If β, β+α 1, …, β+αs are linearly related, and some numbers k0, k 1, …, ks are not all zero, then,

k0β+k 1(β+α 1)+…+ks(β+αs)= 0。

Multiply a by both sides of the above formula at the same time, and note that α 1, …, αs is the basic solution system, and α I = 0 (I = 1, 2, …, s), we get

(k0+k 1+…+ks)Aβ=0。

To k0+k 1+…+ks≠0 (because k0, k 1 …, ks are not all zero), there is a β≠0 from the topic, which is contradictory!

Therefore, β, β+α 1, …, β+αs are linearly independent.