令
0 = x[2a1 + 3a2] + y[a2 + 4a3] + z[a1 + 5a3]
= a1[2x + z] + a2[3x + y] + a3[4y + 5z]
因 a1,a2,a3线性无关.
所以
2x + z = 0 = 3x + y = 4y + 5z,
z = -2x,
y = -3x,
0 = 4y + 5z = 4(-3x) + 5(-2x) = -22x,
x = 0.
y = -3x = 0,
z = -2x = 0.
所以,
2a1+3a2,a2+4a3,5a3+a1线性无关.