试题与答案

设A=aβT+βαT.其中α,β是三维单位正交列向量. (1)求|A|. (2)

题型:问答题

题目:

设A=aβT+βαT.其中α,β是三维单位正交列向量.
(1)求|A|.
(2)验证α+β,α-β是A的特征向量.
(3)证A~Λ,并求Λ.

答案:

参考答案:(Ⅰ)r(A)=r(aβT+βαT)≤r(A)+r(B)≤ 2,故|A3×3|=0。
(Ⅱ)α,β是三维单位正交列向量.故有
(α,β)=αTβ=βTα=0.
(α,α)=(αTα)=1,(β,β)=βTβ=1。
A(α+β)=(aβT+βαT)(α+β)=α(βαT)+β(αTα)+α(βTβ)+β(αTβ)=(α+β).
A(α+β)=(aβT+βαT)(α-β)=α(βαT)-α(βTβ)+β(αTα)-β(αTβ)
=-(α-β).
故α+β,α-β是A的分别属于λ1=1.λ2=-1的特征向量.
(Ⅲ)因|A|=0,故知λ3=0,从而知A有三个互不相同的特征值,故A~Λ,其中
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