则 A=Q ∞(3:)(2,)(京 001 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)2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Q和对角阵B,使得Q-1AQ=Q74Q=B; (3)求A和(A-是E)° 解(1)因为向量a1=(-1,2,-1)2,a2=(0,-1,1)是线性方程组AX=0的两个线性无关的 解.所以A的属于特征值0的特征向量为k1a1+k2a2,其中k1,k2为不全为零的常数 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)2,其中k3为非零常数; (2)对a1,a2做正交化,得到 B2 1,0,1) 单位化得到 (-1,2,-1)7,2=1 (-1,0,1),8=(1,1,1) 令Q=(1,72,3),则 000 Q AQ=Q AQ=B 000 003 (3)因为 A(a1,a2,a3)=(0,0,3a3) 111 A=00,3a3)(a1,a2,a3)-1=111 记P=(a1,a2,a3),则 000 A=P000|P 所以 P 0 00E A = Q   2 1 1   Q 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   . b 3 ￾~N)I A "3+BN: 3, )a α1 = (−1, 2, −1)T , α2 = (0, −1, 1)T &,. W AX = 0 "`2Q (1) w A "JP:J)a (2) wKJSI Q :)KI B, ! Q−1AQ = QT AQ = B; (3) w A : (A − 3 2E) 6 . ` (1) 8)a α1 = (−1, 2, −1)T , α2 = (0, −1, 1)T &,.W AX = 0 "`2&,6" Q6 A ">JP 0 "J)a k1α1 + k2α2, rR k1, k2 y "Æ 8 A "3+BN: 3, 6 A   1 1 1   = 3   1 1 1   . 6 A ">JP 3 "J)a k3α3 = k3(1, 1, 1)T , rR k3 / Æ  (2) ) α1, α2 XKJ<￾! β1 = α1 = (−1, 2, −1)T , β2 = 1 2 (−1, 0, 1)T . <! γ1 = 1 √ 6 (−1, 2, −1)T , γ2 = 1 √ 2 (−1, 0, 1)T , γ3 = 1 √ 3 (1, 1, 1)T . e Q = (γ1, γ2, γ3), E Q −1AQ = Q T AQ = B =   0 0 0 0 0 0 0 0 3   . (3) 8 A(α1, α2, α3) = (0, 0, 3α3). 6 A = (0, 0, 3α3)(α1, α2, α3) −1 =   1 1 1 1 1 1 1 1 1   . D P = (α1, α2, α3), E A = P   0 0 0 0 0 0 0 0 3   P −1 . 6 (A − 3 2 E) 6 = P   −3 2 0 0 0 −3 2 0 0 0 3 2   6 P −1 = (3 2 ) 6E. 5