定理921设A,22,…, Hermite矩阵A的n 个特征值,那么 maX 1<i<n 证:由4=p(A4)=(4)=(p(A)2 因此‖412 maX 1<i<n 又由4e=br(4=m()=∑2 得‖4=∑码定理9.2.1 设 是Hermite矩阵A的n 个特征值，那么 证: 1 2 , ,...,   n 2 1 max i i n A    = 2 1 n F i i A  = =  2 2 2 2 ( ) ( ) ( ( )) H 由 A A A A A = = =    2 1 max i i n A    因此 = 2 2 2 1 ( ) ( ) n H F i i A tr A A tr A  = 又由 = = =  2 1 n F i i A  = 得 = 