(2)对每个特征值A,解线性方程组(AE-A)X=0,得到特征子空间的一组基,施行 Schmidt正交 化,单位化,得到特征子空间的一组标准正交基; (3)将不同特征子空间的标准正交基凑成R的标准正交基1, 令T=(1,72,…,mn) 则T-1AT为对角矩阵,对角元素分别是对应于m1,2,…,mn的全部特征值A1,A2,…,A 例2设A是3阶实对称矩阵,A的特征值为2,1,1.已知属于特征值2的特征向量X1=(1,1,0) 又向量X2=(1,-1,0)是属于特征值1的特征向量,求矩阵A. 解由定理834知A有特征向量与X1,X2都正交.设之为X3=(x1,x2,x3) 解得X3=(0,0,1)2.将X1,X2,X3单位化,得 1.0),=(0,0,1) 记 r=012102=(支 001 A=T )-(-)(2,)(-;) 例3设三阶实对称阵A的各行元素之和为3,向量a1=(-1,2,-1),a2=(0,-1,1)是线性方 程组AX=0的两个解 (1)求A的特征值和特征向量; (2)求正交矩阵T和对角阵B,使得T-1AT=TAT=B 3)求A和(4-E)° 解(1)因为A的各行元素之和为3,所以 所以A的属于特征值3的特征向量为ka3=k3(1,1,1),其中k3为非零常数 又因为向量a1=(-1,2,-1),a2=(0,-1,1)是线性方程组AX=0的两个解.所以A的属于 特征值0的特征向量为k1a1+k2a2,其中k1,h2为不全为零的常数 (2)对a1,a2做正交化,得到 单位化得到 (-1,2,-1) 1,0.1)2,7=方(,1,1)(2) *n3V\ λi , V06/d (λE − A)X = 0, ! Vb^L">dE￾ 5 Schmidt WO A￾&A￾! Vb^L">daWOE (3) N!Vb^L"aWOE R n "aWOE γ1, γ2, · · · , γn, j T = (γ1, γ2, · · · , γn), Q T −1AT $*PXU￾*PN1 Æ*FK γ1, γ2, · · · , γn "￾V\ λ1, λ2, · · · , λn. p 2  A Æ 3 S*ÆXU￾ A "V\$ 2, 1, 1. YKV\ 2 "V3f X1 = (1, 1, 0)T , J3f X2 = (1, −1, 0)T ÆKV\ 1 "V3f￾~XU A. n H'b 8.3.4 Y A I>V3fL X1, X2 (WOZ$ X3 = (x1, x2, x3) T , !  x1 + x2 = 0 x1 − x2 = 0 V! X3 = (0, 0, 1)T . N X1, X2, X3 &A￾! Y1 = ( 1 √ 2 , 1 √ 2 , 0)T , Y2 = ( 1 √ 2 , − 1 √ 2 , 0)T , Y3 = (0, 0, 1)T . I T = (Y1, Y2, Y3) =   √ 1 2 √ 1 2 0 √ 1 2 − √ 1 2 0 0 0 1   , Q A = T   2 1 1   T T =   √ 1 2 √ 1 2 0 √ 1 2 − √ 1 2 0 0 0 1     2 1 1     √ 1 2 √ 1 2 0 √ 1 2 − √ 1 2 0 0 0 1   . p 3 S*ÆU A "45NZ=$ 3, 3f α1 = (−1, 2, −1)T , α2 = (0, −1, 1)T Æ06/ d AX = 0 "e3V (1) ~ A "V\=V3f (2) ~WOXU T =*PU B, ! T −1AT = T TAT = B; (3) ~ A = (A − 3 2E) 6 . n (1) D$ A "45NZ=$ 3, A A   1 1 1   = 3   1 1 1   . A A "KV\ 3 "V3f$ k3α3 = k3(1, 1, 1)T , y_ k3 $0h  JD$3f α1 = (−1, 2, −1)T , α2 = (0, −1, 1)T Æ06/d AX = 0 "e3VA A "K V\ 0 "V3f$ k1α1 + k2α2, y_ k1, k2 $￾$h"  (2) * α1, α2 fWOA￾! β1 = α1 = (−1, 2, −1)T , β2 = 1 2 (−1, 0, 1)T . &A! γ1 = 1 √ 6 (−1, 2, −1)T , γ2 = 1 √ 2 (−1, 0, 1)T , γ3 = 1 √ 3 (1, 1, 1)T . 5