A-E =
1 0 0
0 2 2
0 2 2
r3-r2,r2*(1/2)
1 0 0
0 1 1
0 0 0
(A-E)X=0 的基础解系为 a1=(0,1,-1)'.
A-2E =
0 0 0
0 1 2
0 2 1
r3-2r2
0 0 0
0 1 2
0 0 -3
r3*(-1/3),r2-2r3
0 0 0
0 1 0
0 0 1
(A-2E)X=0 的基础解系为 a2=(1,0,0)'.
A-5E =
-3 0 0
0 -2 2
0 2 -2
r1*(-1/3),r3+r2,r2*(-1/2)
1 0 0
0 1 -1
0 0 0
(A-5E)X=0 的基础解系为 a3=(0,1,1)'.
a1,a2,a3 单位化得
b1=(0,1/√2,-1/√2)'
b2=(1,0,0)'
b3=(0,1/√2,1/√2)'
令 Q = (b1,b2,b3),则 Q 是正交矩阵,且
Q^-1AQ = diag(1,2,5).