Question

若矩阵A、B均为m×n矩阵，如何证明r(A-B)≤r(A)+r(B)

A=(a1,a2,...,an)
B=(b1,b2,...bn)

Answer 1

设r(A)=s，r(B)=t

则a1,a2,...,an的极大无关组为ai1,ai2,...,ais
b1,b2,...,bn的极大无关组为bj1,bj2,...,bjt
因为a1,a2,...,an可由ai1,ai2,...,ais线性表示，b1,b2,...,bn可由bj1,bj2,...,bjt线性表示
所以a1-b1,a2-b2,...,an-bn可由ai1,ai2,...,ais,bj1,bj2,...,bjt线性表示
即a1-b1,a2-b2,...,an-bn的极大无关组也可由ai1,ai2,...,ais,bj1,bj2,...,bjt线性表示
r(a1-b1,a2-b2,...,an-bn)<=r(ai1,ai2,...,ais,bj1,bj2,...,bjt)<=s+t
即r(A-B)<=r(A)+r(B)

Answer 2

取A=(a,a,,an),B=(b,b,bn),其中a,aan为