LA@单位坐标向量乘法@对角阵乘法@矩阵多项式
文章目录
单位坐标向量左右乘
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设 A A A是 m × n m\times{n} m×n的, e i T e_i^{T} eiT第 i i i元为1的 m m m维单位坐标行向量, e j e_j ej是第 j j j元为1的 n n n维单位坐标列向量
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e i T A e_i^{T}A eiTA相当于抽取 A A A的第 i i i行
e i T A = ( 0 ⋯ 1 i ⋯ 0 ) ( a 11 a 12 ⋯ a 1 n a 21 a 22 ⋯ a 2 n ⋮ ⋮ ⋮ a m 1 a m 2 ⋯ a m n ) = ( a i 1 , a i 2 , ⋯ , a i n ) e i T A = ( 0 ⋯ 1 i ⋯ 0 ) ( β 1 T β 2 T ⋮ β m T ) = β i T e_i^{T}A= \begin{pmatrix} 0&\cdots&1_i&\cdots&0 \end{pmatrix} \begin{pmatrix} a_{11} &a_{12} &\cdots &a_{1n} \\ a_{21} &a_{22} &\cdots &a_{2n} \\ \vdots &\vdots & &\vdots \\ a_{m1} &a_{m2} &\cdots &a_{mn} \\ \end{pmatrix} =\begin{pmatrix} a_{i1},a_{i2},\cdots,a_{in} \end{pmatrix} \\ e_i^{T}A= \begin{pmatrix} 0&\cdots&1_i&\cdots&0 \end{pmatrix} \begin{pmatrix} \beta_1^{T}\\ \beta_2^{T}\\ \vdots\\ \beta_m^{T} \end{pmatrix}=\beta_i^{T} eiTA=(0⋯1i⋯0) a11a21⋮am1a12a22⋮am2⋯⋯⋯a1na2n⋮amn =(ai1,ai2,⋯,ain)eiTA=(0⋯1i⋯0) β1Tβ2T⋮βmT =βiT -
A e j Ae_{j} Aej相当于抽取 A A A的第 j j j列
A e j = ( a 11 a 12 ⋯ a 1 n a 21 a 22 ⋯ a 2 n ⋮ ⋮ ⋮ a m 1 a m 2 ⋯ a m n ) ( 0 ⋮ 1 j ⋮ 0 ) = ( a 1 j a 2 j ⋮ a n j ) A e j = ( α 1 , ⋯ , α n ) ( 0 ⋮ 1 j ⋮ 0 ) = α j Ae_j=\begin{pmatrix} a_{11} &a_{12} &\cdots &a_{1n} \\ a_{21} &a_{22} &\cdots &a_{2n} \\ \vdots &\vdots & &\vdots \\ a_{m1} &a_{m2} &\cdots &a_{mn} \\ \end{pmatrix} \begin{pmatrix} 0\\\vdots\\1_j\\\vdots\\0 \end{pmatrix} =\begin{pmatrix} a_{1j}\\a_{2j}\\\vdots\\a_{nj} \end{pmatrix} \\ Ae_j=(\alpha_1,\cdots,\alpha_n) \begin{pmatrix} 0\\\vdots\\1_j\\\vdots\\0 \end{pmatrix} =\alpha_j Aej= a11a21⋮am1a12a22⋮am2⋯⋯⋯a1na2n⋮amn 0⋮1j⋮0 = a1ja2j⋮anj Aej=(α1,⋯,αn) 0⋮1j⋮0 =αj
单位坐标向量和二次型
- 设 n n n元二次型为 f ( x ) = x T A x f(\bold{x})=\bold{x^{T}A{x}} f(x)=xTAx,那么 f ( e k ) f(e_{k}) f(ek)= e k T A e k \bold{e}_k^T\bold{Ae}_{k} ekTAek= β k T e k \beta_k^{T}\bold{e}_k βkTek= a k k a_{kk} akk
- 这个结论在证明和推导某些正定二次型的性质时很有用
对角阵
- Λ = d i a g ( λ 1 , λ 2 , ⋯ , λ n ) = ( λ 1 λ 2 ⋱ λ n ) \Lambda =\mathrm{diag}(\lambda_{1},\lambda_{2},\cdots,\lambda_{n}) =\begin{pmatrix} {{\lambda _1}} & {} & {} & {} \cr {} & {{\lambda _2}} & {} & {} \cr {} & {} & \ddots & {} \cr {} & {} & {} & {{\lambda _n}} \cr \end{pmatrix} Λ=diag(λ1,λ2,⋯,λn)= λ1λ2⋱λn
对角阵乘以一般矩阵
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n n n阶对角阵 A = Λ = d i a g ( λ 1 , ⋯ , λ n ) A=\Lambda=\rm{diag}(\lambda_1,\cdots,\lambda_n) A=Λ=diag(λ1,⋯,λn),其中 λ i = a i i \lambda_{i}=a_{ii} λi=aii
