Question

线性无关证明设A是n阶矩阵,b1、b2、b3是n维列向量,若Ab1=b1≠0,Ab2=b1+b2,Ab3=b2+b3,证

Answer 1

设k1b1+k2b2+k3b3=0 (1)

令B=A-E

由已知等式,得

Bb1=0,Bb2=b1, Bb3=b2

(1)式两端同时左乘B,有

k1Bb1+k2Bb2+k3Bb3=0,即 k2b1+k3b2=0 (2)

(2)式两端同时左乘B,有

k3Bb2=0,即k3b1=0

∵b1≠0

∴k3=0

k3=0代入(2),得k2=0

k2=0,k3=0代入(1)中,得k1=0

∴b1.b2,b3线性无关