n阶方阵A,B满足矩阵方程AB=0,则秩R(A)+R(B)____n(≤,≥,<,>,=)

问题描述:

n阶方阵A,B满足矩阵方程AB=0,则秩R(A)+R(B)____n(≤,≥,<,>,=)

答案是 ≤
假设 A = O ,B = O
显然满足题意 AB = O
此时 R(A) + R(B) = 0 假设 A = E,B = O
显然也满足题意 AB = O
此时 R(A)+R(B) = n
综上 R(A)+R(B) ≤ n