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矩阵乘积AB和BA的相关性质80
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矩阵乘积AB和BA的相关性质80
第24卷第2期长春工业大学学报Vol124No.;矩阵乘积AB和BA的相关性质;王秀玉1,姜兴武2,曲波1;(1.长春工业大学基础科学学院,吉林长春1300;摘要:讨论了矩阵乘积AB与BA的特征值、特征向量;定义1设A∈Cm×n,B∈Cn×m,不妨设m;≤n;零解ΖA;-μ1Im;-μ2B;=0Ζ-μ10;-In;-μ1Im;=0Ζ;-1;-μB2In+μ1
 第24卷 第2期          长 春 工 业 大 学 学 报         Vol124 No.2   2003年6月   JOURNALOFCHANGCHUNUNIVERSITYOFTECHNOLOGY         Jun12003文章编号:03)  ①矩阵乘积AB和BA的相关性质王秀玉1, 姜兴武2, 曲 波1(1.长春工业大学基础科学学院,吉林长春 .吉林商业高等专科学校,吉林长春 130062)摘 要:讨论了矩阵乘积AB与BA的特征值、特征向量及秩等的关系,并得到了矩阵的奇异值分解。关键词:特征值;特征向量;奇异值中图分类号:O151.2   文献标识码:A  定义1 设A∈Cm×n,B∈Cn×m,不妨设mn≤n。对μ1,μ2∈C,若存在非零向量ξ∈C,η∈ξ=μ1η,Bη=μξ,则称(μC使得A21,μ2)为矩阵对(A,B)的一个特征值对,(ξ,η)A,于(μ1,μ2)=若AB1μ2,ξ=μBA121μ2AB与BA的特征值,η为AB属于μ1μ2的特征向量,ξ为BA属于μ1μ2的特征向量。m零解ΖA-μ1ImB-μ2B=0Ζ-μ10-In-μ1Im=0Ζ-1-μB2In+μ1BAμ  |μ12In-BA|=0。ξ引理3若μ0是AB的特征值,BA1μ2≠i=μξ1μ2i,i=1,2,…,s,而且ξ1,ξ2,…,ξs线性无-1ξ关,记ηi=μ1Ai,i=1,2,…,s,则η1,η2,…,ηs也线性无关。-1ηξξ证明Bi=μ1BAi=μ2i,i=1,2,…,s,若k1η1+k2η2+…+kηss=0,两边用B作用得:ηηηξk1B1+k2B2+…+ksBs=0,也就是:μ2k11+1AB与BA的特征值与特征向量的关系mn-In-BA|=λ引理1f(λ)=|λ)=|λg(λIm-AB|。g(λ),证明AλIm-ABIn(-当λ≠0时,有:=(-1)-mn+n(n+1)AλImInB=μξ0,则k1=k2=…=2k22+…+μ2kξss=0,μ2≠ks=0。01) mn|λIm-AB|=AλImInB|λImλImBAB|,又mn+m(m+1)=ηξ在引理3中,ABi=μ2Ai=μ1μ2ηi,i=1,2,…,s,由引理1和引理3可得。定理1设A∈Cm×n,B∈Cn×m,则λ≠0作为AB与BA的特征值,有相同的代数重数,同时也有相同的几何重数。-1In-λBA=(-1)mλIn-λ-1BA=n-=λmm-n(-1)mnλIn-BA|λIn-BA|,从而有:λ|λImm)=-AB|,当λ=0时,若m&n,由于|BA|=0,亦有f(λn-mn-λ),若m=n,则有|BA|=|AB|,也有f(λ)=λg(λ2 AB与BA的秩与最小多项式引理4μ0,(μ1≠1,0)为(A,B)的特征值对ΖdimN(A)dimN(BA)Ζrank(BA&rank(A)。ξ=μξ=0ABA1η证明有非零解Ζ有非ξμηA=η1B=0零解ΖN(A)≠N(BA)ΖdimN(A)&dimN(BA)Ζrank(BA)&rank(A)。)。g(λ若μ0,(μ1μ2≠1,μ2)为矩阵对(A,B)的特征值对Ζμ1μ2为BA的特征值。引理2证明ξ=μηA1η=μξB2有非零解η=μμηAB12η=μξB2有非①收稿日期:作者简介:王秀玉(1965-),女,吉林蛟河人,长春工业大学副教授,理学硕士,主要从事矩阵理论研究.第2期             王秀玉等:矩阵乘积AB和BA的相关性质65引理5μ0,(0,μ2≠2)为(A,B)的特征值对Ζrank(AB)rank(B)。ξ=0η=0AAB证明有非零解Ζ有非η=μ2ξη=μξBB2零解ΖN(AB)≠N(B)ΖdimN(B)&dimN(AB)Ζrank(AB)&rank(B)。