Question

矩阵的秩A是m×n矩阵,B是n×s矩阵,AB=O 则r(A)+r(B)≤n怎么证明的?

Answer 1

因为AB = O, 所以B的列向量都是Ax = 0的解.

于是r(B) ≤ n - r(A),

从而r(A)+r(B)≤n.

注:

1. 设B = (B1, B2, ... Bs),

由AB = O得 A(B1, B2, ... Bs) = O,

即(AB1, AB2, ... ABs) = O,

可见B的列向量都是Ax = 0的解.

2. 设r(A) = r, 则AX = 0的基础解系中含有 n - r 个线性无关的解向量,

比如说X_1, X_2, ... X_(n-r)为AX = 0的基础解系.

则B1, B2, ... Bs能由X_1, X_2, ... X_(n-r)线性表示,

因而r(B1, B2, ... Bs) ≤ n-r,

即r(B) ≤ n - r(A).