求2个矩阵的特征值和特征向量

2025-04-07 01:59:10
推荐回答(1个)
回答1:

1,
0 = det[A - aI] =
[-a,1/2,-3/2]
[1/2,2-a,-1/2]
[-3/2,-1/2,-a]
=
[-a,1/2,-3/2]
[1/2,2-a,-1/2]
[-a-3/2,0,-a-3/2]
=-(a+3/2)*
[-a,1/2,-3/2]
[1/2,2-a,-1/2]
[1,0,1]
=-(a+3/2)*
[0,1/2,a-3/2]
[0,2-a,-1]
[1,0,1]
=-(a+3/2)*[-1/2-(2-a)(a-3/2)] = (a+3/2)[1/2+(1-a+1)(a-1-1/2)]
=(a+3/2)[1/2 - (1-a)^2 + a-1 - 1/2 - 1/2(1-a)]
=(a+3/2)(a-1)[-a+1 + 1 + 1/2]
=(a+3/2)(a-1)(5/2-a)
a= -3/2, 1, 5/2.

0 = [A + 3/2I]X,
3x/2 + y/2 - 3z/2 = 0,
x/2 + y/2 - z/2 = 0,
-3x/2 - y/2 + 3z/2 = 0.
x=z,y=0. [1,0,1]^T是对应于特征值-3/2的特征向量。

0 = [A-I]X,
-x + y/2 - 3z/2 = 0,
x/2 + y - z/2 = 0,
-3x/2 - y/2 - z = 0.
y=z=-x. [-1,1,1]^T是对应于特征值1的特征向量。

0 = [A-5/2I]X,
-5x/2 + y/2 - 3z/2 = 0,
x/2 - y/2 - z/2 = 0,
-3x/2 - y/2 - 5z/2 = 0.
y=2x,z=-x. [1,2,-1]^T是对应于特征值1的特征向量。

2,
0 = det[A - aI] =
[-a,-1,0]
[1,-a,1]
[0,-1,-a]
= -a^3 - a - a = -a[a^2 + 2]
a = 0,i2^(1/2),-i2^(1/2).

0 = [A-0I]X,
-y = 0,
x + z = 0,
-y=0,
x=-z,y=0.[1,0,-1]^T是对应于特征值0的特征向量。

0 = [A-i2^(1/2)I]X,
- i2^(1/2)x -y = 0,
x - i2^(1/2)y + z = 0,
-y - i2^(1/2)z =0,
x=z,y=-i2^(1/2)z.[1,-i2^(1/2),1]^T是对应于特征值i2^(1/2)的特征向量。

0 = [A+i2^(1/2)I]X,
i2^(1/2)x -y = 0,
x + i2^(1/2)y + z = 0,
-y + i2^(1/2)z =0,
x=z,y=i2^(1/2)z.[1,i2^(1/2),1]^T是对应于特征值-i2^(1/2)的特征向量。

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