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
问题描述：
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
答
|E-A|=|E-A||A^T|=|A^T-E|=|A-E|=-|E-A|
答
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.