设x·α1+y·α2+z·α3+w(kβ1+β2) = 0.
由β1可由α1,α2,α3线性表示,可设β1 = a·α1+b·α2+c·α3,代入得
(x+awk)α1+(y+bwk)α2+(z+cwk)α3+w·β2 = 0.
于是w = 0,否则β2 = -(x/w+ak)α1-(y/w+bk)α2-(z/w+ck)α3被α1,α2,α3线性表示.
带回得x·α1+y·α2+z·α3 = 0.
而由α1,α2,α3线性无关,有x = y = z = 0.
组合系数只有零解,即α1,α2,α3,kβ1+β2线性无关.