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定义2个矩阵 A , B A,B A,B
- A = ( a 11 a 22 ⋱ a n n ) = ( λ 1 λ 2 ⋱ λ n ) B n × s = ( b 11 b 12 ⋯ b 1 s b 21 b 22 ⋯ b 2 s ⋮ ⋮ ⋱ ⋮ b n 1 b n 2 ⋯ b n s ) A= \begin{pmatrix} {{a _{11}}} & {} & {} & {} \cr {} & {{a _{22}}} & {} & {} \cr {} & {} & \ddots & {} \cr {} & {} & {} & {{a _{nn}}} \cr \end{pmatrix}= \begin{pmatrix} {{\lambda _1}} & {} & {} & {} \cr {} & {{\lambda _2}} & {} & {} \cr {} & {} & \ddots & {} \cr {} & {} & {} & {{\lambda _n}} \cr \end{pmatrix} \\ B_{n\times{s}}= \begin{pmatrix} b_{11}& b_{12}& \cdots & b_{1s} \\ b_{21}& b_{22}& \cdots & b_{2s} \\ \vdots & \vdots & \ddots & \vdots \\ b_{n1}& b_{n2}& \cdots & b_{ns} \end{pmatrix} A= a11a22⋱ann = λ1λ2⋱λn Bn×s= b11b21⋮bn1b12b22⋮bn2⋯⋯⋱⋯b1sb2s⋮bns
左乘对角阵
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对于 C = A B C=AB C=AB
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C = A B = ( λ 1 λ 2 ⋱ λ n ) ( b 11 b 12 ⋯ b 1 s b 21 b 22 ⋯ b 2 s ⋮ ⋮ ⋱ ⋮ b n 1 b n 2 ⋯ b n s ) C=AB=\begin{pmatrix} {{\lambda _1}} & {} & {} & {} \cr {} & {{\lambda _2}} & {} & {} \cr {} & {} & \ddots & {} \cr {} & {} & {} & {{\lambda _n}} \cr \end{pmatrix} \begin{pmatrix} b_{11}& b_{12}& \cdots & b_{1s} \\ b_{21}& b_{22}& \cdots & b_{2s} \\ \vdots & \vdots & \ddots & \vdots \\ b_{n1}& b_{n2}& \cdots & b_{ns} \end{pmatrix} C=AB= λ1λ2⋱λn b11b21⋮bn1b12b22⋮bn2⋯⋯⋱⋯b1sb2s⋮bns
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a i j = { 0 i ≠ j a k k = λ k i = j = k i , j , k = 1 , 2 , ⋯ , n a_{ij}=\begin{cases} 0&i\neq{j}\\ a_{kk}=\lambda_{k}&i=j=k \end{cases} \quad i,j,k=1,2,\cdots,n aij={0akk=λki=ji=j=ki,j,k=1,2,⋯,n
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将 a i j a_{ij} aij代入到 c i j c_{ij} cij中
c i j = ∑ k = 1 n a i k b k j = a i i b i j = λ i b i j , ( k = i ) c_{ij}=\sum_{k=1}^n a_{ik}b_{kj}=a_{ii}b_{ij}=\lambda_{i}b_{ij},(k=i) cij=k=1∑naikbkj=aiibij=λibij,(k=i)- 可见,对角矩阵 Λ ( λ 1 , ⋯ , λ 2 ) \Lambda(\lambda_{1},\cdots,\lambda_{2}) Λ(λ1,⋯,λ2),左乘B相当于对B的第 i i i行乘以 λ i \lambda_{i} λi
- 也就是说,起到每行倍乘一个独立的数的作用
- 特别的,当 λ i = 1 , i = 1 , 2 , ⋯ , n \lambda_{i}=1,i=1,2,\cdots,n λi=1,i=1,2,⋯,n,也就是 A = E A=E A=E,那么 A B AB AB= E B EB EB= B B B
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右乘对角阵
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下面我们讨论另一侧的情况 D = B F D=BF D=BF
