求向量组 $$ \alpha_{1}=(2,3,-1,4), \alpha_{2}=(-4,-6,2,-8), \alpha_{3}=(0,1,3,2), \alpha_{4}=(4,5,-5,6) $$ 的秩, 并求一个极大线性无关组.
解：
$$ \begin{aligned} A &=\left(\begin{array}{cccc} 2 & -4 & 0 & 4 \\ 3 & -6 & 1 & 5 \\ -1 & 2 & 3 & -5 \\ 4 & -8 & 2 & 6 \end{array}\right) \rightarrow\left(\begin{array}{cccc} 2 & -4 & 0 & 4 \\ 0 & 0 & 1 & -1 \\ 0 & 0 & 3 & -3 \\ 0 & 0 & 2 & -2 \end{array}\right) \\ & \rightarrow\left(\begin{array}{cccc} 2 & -4 & 0 & 4 \\ 0 & 0 & 1 & -1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right) \end{aligned} $$求齐次线性方程组 $\left\{\begin{array}{c}x_{1}-x_{3}+x_{5}=0 \\ x_{2}-x_{4}=0 \\ x_{1}-x_{2}+x_{5}=0 \\ x_{1}-x_{4}+x_{5}=0\end{array}\right.$ 的解空间的一个基.
解: $A=\left(\begin{array}{ccccc}1 & 0 & -1 & 0 & 1 \\ 0 & 1 & 0 & -1 & 0 \\ 1 & -1 & 0 & 0 & 1 \\ 1 & 0 & 0 & -1 & 1\end{array}\right) \rightarrow\left(\begin{array}{ccccc}1 & 0 & 0 & -1 & 1 \\ 0 & 1 & 0 & -1 & 0 \\ 0 & 0 & 1 & -1 & 0 \\ 0 & 0 & 0 & 0 & 0\end{array}\right), r(A)=3$ $$ \left\{\begin{array} { c } { x _ { 1 } - x _ { 4 } + x _ { 5 } = 0 } \\ { x _ { 2 } - x _ { 4 } = 0 } \\ { x _ { 3 } - x _ { 4 } = 0 } \\ { x _ { 5 } = x _ { 5 } } \end{array} \Rightarrow \left\{\begin{array}{l} x_{1}=x_{4}-x_{5} \\ x_{2}=x_{4} \\ x_{3}=x_{4} \\ x_{4}=x_{4} \\ x_{5}=x_{5} \end{array} \Rightarrow\left(\begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \\ x_{5} \end{array}\right)=\left(\begin{array}{c} 1 \\ 1 \\ 1 \\ 1 \\ 0 \end{array}\right) x_{4}+\left(\begin{array}{c} -1 \\ 0 \\ 0 \\ 0 \\ 1 \end{array}\right) x_{5}\right.\right. $$ 解空间一个向量基为 $\xi_{1}=(1,1,1,1,0)^{T}, \quad \xi_{2}=(-1,0,0,0,1)^{T}$.
设 $\beta_{1}=\alpha_{1}+\alpha_{32} \beta_{2}=\alpha_{1}+\alpha_{4} \quad \beta_{3}=\alpha_{3}+\alpha_{4} \beta_{4}=\alpha_{2}+\alpha_{3}$, 证明 $\beta_{1}, \beta_{2}, \beta_{3}, \beta_{4}$ 线性相关.
证: $\because \beta_{2}-\beta_{1}=\beta_{3}-\beta_{4}$
$\therefore-\beta_{1}+\beta_{2}-\beta_{3}+\beta_{4}=0$
$\therefore$ 存在一组不全为零数 $-1,1,-1,1$ 使上式成立 由线性相关定义知 $\beta_{1}, \beta_{2}, \beta_{3}, \beta_{4}$ 线性相关.