Question

线性代数问题,已知A为2n+1阶正交矩阵且|A|=1,证A必有特征值1

Answer 1

用A'表示A的转置,E表示单位阵.

由A为正交阵,有A'A = E.

于是|E-A| = |A'A-A|

= |(A'-E)A|

= |A'-E|·|A|

= |A'-E| (∵|A| = 1)

= |(A-E)'| (∵E' = E)

= |A-E|

= |-(E-A)|

= (-1)^(2n+1)·|E-A|

= -|E-A|.

因此|E-A| = 0,即1是A的一个特征值.