记rank(A)=rA,rank(B)=rB,rank(AB)=rAB,rank(BA)=rBA,则有:N(A)&N(BA),N(B)证明设BA的最小多项式为:ψ(λ)=lλ),由于φ(λ)为AB的最小多项式,故有p(λφ(AB)=0,即(AB)k[(AB)t+qt-1(AB)t-1+…+q1(AB)+q0I]=0,左乘B,右乘A得:B(AB)[(AB)+qt-1(AB)tt-1k+…+q1(AB)+q0I]A=[(BA)+qt-1(BA)tt-10,即(BA)k[(BA)t+1+qt-1(BA)t+…+q1(BA)2k+1+q0(BA)]=0,(BA)+&N(AB)。引理6rA+rB≤m+rBA。证明 当rBA=rA时,有rA+rB=rBA+rB≤m+rBA,当rBA&rA时,取N(A)的一组基为:ζ1,ζζ2,…,ζn-rA,把它扩充成N(BA)的一组基为:1,ζ2,…,ζn-rA…+q1(BA)+q0I]=0。)整除λφ(λ),而  由最小多项式的性质得ψ(λλ不能整除q(λ),所以必有p(λ)整除q(λ),反)整除p(λ),因而有p(λ)=q过来同理可证q(λl(λ),即得ψ(λ)=λ)。q(λ推论 AB与BA的Smithλ,,BA以λ≠0为块均相同。,ζn-rA+1,…,ζn-rBA,取μ0,μ1≠1∈-1ζC,(μ1,0)为(A,B)的特征值对,记βi=μ1Ai,i=1,2,…,n-rBA,则β1=β2==n-rA=,βn-rA+1,+2,,rA+2rA+1rBA关rBA×nmn设A∈Cm,μ∈C,υ∈C,σ≥0rk1βn-rA+2n-+krA-βn-rBA=0,也  定义2-1ζ就是k1μ1An--1ζ+k2μ1An-rA+1rA+2+…+rA+2rBAkrA-rBA-1μζ1An-rBArBA=0,k1ζn-+k2ζn-+…=0,ζ+krA-rBAn-故βn-rA+1∈N(A),k1=k2=…=krA-rA+2,βn-,…,βn-rBAβ线性无关。又Bi=0,i=n-rA+1,…,n-rBA,故有n-rBA-(n-rA)≤m-rB,rA+rB≤m+rBA。υ=σμ,A3μ=συ,则称σ为A的奇异值,μ和使Aυ分别称为A的左、右奇异向量。由引理2知,σ≠0为A的奇异值,当且仅当2σ为AA3的特征值,且A的关于σ的线性无关的左奇异向量的个数与A的关于σ的线性无关的右奇异向量的个数相同。υ=σμA3υAμ=σ引理7证明rA+rB≤n+rAB。Ζ32AAμ=σμ3υAμ=σ,因此有引理8。当rAB=rB时,rA+rB=rA+rAB≤n+rAB,当rAB&rB时,取μ0,(0,μ2≠2)为(A,B)的特征值对,取N(B)的一组基为:η1,η2,…,ηm-rB,把它扩充成N(AB)的一组基为:η1,η2,…,ηm-γm-rB引理8σ≠0为A的奇异值,AA3μi=2σμi,i=1,2,…,s,且有(μi,μj)=1 i=j0 i≠j-13,υ,s,i=σAμi,i=1,2,…,ηm-rB+1,…,ηm-rAB,取γi=μ2-1Bηi,rB+1i,则(υi,υj)==m-rB+1,m-rB+2,…,m-rAB,则γm-rB+21 i=j0 i≠j,…,γm-rABγi=0,i=m-线性无关,又A证明rB+1,m-rB+2,…,m-rAB,从而有m-rAB-(m-rB)≤n-rA,即rA+rB≤n+rAB。-13-13(υAμAμi,υj)=(σi,σj)=σ-2μj3AA3μi=μj3μi=(μi,μj)。2设AA3的非零且互不相同的特征值为σ1&2232σμ2&…&σs&0,AAμij=σiij,j=1,2,…,ti,i=由引理6及引理7可得。定理2定理3设A∈Cm×n,B∈Cn×m,则有rA+rB≤min{m+rBA,n+rAB}。设A∈Cm×n,B∈Cn×m,Ind(AB)=)k,Ind(BA)=l,矩阵AB的最小多项式为:φ(λkktt-1)=λ(λ=λq(λ+qt-1λ+…q1λ+q0),q0≠l)=λ)。0,则BA的最小多项式为:ψ(λq(λ1,2,…,s,且(μij,μik)=1 j=k0 j≠ks,i=1∑ti=r,则有引理9。  引理9 μ11,μ12,…,μ1t1,μ21,μ22,…,μ2t2,…,μs1,μs2,…,μsts为标准正交向量组。