Question

A是n阶正交矩阵,若A的行列式为1,证明当n为奇数时,E—A的行列式为0

Answer 1

证明:由已知,AA' = E

所以 |E-A|=|AA'-A|

= |A(A'-E)|

= |A||A'-E|

= 1* |(A-E)'|

= |A-E|

= |-(E-A)|

= (-1)^n|E-A|

= - |E-A|.

故 |E-A| = 0.