Question

线性代数高手求教

设a是n阶正交阵，证明，当n为奇数且｜a｜等于1｜e－a｜=0

Answer 1

|A'|=| A*/|A| | =| A* | |E-A| =| A |^(n-1)=|A|=1
|E-A|=|AA'-A|=|(A-E)A'|=|A'||A-E|=|A||A-E|=|A-E|=|-1*(E-A)|=(-1)^n|E-A|= -|E-A|=0