则a=aa=.a(1) 又∵A是实对称矩阵,∴A=A且AT=A, ∴Aa=A·a=4.a(2) 由(12)有·a=4·a,等号两边同时左乘a 左边=a·(几·a)=元.a·a 右边=a1(A.a)=a!·Ar.a=(a)1·a =(4a)·a=4.a1.a ·a·c=A·a1:a 即(2-元)·a·a=0则 A    = =  (1) 又 A 是实对称矩阵，  = A A 且 . T A A =  =   A A A    = (2) 由(1)(2)有      = , A 等号两边同时左乘 T  左边 ( ) T T =   =         右边 ( ) ( ) ( ) T T T T T T       A A A      =   =   =  =  =   T T    =         即 ( ) 0 T     −   =