设A为n阶方阵,证明：det(E-A*A)=0,则1或-1至少有一个是A的特征值.

E-A*A=（E-A)*(E+A)det(E-A*A)=det[E-A)*(E+A)]=detE-A)*det(E+A)=0so detE-A)=0 or det(E+A)=0if detE-A)=0,1 is a eigenvalue of matrix Aif detE+A)=0,-1 is a eigenvalue of matrix A