第7章矩陣的特征值和特征向量(完整版)
【正文】 ????????????????0,qppqpqqqpppqppppqipiqiiqqipipiipbbtaabtaabqpicadabbqpidacabb數(shù) 學 系 University of Science and Technology of China DEPARTMENT OF MATHEMATICS Jacobi迭代 取 p,q使 ijjipq aa ?? m a x，則 ),(),( )()1( ?? qpQAqpQA kTk ??定理： 若 A對稱，則 },{ 1)1( nk d ia gA ?? ???數(shù) 學 系 University of Science and Technology of China DEPARTMENT OF MATHEMATICS 解 記 A(0)=A,取 p=1,q=2,apq(0)=a12(0)=2,于是有 例 用 Jacobi 方法計算對稱矩陣 的全部特征值 . ???????????612152224A )0(12)0(22)0(11 ????aaa? )1|/ ( |)s g n (, 2 ????? ???t7 8 8 2 0 )1(c o s 212 ??? ?t? 61 541 o ssi n, ??? ?? t從而有 數(shù) 學 系 University of Science and Technology of China DEPARTMENT OF MATHEMATICS 所以 再取 p=2,q=3,apq(1)=a23(1)=,類似地可得 ?????????????????????? ???10007 8 8 2 0 1 5 4 1 06 1 5 4 1 8 8 2 0 1000c o ssin0sinc o s)(1 ?????pqRR( 1 ) ( 0 )11 0 0 6. 56 15 52 2. 02 01 900. 96 1 2. 02 01 90 6T????????A R A R???????????2 4 1 1 6 2 4 7 9 03 2 0 3 8 3 1 0 2 7 2 4 7 9 3 1 0 2 3 8 4 4 )2(A數(shù) 學 系 University of Science and Technology of China DEPARTMENT OF MATHEMATICS ???????????4 9 6 4 2 0 9 6 1 2 0 9 6 1 2 0 3 8 9 5 1 9 05 9 5 1 9 8 3 1 8 )3(A?????????????4 9 6 4 2 0 8 6 5 2 0 0 4 2 0 8 6 5 7 7 5 7 0 2 0 0 4 2 5 9 9 )4(A???????????????4 8 5 2 3 2 0 0 1 03 8 8 7 6 0 1 0 7 0 2 0 0 1 0 1 0 7 2 5 9 9 )5(A數(shù) 學 系 University of Science and Technology of China DEPARTMENT OF MATHEMATICS ?????????????4 8 5 4 0 0 0 0 0 0 1 0 7 0 0 0 0 0 8 8 7 6 00 0 1 0 7 2 5 8 2 )6(A???????????4 8 5 4 0 0 0 0 0 0 0 0 0 0 8 8 7 6 001 2 5 8 2 )7(A從而 A的特征值可取為 ?1?, ?2?, ?3? 數(shù) 學 系 University of Science and Technology of China DEPARTMENT OF MATHEMATICS 為了減少搜索非對角線絕對值最大元素時間 , 對經(jīng)典的 Jacobi方法可作進一步改進 . Jacobi方法 : 按 (1,2),(1,3),…,(1,n),(2,3),(2,4),…,(2,n),…,(n 1,n)的順序 , 對每個 (p,q)的非零元
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