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D = B F = ( b 11 b 12 ⋯ b 1 s b 21 b 22 ⋯ b 2 s ⋮ ⋮ ⋱ ⋮ b n 1 b n 2 ⋯ b n s ) ( λ 1 λ 2 ⋱ λ s ) D=BF =\begin{pmatrix} b_{11}& b_{12}& \cdots & b_{1s} \\ b_{21}& b_{22}& \cdots & b_{2s} \\ \vdots & \vdots & \ddots & \vdots \\ b_{n1}& b_{n2}& \cdots & b_{ns} \end{pmatrix} \begin{pmatrix} {{\lambda _1}} & {} & {} & {} \cr {} & {{\lambda _2}} & {} & {} \cr {} & {} & \ddots & {} \cr {} & {} & {} & {{\lambda _s}} \cr \end{pmatrix} D=BF= b11b21⋮bn1b12b22⋮bn2⋯⋯⋱⋯b1sb2s⋮bns λ1λ2⋱λs
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分析方法类似, c i j = λ j b i j c_{ij}=\lambda_{j}b_{ij} cij=λjbij,其作用相当于将 B B B的第 j j j列乘以 λ j \lambda_{j} λj
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每行不超过一个非0元素的矩阵和对角阵的乘法
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假设矩阵 A A A是 m × n m\times{n} m×n的,其每一行中的非零元素不超过1个(或者简单地假设每行就有一个非0元素)
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将该矩阵中每行的非0元素所在的列记为 j 1 , ⋯ , j m j_1,\cdots,j_m j1,⋯,jm,即每行的非零元素表示为: a i , j i = ξ i \large{a_{i,j_{i}}}=\xi_{i} ai,ji=ξi,其中 i = 1 , 2 , ⋯ , m i=1,2,\cdots,m i=1,2,⋯,m; j 或 j i = 1 , 2 , ⋯ , n j或j_{i}=1,2,\cdots,n j或ji=1,2,⋯,n
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我们可以将此时的矩阵A记为 A = η ( ξ 1 , ⋯ , ξ m ) A=\eta(\xi_1,\cdots,\xi_m) A=η(ξ1,⋯,ξm)
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a i j = { 0 j ≠ j i ξ i j = j i c i j = ∑ k = 1 n a i k b k j = a i , j i b j i , j = ξ i b j i , j a_{ij}= \begin{cases} 0&j\neq j_i\\ \xi_i&j=j_{i} \end{cases} \\ c_{ij} =\sum_{k=1}^{n}a_{ik}b_{kj} =a_{i,j_i}b_{j_i,j}=\xi_ib_{j_i,j} aij={0ξij=jij=jicij=k=1∑naikbkj=ai,jibji,j=ξibji,j
C = A B = η ( ξ 1 , ⋯ , ξ m ) ( b 11 b 12 ⋯ b 1 s b 21 b 22 ⋯ b 2 s ⋮ ⋮ ⋱ ⋮ b n 1 b n 2 ⋯ b n s ) = ( a 1 , j 1 b j 1 , 1 a 1 , j 1 b j 1 , 2 ⋯ a 1 , j 1 b j 1 , s a 2 , j 2 b j 2 , 1 a 2 , j 2 b j 2 , 2 ⋯ a 2 , j 2 b j 2 , s ⋮ ⋮ a m , j m b j m , 1 a m , j m b j m , 2 ⋯ a m , j m b j m , s ) = ( ξ 1 b j 1 , 1 ξ 1 b j 1 , 2 ⋯ ξ 1 b j 1 , s ξ 2 b j 2 , 1 ξ 2 b j 2 , 2 ⋯ ξ 2 b j 2 , s ⋮ ⋮ ξ m b j m , 1 ξ m b j m , 2 ⋯ ξ m b j m , s ) \begin{aligned} C &=AB\\ &=\eta(\xi_1,\cdots,\xi_m) \begin{pmatrix} b_{11}& b_{12}& \cdots & b_{1s} \\ b_{21}& b_{22}& \cdots & b_{2s} \\ \vdots & \vdots & \ddots & \vdots \\ b_{n1}& b_{n2}& \cdots & b_{ns} \end{pmatrix} \\ &=\begin{pmatrix} a_{1,j_1}b_{j_1,1}&a_{1,j_1}b_{j_1,2}&\cdots&a_{1,j_1}b_{j_1,s}\\ a_{2,j_2}b_{j_2,1}&a_{2,j_2}b_{j_2,2}&\cdots&a_{2,j_2}b_{j_2,s}\\ \vdots&&&\vdots\\ a_{m,j_m}b_{j_m,1}&a_{m,j_m}b_{j_m,2}&\cdots&a_{m,j_m}b_{j_m,s} \end{pmatrix} \\ &=\begin{pmatrix} \xi_1b_{j_1,1}&\xi_1b_{j_1,2}&\cdots&\xi_1b_{j_1,s}\\ \xi_2b_{j_2,1}&\xi_2b_{j_2,2}&\cdots&\xi_2b_{j_2,s}\\ \vdots&&&\vdots\\ \xi_mb_{j_m,1}&\xi_mb_{j_m,2}&\cdots&\xi_mb_{j_m,s} \end{pmatrix} \end{aligned} C=AB=η(ξ1,⋯,ξm) b11b21⋮bn1b12b22⋮bn2⋯⋯⋱⋯b1sb2s⋮bns = a1,j1bj1,1a2,j2bj2,1⋮am,jmbjm,1a1,j1bj1,2a2,j2bj2,2am,jmbjm,2⋯⋯⋯a1,j1bj1,sa2,j2bj2,s⋮am,jmbjm,s = ξ1bj1,1ξ2bj2,1⋮ξmbjm,1ξ1bj1,2ξ2bj2,2ξmbjm,2⋯⋯⋯ξ1bj1,sξ2bj2,s⋮ξmbjm,s
