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1、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, Norman Foster and Partners and Arup Associates),I decide that I have to write something today, otherwise I would not know how to speak English here. Th
2、is is a very quick story about a bridge. London launched three major construction projects to celebrate the arrival of the Millennium. After all, Greenwich (pronounced green-ich) is supposed to be (supposed to be?!) where the prime meridian lies, and the place where the Millennium officially starts
3、in the world. The three projects are the Millennium Dome in North Greenwich, so far the largest single roofed structure in the world, London Eye right across Westminster, which becomes so far the largest observation wheel in the world, and the Millennium Bridge that links Southeast London with St. P
4、auls Cathedral, which is currentlywell.not swinging any more, it is said.,The bridge was designed by Imperial College, a college of my former university. On the very first day that the bridge was open to public, there were simply so many people going there to walk from the south bank to St. Pauls th
5、at the weight completely exceeded the architects expectation. The slender steel truss bridge began to vibrate with a million people on there. The opening ceremony ended up in an embarrassing vertigo. Millennium left Londoners a happy adage about swinging bridge, meaning fancy technology that looks g
6、ood but functions in a funny fashion. Am I using too many Fs here? Or is it simply because my tongue starts to swing in the same direction when I am writing about this wobbly bridge? Next time you visit London, I strongly recommend this place. After all, with a little swing, this is a shortcut to da
7、sh into St. Pauls directly from the southeast!,G: Google Matrix, “the worlds largest matrix computation”. 4,300,000,000 x: PageRank（网页级别） vector “The $25,000,000,000 Eigenvector”,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，
8、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 1 2 n ，且 | 2 | | n |。,p = ( 2 + n ) / 2,思路,令 B = A pI ，则有 | IA | = | I(B+pI) | = | (p)IB | A p = B 。而 ，所以求B的特征根收敛快。,3. 反幂法,所以，A和A1的特征值互为倒数,求A1的按模最大特征值及相应的特征向量， 就可以求出A的按模最小特征值及相应的特征向量。,为避免求逆的运算，可以解线性方程组,反幂法：,