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阜师院数科院2020/3/93.2Jacobi方法J-方法用于求实对称阵的全部特征值、特征向量.对于实对称阵A,必有正交阵U,使UTAU=D。其中D是对角阵,其主对角线元i是A的特征值.正交阵U的第j列是A的属于i的特征向量.原理:Jacobi方法用平面旋转对矩阵A做相似变换,化A为对角阵,进而求出特征值与特征向量.阜师院数科院2020/3/9平面旋转矩阵对于p≠q,下面定义的n阶矩阵Upq是平面旋转矩阵。10001cossin11sincos10001pqpqpqU第列第列第行第行阜师院数科院2020/3/9容易验证Upq是正交阵。对于向量x,Upqx相当于把坐标轴Oxp和Oxq于所在的平面内旋转角度.变换过程:在保证相似条件下,使主对角线外元素趋于零!记n阶方阵A=[aij],对A做下面的变换:A1=UpqTAUpq,(3.12)A1仍然是实对称阵,因为,UpqT=Upq-1,知A1与A的特征值相同.下面,以4阶矩阵为例,来计算(3.12)阜师院数科院2020/3/92323100010000cossin00cossin0,0sincos00sincos000010001TUU1.TpqpqAUAU2,3,pq取U与A的行、列分块记法如下111213142122232413132333441424344writebbbbbbbbAbbbbbbbb=,ijAa00cossin,,sincos00TTpqpqIIxxUCSUCSCSxxII111213212223313233,AAAAAAAAAA阜师院数科院2020/3/9111213121222331323300,00TIAAAIACSAAACSIAAAI1.TpqpqAUAU2,3,pq1ijAb令1112132122233132330,0TTTAAAICSACSACSACSAAAI111213212223313233,TTTAACSACSACSACSCSAAACSA阜师院数科院2020/3/91114111441444144,,,(1)bbaaijpqbbaa1112141212223413244TTTaACSaACSACSACSCSAaACSa2223222332333233cossincossin(4)sincossincos;bbaabbaaipjq2,3,pq2124212431343134cossin,;,(2)sincosbbaaipqjpqbbaa1213121342434243cossin,,;,(3)sincosbbaaipqjpqbbaa比较元素,有由此,乘开各式,我们得到下面的关系式:111213142122232413132333441424344bbbbbbbbAbbbbbbbb阜师院数科院2020/3/92222cossin2cossin(1)sincos2cossin1sin2cos22cossin(2),,sincoscossin(3)sincppppqqpqqqppqqpqpqqpqqpppqpipiqiqijpjqbaaabaaabbaaabaipqabbb(3.13),,os(4),,;,jpjqijjiijajpqabbaipqjpq•由此见到,矩阵A1的第p行、列与第q行、列中的元素发生了变化,其它行、列中的元素不变。•特别,取旋转角满足下面的条件:阜师院数科院2020/3/9则可以得到0.pqqpbb2tan2pqqqppaaa(3.14)1sin2cos221sin2221sin2022pqqqpppqppqqqqpppqpqppqqqqppbaaaaaaaaaaaaa这是因为,为保证单值,限定||/4。阜师院数科院2020/3/9Jacobi算法(1)在A的非主对角线元素中,找到最大元apq.(2)用式(3.14)计算tan2,求cos,sin及矩阵Upq.(3)用公式(3.13)求A1。(4)若,停止计算.否则,令A=A1,重复执行(1)~(4).max||ijijb停止计算时,得特征值i≈bii,i=1,2,…,n.2tan2pqqqppaaa特征向量的求解设经过N次迭代,得到对角阵,则做了下面的变换一般,记,则U为正交阵,且U的列向量为A的特征向量。1111222222111122121,TTTTpqpqpqpqpqpqpqpqAUAUAUAUUUAUU22111122NNNNTTTNpqpqpqpqpqpqAUUUAUUU1122NNpqpqpqUUUU.TUAU阜师院数科院2020/3/9注意到22,{,}{,}(1)ijijbaijpqcossin,,,(3)sincospipjpipjqiqjqiqjbbaaijpqbbaacossincossin(4)sincossincospppqpppqqpqqqpqqbbaabbaa正交阵正交阵再注意,相似阵的迹不变,我们得到下面的关系式:2222,,(2)piqipiqibbaaipq由矩阵A1,A均是对称阵,得2222,,(3)ipiqipiqbbaaipq22222cossincossin(5)sincossincospppqqpqqbbBAbb2222tr()tr()BA22222222(4)ppqqpqppqqpqbbbaaa再注意到正交变换不改变向量的内积,故有(2o)阜师院数科院2020/3/9定理3.1设A是实对称阵,由J-方法,第k次得到的矩阵记为,又记()kkijnnAa2().kkijija则有lim0.kkJacobi算法收敛性判别使用式子,可以得到进而有1~422221,,ijijFFijijAbaA阜师院数科院2020/3/9旋转矩阵Upq的计算sin和cos的计算公式:22222(1),tan2,sgn().422tan1(2),tan21tantan2tan10,2441tan121,tan1,4ppqqpqpqppqqppqqaaaaaaaacccccccc当时当时限定故2222sgn()tan,ˆ111cos,sincos1tan11ctcctttt故可取则阜师院数科院2020/3/9例子试用Jacobi方法计算矩阵的全部特征值和相应的特征向量.(误差为ε=10-3)5.89595.86565.3A23223323:9(2,3),,tan2ππsgn()4411cos,sin22apqaaa解第一步选非对角线元素中的主元素阜师院数科院2020/3/93.57.77820.70717.778217.500.707100.51003.5651000cossin68.590cossin0sincos598.50sincos12323TAUAU1(1)(1)1212(1)(1)1122222:217.7782(1,2),tan23.517.510.899950.9,0.445427.77820.90.911cos0.9135,sin0.406911Aaapqaacctttt第二步在中选非对角线元素中的主元素阜师院数科院2020/3/9212112cossin03.57.77820.7071cossin0sincos07.778217.50sincos00010.707100.0010.035800.6459020.96530.28770.64590.28770.5TAUAU2(2)(2)1313(2)(2)1133222:210.6459(1,3),tan20.03580.510.4148,0.667820.64590.41480.414811cos0.8316,sin0.555411Aaapqaacctttt第三步在中选非对角线元素中的主元素阜师院数科院2020/3/9313213cos0sin0.035800.6459cos0sin010020.96530.2877010sin0cos0.64590.28770.5sin0cos0.46720.159800.159820.96530.239200.23920.9312TAUAU3(3)(3)2323(3)(3)2233222:210.2392(2,3),tan220.96530.9312145.7703,0.010920.239245.770345.770311cos0.9999,sin0.010711Aaapqaacctttt第四步在中选非对角线元素中的主元素阜师院数科院2020/3/94233231000.46720.159801000cossin0.159820.96530.23920cossin0sincos00.23920.93120sincos0.46720.15980.00170.15980.934000.0017020.9669TAUAU4(4)(4)1212(4)(4)1122222:210.1598(1,2),tan20.46720.934014.3842,0.112620.15984.38424.384211cos0.9937,sin0.111911Aaapqaacctttt第五步在中选非对角线元素中的主元素阜师院数科院2020/3/9512412cossin00.46720.15980.0017cossin0sincos00.15980.93400sincos00010.0017020.966900
本文标题:数值分析-Jacobi矩阵特征值和特征向量
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