设A为2n+1阶方阵,且满足AA^T =E,|A|>0,证明行列式|A-E|=_数学解答

设A为2n+1阶方阵,且满足AA^T =E,|A|>0,证明行列式|A-E|=

人气：224 ℃　时间：2019-08-22 14:58:34

解答

|A-E|
= |A-AA^T|
= |A(E-A^T)|
= |A||E-A^T|
= |A||E-A| --- (E-A^T)^T = E-A
= |A| (-1)^(2n+1) |A-E|
= -|A||A-E|
所以 |A-E|(1+|A|)=0
因为 |A|>0
所以 1+|A|≠0
所以 |A-E| = 0.