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观察可以发现,C的第 i i i行是通过对B中的第 j i j_i ji行乘以一个系数 ξ i \xi_{i} ξi得到
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因此,可以对矩阵B左乘 A A A来间接调整(覆盖)B中各行位置并乘以某个倍数的效果(将系数都设为1)
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此外,当 j p = j q , ξ p = ξ q , p ≠ q j_p=j_q,\xi_{p}=\xi_{q},p\neq{q} jp=jq,ξp=ξq,p=q时,结果矩阵 C = A B C=AB C=AB中第 p , q p,q p,q行会是相等的
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基于上述基本结论,如果我们将A设定为某个初等矩阵,也就是第 i , j i,j i,j行交互位置,那么 C = A B C=AB C=AB等价于对矩阵B的第 i , j i,j i,j行对调
使用分块矩阵来描述
- 对角阵左乘(右乘)一个分块矩阵
左乘对角阵
- Λ P = ( λ 1 λ 2 ⋱ λ n ) ( β 1 β 2 ⋮ β n ) = ( λ 1 β 1 λ 2 β 2 ⋮ λ n β n ) \Lambda{P} =\begin{pmatrix} {{\lambda _1}} & {} & {} & {} \cr {} & {{\lambda _2}} & {} & {} \cr {} & {} & \ddots & {} \cr {} & {} & {} & {{\lambda _n}} \cr \end{pmatrix} \begin{pmatrix} \beta_{1}\\ \beta_{2}\\ \vdots \\ \beta_{n} \\ \end{pmatrix} =\begin{pmatrix} \lambda_1\beta_{1}\\ \lambda_2\beta_{2}\\ \vdots \\ \lambda_n\beta_{n} \\ \end{pmatrix} ΛP= λ1λ2⋱λn β1β2⋮βn = λ1β1λ2β2⋮λnβn
右乘对角阵
- Λ = ( α 1 , ⋯ , α n ) ( λ 1 λ 2 ⋱ λ n ) = ( λ 1 α 1 , ⋯ , λ n α n ) \Lambda =(\alpha_1,\cdots,\alpha_n) \begin{pmatrix} {{\lambda _1}} & {} & {} & {} \cr {} & {{\lambda _2}} & {} & {} \cr {} & {} & \ddots & {} \cr {} & {} & {} & {{\lambda _n}} \cr \end{pmatrix} =(\lambda_{1}\alpha_1,\cdots,\lambda_n\alpha_n) Λ=(α1,⋯,αn) λ1λ2⋱λn =(λ1α1,⋯,λnαn)
对角阵左右乘与矩阵初等变换
- 上述结论可作为矩阵初等变换的基础
对角阵之间的乘法
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根据上一节的结论,对角阵左 A A A乘另一个矩阵 B B B,相当于将 B B B的第 i i i行乘以 a i a_i ai;或者理解为矩阵A右乘对角阵B,相当于对 A A A的第 i i i列乘以系数 b i b_i bi
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也就是说,两个同型对角阵的矩阵乘法和hadamard运算结果一致 Λ 1 Λ 2 = Λ ⊙ Λ 2 \Lambda_1\Lambda_2=\Lambda\odot\Lambda_2 Λ1Λ2=Λ⊙Λ2
( a 1 a 2 ⋱ a n ) ( b 1 b 2 ⋱ b n ) = ( a 1 b 1 a 2 b 2 ⋱ a n b n ) \begin{pmatrix} {{a _1}} & {} & {} & {} \cr {} & {{a _2}} & {} & {} \cr {} & {} & \ddots & {} \cr {} & {} & {} & {{a _n}} \cr \end{pmatrix} \begin{pmatrix} {{b_1}} & {} & {} & {} \cr {} & {{b _2}} & {} & {} \cr {} & {} & \ddots & {} \cr {} & {} & {} & {{b _n}} \cr \end{pmatrix} =\begin{pmatrix} {{a_1b _1}} & {} & {} & {} \cr {} & {{a_2b _2}} & {} & {} \cr {} & {} & \ddots & {} \cr {} & {} & {} & {{a_nb _n}} \cr \end{pmatrix} a1a2⋱an b1b2⋱bn = a1b1a2b2⋱anbn -
