设A,B为n阶方阵,且AA-AB=I,求秩R(AB-BA+2A)

问题描述:

设A,B为n阶方阵,且AA-AB=I,求秩R(AB-BA+2A)

首先明确一点A是可逆的,如果A不可逆,AA-AB=A(A-B)的秩小于A,那么AA-AB≠E.AA-AB=A(A-B)=E;AAA-ABA=A,所以AA-BA=E.AB-BA+2A=(AB-AA)+(AA-BA)+2A=-E+E+2A=2A.所以R(AB-BA+2A)=R(2A)=R(A)=n.