数值分析特征值问题

Ch5矩阵特征值与特征向量的计算1.引言,工程实践中有多种振动问题，如桥梁或建筑物的振动，机械机件、飞机机翼的振动，及一些稳定性分析和相关分析可转化为求矩阵特征值与特征向量的问题。,London,England:Millennium(Wobbly)Bridge(1998-2002,NormanFosterandPartnersandArupAssociates),IdecidethatIhavetowritesomethingtoday,otherwiseIwouldnotknowhowtospeakEnglishhere.Thisisaveryquickstoryaboutabridge.LondonlaunchedthreemajorconstructionprojectstocelebratethearrivaloftheMillennium.Afterall,Greenwich(pronouncedgreen-ich)issupposedtobe(supposedtobe?!)wheretheprimemeridianlies,andtheplacewheretheMillenniumofficiallystartsintheworld.ThethreeprojectsaretheMillenniumDomeinNorthGreenwich,sofarthelargestsingleroofedstructureintheworld,LondonEyerightacrossWestminster,whichbecomessofarthelargestobservationwheelintheworld,andtheMillenniumBridgethatlinksSoutheastLondonwithSt.PaulsCathedral,whichiscurrentlywell.notswinginganymore,itissaid.,ThebridgewasdesignedbyImperialCollege,acollegeofmyformeruniversity.Ontheveryfirstdaythatthebridgewasopentopublic,thereweresimplysomanypeoplegoingtheretowalkfromthesouthbanktoSt.Paulsthattheweightcompletelyexceededthearchitectsexpectation.Theslendersteeltrussbridgebegantovibratewithamillionpeopleonthere.Theopeningceremonyendedupinanembarrassingvertigo.MillenniumleftLondonersahappyadageaboutswingingbridge,meaningfancytechnologythatlooksgoodbutfunctionsinafunnyfashion.AmIusingtoomanyFshere?OrisitsimplybecausemytonguestartstoswinginthesamedirectionwhenIamwritingaboutthiswobblybridge?NexttimeyouvisitLondon,Istronglyrecommendthisplace.Afterall,withalittleswing,thisisashortcuttodashintoSt.Paulsdirectlyfromthesoutheast!,G:GoogleMatrix,“theworldslargestmatrixcomputation”.4,300,000,000 x:PageRank（网页级别）vector“The$25,000,000,000Eigenvector”,搜索引擎,2.幂法,设A是n阶矩阵，x是非零列向量.如果有数存在，满足，(1)那么，称x是矩阵A关于特征值的特征向量.,幂法就是一种求矩阵按模最大特征值的方法.,幂法要求A有完备的特征向量系。即A有n个线性无关的特征向量。在实践中，常遇到的实对称矩阵和特征值互不相同的矩阵就具有这种性质。设A的特征值和特征向量如下：,特征值：,特征向量：,幂法可以求,，基本思想很简单.,设,线性无关，取初值,，作迭代,设：,则有：,利用,则k足够大时，有,可见,几乎仅差一个常数,所以,任意分量相除,特征向量乘以任意数，仍是特征向量,于是，可得到算法：,1.给出初值，计算序列,2.若序列相邻两个向量各个分量比趋向于常数，则,求矩阵A的按模最大的特征值,解取x(0)=(1,0)T,计算x(k)=Ax(k-1),结果如下,例,可取0.41263,x1(0.017451,0.014190)T.,在幂法中，我们构造的序列,可以看出,因此，若序列收敛慢的话，可能造成计算的溢出或归0.,定理,规范化,决定收敛的速度，特别是|2/1|,希望|2/1|越小越好。,不妨设12n，且|2|n|。,p=(2+n)/2,思路,令B=ApI，则有|IA|=|I(B+pI)|=|