66长 春 工 业 大 学 学 报               第24卷22333σμμij=μpkAAμi(μij,μpk)=σipkij=证明(AA3μpk)32μij=(μij,AA3μpk)=(μij,σμpk)=pυr+1,υr+2,…,υn,则υ11,υ12,…,υ1t1,…,υs1,υs2,…,υsts,υr+1,υr+2,…,υn为标准正交向量组,记V=(υ11,υ12,…,υ1t1,…,υs1,υs2,…,υsts,υr+1,υr+2,2σp(μij,μpk),当i≠p时,(μij,μpk)=0。取N(A3)=N(AA3)的标准正交基为:μr+1,μr+2,…,μn,则μ11,μ12,…,μ1t1,μ21,μ22,…,μ2t2,…,μs1,μs2,…,μsts,μr+1,μr+2,…,μn为标准正交向量组,记U=(μ11,μ12,…,μ1t1,…,μs1,μs2,…,μsts,μr+1,μr+2,…,μn)为酉矩阵,记-13υij=σiAμij,j=1,2,…,ti,i=1,2,…,s,则…,υn),则V为酉矩阵,则有定理4。∑0定理4U3AV=,∑=diag(σ1,σ2,0…,σr)。参考文献:[1] 程云鹏.矩阵论[M].西安:西北工业大学出版υ11,υ12,…,υ1t1,…,υs1,υs2,…,υsts为标准正交向量组,取N(A)=N(A3A)的一组标准正交基为社,20001SomePropertiesoftheProducts212 QUBo1(1.Basic,UniversityofTechnology,Changchun130012,C2.JilinCommercialCollege,Changchun130062,China)Abstract:Therelationshipbetweentheeigenvalue,eigenvectorandtherankoftheproductofmatricesABandBAisdiscussed,andthefactorizationofthesingularvaluesofmatrixAisobtained.Keywords:singularvalue.包含各类专业文献、行业资料、中学教育、应用写作文书、高等教育、文学作品欣赏、幼儿教育、小学教育、矩阵乘积AB和BA的相关性质80等内容。
 了矩阵 A、B 在什么条件下其特征值的乘积是 AB 与 BA 的特征值, 以及矩阵...给出 了矩阵 AB 与 BA 相似的情况下的一些相关性质的讨论与推广,我相信这些...  Hadamard乘积矩阵的一些性... 5页 5财富值 矩阵乘积AB和BA的相关性质 3页 1...? 2.收敛的无穷乘积的性质 2.1 收敛的无穷乘积的性质定理 2 定理 3 [ 4]...  回顾导入(20 分钟)――复习矩阵的运算求乘积 AB 与 BA 及 AI 与 IA 引入...二、主要教学过程(60 分钟,其中学生练习 20 分钟)三:矩阵乘法的性质: 1) ...  比如乘法 AB 一、1)用 A 的第 1 行各个数与 B 的第 1 列各个数对应...(A)是环中的可逆元那么公式(*)表明 性质对 n×n 的矩阵 A 和 B,有: 1...  Hadamard 乘积及 Kronecker 乘积的定义,证明 了它们的一些运算性质,并介绍了它们...2)交换律 矩阵的乘法不适合交换律,即一般说来, AB ? BA 。 这是由于,首先...  ,计算 AB,BA;(AB)C.A(BC ?0 2 ? ? ? ? ?2 3 ? ? ? (2)已知 ...第10课时矩阵乘法的的性... 3页 免费 2.线性变换的基本性质与... 暂无评价...  ? 定理 2.1.1 设 A、B 为 n 阶循环矩阵,则有: (1)乘积 AB 仍是循环矩阵,且满足乘法交换律,即 AB ? BA ;(2)若 A 可逆,则 A 的逆矩阵也是循环...  4、矩阵乘法的定义和性质(1)当矩阵 A 的列数和 B 的行数相等时,则 A 和...② 矩阵乘法无交换律. 即 AB ? BA 乘法公式 一般地,由于交换性的障碍,小...  a0 I . 性质 1.1 证明 设 两个循环矩阵 A, B 的乘积仍为循环矩阵,且 AB ? BA . A ? a0 I +a1D ? a2 D 2 ? ?? ? an ?1D n ?1 ? f (...In-GameDota 2 - Join
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