4-4-单纯矩阵的谱分解

矩阵论电子教程哈尔滨工程大学理学院应用数学系DepartmentofMathematics矩阵的分解第四章DepartmentofMathematics§4.4单纯矩阵的谱分解定理1:设是一个阶可对角化的矩阵，相异特征值为，则使得:,,,21An),2,1(iCEnni1iiiEA此式称为A的谱分解称为A的谱族EEE,,,21且满足:niiiijjiIEEEE1)2(,)1(iiiiEAEAE)3(iimrankE)4(的谱族是唯一的即是唯一的AEi,)5(DepartmentofMathematics分析:设是的代数重复度imi),2,1(i则:12211),,,,,,,(PPdiagA1m2mmnmimniiiCPCPPPPPPPPP~,:,~~~,),,(21121其中设,2,1i1212121~~~~000000),,(21iiiimmmPPPPPIIIPPPAiiiPPE~1iiiEADepartmentofMathematics证明(1)因为11212(,,),imnijPIijPPPPPPIPPOijP所以:iijjjiijjiijiEPPPPPPPPEE~)~(~~证明(2)niiIPPPPPPPPE121211~~~),,(iijjiEEEDepartmentofMathematics(3)由得同理可得1iiiiiEEAE1iiiEAiiiEAE证明:而:,所以:nmrankEErankrankIniiiiiin111)4(因为11iiiimErankiimrankEiimrankE证明:证明:(5)假设A有谱分解和1iiiGA1iiiEADepartmentofMathematics则由(3)知:iiiEAE)(jiGAGjjjjiijiGEGAE)(jiGEGAEjijjijiijiiGEGE由于,所以:jiOGEji同理可得:OGEijiiiiiiniGEGEGIG)(1iiiiiniiGEGEIEE)(1OGEji因为OGEij因为所以,唯一性得证iiGEDepartmentofMathematics可对角化矩阵的谱分解步骤：（1）首先求出矩阵的全部互异特征值及每个特征值所决定的线性无关特征向量Ai,,,21),,,(,,,,2121iiimiiimiipppPppp（3）令:iiimTimiTiiTiipppE2211（2）写出),,,()(211iimiiTP（4）最后写出12211iiiEEEEADepartmentofMathematics460350360A例1：已知矩阵为一个可对角化矩阵，求其谱分解表达式。解:首先求出矩阵的特征值与特征向量。容易计算A2(1)(2)IA从而的特征值为A1231,2可以求出分别属于这三个特征值的三个线性无关的特征向量:DepartmentofMathematics1122332,1,00,0,11,1,1TTT123,,201101011P于是111231,,110121120111()122010TPPDepartmentofMathematics1231,1,01,2,11,2,0TTT取11122220110121TTG令233120120120TG那么其谱分解表达式为122AGGDepartmentofMathematics正规阵的谱分解:设为正规矩阵，那么存在AnnUU使得:112212111222,,,HHnHnnHHHnnnA其中是矩阵的特征值所对应的单位特征向量。我们称上式为正规矩阵的谱分解表达式。iAiADepartmentofMathematics111inrrHiijijiiijiAG设正规矩阵有个互异的特征值，特征值的代数重数为，所对应的个两两正交的单位特征向量为,则的谱分解表达式又可以写成Ar12,,,riini12,,,iiiinA1inHiijijiG其中，并且显然有:2,0()HiiiikGGGGGikDepartmentofMathematics211(1);(2)(3)0();(4)(5)()rHiiiiiirikiiiiAGGGGGGikGIrankGn（6）满足上述性质的矩阵是唯一的。我们称为正交投影矩阵。iGiG即对于正规阵,满足如下6条:推论1设是一个阶可对角化的矩阵，谱分解为:,若:则有An1iiiEA,)(0mkkkaf1)()(iiiEfAfDepartmentofMathematics解：首先求出矩阵的特征值与特征向量。容易计算A3(1)(3)IA0111101111011110A例2：求正规矩阵的谱分解表达式。12341,3从而的特征值为ADepartmentofMathematics当时，求得三个线性无关的特征向量为11231,1,0,01,0,1,01,0,0,1TTT当时，求得一个线性无关的特征向量为将正交化与单位化可得341,1,1,1T123,,DepartmentofMathematics12311,,0,022112,,,06663111,,,12121212TTT将单位化可得：441111,,,2222TDepartmentofMathematics111223331114444311144443111444431114444HHHG于是有DepartmentofMathematics24411114444111144441111444411114444HG这样可得其谱分解表达式为123AGGDepartmentofMathematics解：首先求出矩阵的特征值与特征向量。A2(2)IA0110000iAi求正规矩阵的谱分解表达式。练习从而的特征值为A1232,2,0ii可以求出分别属于这三个特征值的三个线性无关的特征向量:DepartmentofMathematics1122332,,12,,10,,1TTTiii再将其单位化可得三个标准正交的特征向量12311,,22211,,22210,,22TTTiiiDepartmentofMathematics1112212442144421444HGiiii于是有:DepartmentofMathematics2222212442144421444HGiiiiDepartmentofMathematics33300010221022HGii这样可得其谱分解表达式为123220AiGiGGDepartmentofMathematics