证明; ∵r(Ⅰ)=r(Ⅱ)=r(Ⅲ)=s
∴a1,a2,..,as线性无关③;β可以a1,a2,..,as表示,即存在不全为0的ki使β=k1a1+k2a2+...+ksas①;a1,a2,..,as,γ线性无关④.
假设m1a1+m2a2+..+msas+m(s+1)*(γ-β)=0②
①代入②变形得,n1a1+n2a2+...+nsas+m(s+1)*γ=0
∵④
∴n1=n2=...=ns=m(s+1)=0
代入②得,m1a1+m2a2+...+msas=0
∵③
∴m1=m2=...=ms=0
∴对于②,m1=m2=...=ms=m(s+1)=0
∴向量组α1,……αs,γ-β线性无关