Question

线性代数 矩阵证明已知AB=A+B,证：1.(A-I)可逆；2.AB=BA .

Answer 1

有 AB-A-B=0

(A-I)B - A =0

(A-I)B - (A-I) =I

即 (A-I)(B-I)=I

所以A-I,可逆.

故 (A-I)(B-I) = (B-I)(A-I) =I

即有 AB -A -B +I =BA-B-A+I

整理一下有 AB=BA