由此可知,对角阵乘方的结果仍然是一个对角阵
对角阵方幂:
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设对角阵 d i a g ( λ 1 , λ 2 , ⋯ , λ n ) \mathrm{diag}(\lambda_{1},\lambda_{2},\cdots,\lambda_{n}) diag(λ1,λ2,⋯,λn),则 Λ n \Lambda^n Λn= d i a g ( λ 1 n , λ 2 n , ⋯ , λ n n ) \mathrm{diag}(\lambda_{1}^n,\lambda_{2}^n,\cdots,\lambda_{n}^n) diag(λ1n,λ2n,⋯,λnn)
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c i j = ∑ k = 1 j a i k b k j c_{ij}=\sum_{k=1}^{j}a_{ik}b_{kj} cij=∑k=1jaikbkj
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a i k = δ ( i , k ) a i k a_{ik}=\delta{(i,k)}a_{ik} aik=δ(i,k)aik
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b k j = δ ( k , j ) b k j b_{kj}=\delta(k,j)b_{kj} bkj=δ(k,j)bkj
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a i k b k j = δ ( i , k ) a i k δ ( k , j ) b k j a_{ik}b_{kj}=\delta(i,k)a_{ik}\delta(k,j)b_{kj} aikbkj=δ(i,k)aikδ(k,j)bkj
- 可见, c i j = 0 , i ≠ j c_{ij}=0,i\neq{j} cij=0,i=j
- 当 i = j = k i=j=k i=j=k, c k k = δ ( k , k ) a k k δ ( k , k ) b k k = a k k b k k c_{kk}=\delta(k,k)a_{kk}\delta(k,k)b_{kk}=a_{kk}b_{kk} ckk=δ(k,k)akkδ(k,k)bkk=akkbkk
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其他写法:
- a i j = { a i j i = j 0 i ≠ j b i j = { b i j i = j 0 i ≠ j c i j = ∑ i = 1 n a i k b k j = { 0 i ≠ j a k k b k k i = j = k a_{ij}= \begin{cases} a_{ij}&i=j\\ 0&i\neq{j} \end{cases} \\ b_{ij}= \begin{cases} b_{ij}&i=j\\ 0&i\neq{j} \end{cases} \\ c_{ij}=\sum_{i=1}^{n}a_{ik}b_{kj} =\begin{cases} 0&i\neq{j} \\ a_{kk}b_{kk}&i=j=k \end{cases} aij={aij0i=ji=jbij={bij0i=ji=jcij=i=1∑naikbkj={0akkbkki=ji=j=k
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对角阵求逆
- 设对角阵 Λ = d i a g ( λ 1 , λ 2 , ⋯ , λ n ) \Lambda=\mathrm{diag}(\lambda_{1},\lambda_{2},\cdots,\lambda_{n}) Λ=diag(λ1,λ2,⋯,λn),则 Λ − 1 \Lambda^{-1} Λ−1= d i a g ( λ 1 − 1 , λ 2 − 1 , ⋯ , λ n − 1 ) \mathrm{diag}(\lambda_{1}^{-1},\lambda_{2}^{-1},\cdots,\lambda_{n}^{-1}) diag(λ1−1,λ2−1,⋯,λn−1)
- 证明:由对角阵乘法的性质可知上述结论
矩阵多项式
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设关于 x x x的m次多项式: f ( x ) = ∑ i = 0 m a i x i f(x)=\sum\limits_{i=0}^{m}a_ix^i f(x)=i=0∑maixi
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则 f ( A ) = ∑ i = 0 m a i A i f(\bold{A})=\sum\limits_{i=0}^{m}a_i\bold{A}^i f(A)=i=0∑maiAi
<0>称为矩阵A的 m m m次多项式,其中 A 0 = E \boldsymbol{A}^0=\bold{E} A0=E
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