Question

如果A是数域K上n级矩阵,且满足 AA' = I,|A| = -1.证明|I - A| = 0

Answer 1

结论不成立：

取K为实数或有理数域,n=3,A是3阶对角阵A=diag(-1.-1,-1),AA' = I,|A| = -1.

I – A= diag(2,2,2),但|I - A| ≠ 0