请问在以下矩阵中，特征值t1=-根号（-ab-ac-bc）和t2=根号（-ab-ac-bc）的特征向量怎么求？

矩阵
a b c
b c a
c a b
特征值为

b+a+c
√(b²-ab-bc-ac+a²+c²)
-√(b²-ab-bc-ac+a²+c²)
对应的特征向量为
V =

[ 1, -(a+(b^2-b*a-c*b-c*a+a^2+c^2)^(1/2)-b)/(a-c), -(a-(b^2-b*a-c*b-c*a+a^2+c^2)^(1/2)-b)/(a-c)]
[ 1, -(b-c-(b^2-b*a-c*b-c*a+a^2+c^2)^(1/2))/(a-c), -(b-c+(b^2-b*a-c*b-c*a+a^2+c^2)^(1/2))/(a-c)]
[ 1, 1, 1]

[ 1, -(a+(b^2-b*a-c*b-c*a+a^2+c^2)^(1/2)-b)/(a-c), -(a-(b^2-b*a-c*b-c*a+a^2+c^2)^(1/2)-b)/(a-c)]
[ 1, -(b-c-(b^2-b*a-c*b-c*a+a^2+c^2)^(1/2))/(a-c), -(b-c+(b^2-b*a-c*b-c*a+a^2+c^2)^(1/2))/(a-c)]
[ 1, 1, 1]