝ ⎛β1β2β3⎠ ⎞,BH=(β1T,β2T,β3T)ABH=⎝ ⎛