Question

证明:如果向量组α1α2α3线性无关，那么向量组a1α1+b2α2，a2α2+b3α3，a3α3+b1α1线性无关的充分必要条件是:a1a2a3≠b1b2b3

Answer 1

问：证明:如果向量组α1α2α3线性无关，那么向量组a1α1+b2α2，a2α2+b3α3，a3α3+b1α1线旦裤性无模竖简关的充分必要纤渣条件是:a1a2a3≠b1b2b3

答：设向量组 $\begin{aligned} \alpha_1a_1 + \alpha_2b_2, \alpha_2a_2 + \alpha_3b_3, \alpha_3a_3 + \alpha_1b_1 \end{aligned}$的任何非零线弯档性组合都等于零向量，即存在不全为零的常数$k_1, k_2, k_3$，使得： $\begin{aligned} k_1(\alpha_1a_1 + \alpha_2b_2) + k_2(\alpha_2a_2 + \alpha_3b_3) + k_3(\alpha_3a_3 + \alpha_1b_1) = 0 \end{aligned}$将上式展开，得到： $\begin{aligned} (k_1a_1 + k_3b_1)\alpha_1 + (k_1b_2 + k_2a_2)\alpha_2 + (k_2b_3 + k_3a_3)\alpha_3 = 0 \end{aligned}$由于向量组$\alpha_1, \alpha_2, \alpha_3$线性无关，因此上式中所有系数都应该为零。所以我们可以得枣闹没到如下三个等式： $\begin{aligned} k_1a_1 + k_3b_1 = 0 \\ k_1b_2 + k_2a_2 = 0 \\ k_2b_3 + k_3a_3 = 0 \end{aligned}$解得： $\begin{aligned} k_1 = -\frac{b_2b_3}{a_1a_2a_3} \\ k_2 = -\frac{a_1a_3}{b_1b_2b_3} \\ k_3 = -\frac{b_1a_2}{a_1a_2a_3} \end{aligned}$由于向量组$a_1\alpha_1+b_2\alpha_2, a_2\alpha_2+b_3\alpha_3, a_3\alpha_3+b_1\alpha_1$是线性无关的，因此$k_1, k_2, k_3$必须全为零。代入上式可以得到： $\begin{aligned} -\frac{b_2b_3}{a_1a_2a_3} = 0 \\ -\frac{a_1a_3}{b_1b_2b_3} = 0 \\ -\frac{b_1a_2}{a_1a_2a_3} = 0 \end{aligned}$因此，$a_{1}a_{2}a_{3} \neq b_{1}b_{2}b_{3}$是向量组$a_{1}\alpha_{1}+b_{2}\alpha_{2}, a_{2}\alpha_{2}+b_{3}\alpha_{3}, a_{3}\alpha_{3}+b_{1}\alpha_{1}$线性无关的凳纳充分必要条件。