证明:由已知|A| = 1,n为奇数阶,且 AA'=E.
所以有
|A-E| = |A-AA'| = |A(E-A')| = |A||E-A'| = |(E-A)'| = |E-A|
= |-(A-E)| = (-1)^n|A-E| = -|A-E|.
所以 |A-E| = 0.
所以 1是A的一个特征值.