高等代数 | 向量组、方程组与线性空间 | 由基础解系反解方程组

向量组、方程组与线性空间

由基础解系反解方程组

(东华大学,2021)已知向量组
${ \alpha_{1}=(1,3,-2,2,0)^{\prime},\alpha_{2}=(1,-3,2,0,4)^{\prime},\alpha_{3}=(3,3,-2,4,4)^{\prime} . }$
记 ${M=L\left(\alpha_{1},\alpha_{2},\alpha_{3}\right)}$ 为 ${\alpha_{1},\alpha_{2},\alpha_{3}}$ 生成的子空间.

求一个以 ${M}$ 为解空间的齐次线性方程组 (I);

求一个导出组为 (I),有一个特解为 ${\alpha_{0}=(1,-3,3,0,0)^{\prime}}$ 的非齐次线性方程组 (II)

solution

首先记 ${A=\left(\alpha_{1},\alpha_{2},\alpha_{3}\right)}$ ,对 ${A^{\prime}}$ 进行初等行变换,化为阶梯形,有
$A^{\prime}=\left(\begin{array}{ccccc} 1 & 3 & -2 & 2 & 0 \\ 1 & -3 & 2 & 0 & 4 \\ 3 & 3 & -2 & 4 & 4 \end{array}\right) \rightarrow\left(\begin{array}{ccccc} 1 & 3 & -2 & 2 & 0 \\ 0 & -6 & 4 & -2 & 4 \\ 0 & -6 & 4 & -2 & 4 \end{array}\right) \rightarrow\left(\begin{array}{ccccc} 1 & 0 & 0 & 1 & 2 \\ 0 & 3 & -2 & 1 & -2 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right)$
由此可知 ${r(A)=2}$ ,而明显 ${\alpha_{1},\alpha_{2}}$ 线性无关,所以 ${\alpha_{1},\alpha_{2}}$ 为 ${M}$ 的一组基.另外,根据上述阶梯形可知方程组 ${A^{\prime} X=0}$ 的基础解系为
${ \eta_{1}=(0,2,3,0,0)^{\prime},\eta_{2}=(-3,-1,0,3,0)^{\prime},\eta_{3}=(-6,2,0,0,3)^{\prime} }$
记 ${B=\left(\eta_{1},\eta_{2},\eta_{3}\right)}$ ,则有 ${A^{\prime} B=O}$ ,从而 ${B^{\prime} A=O}$ ,这说明 ${A}$ 的列向量 ${\alpha_{1},\alpha_{2},\alpha_{3}}$ 均为方程组 ${B^{\prime} X=0}$ 的解,而明显 ${r(B)=3}$ ,所以方程组 ${B^{\prime} X=0}$ 的基础解系当中含有 ${5-3=2}$ 个向量,而 ${\alpha_{1},\alpha_{2}}$ 线性无关,所以它们构成 ${B^{\prime} X=0}$ 的基础解系,也就是说方程组 ${(I): B^{\prime} X=0}$ ,即
${ (I)\left\{\begin{array}{l} 2 x_{2}+3 x_{3}=0 \\ -3 x_{1}-x_{2}+3 x_{4}=0 \\ -6 x_{1}+2 x_{2}+3 x_{5}=0 \end{array}\right. }$
的解空间为 ${M}$ .
当 ${x_{1}=1,x_{2}=-3,x_{3}=3,x_{4}=x_{5}=0}$ 时,有
${ 2 x_{2}+3 x_{3}=3 ; -3 x_{1}-x_{2}+3 x_{4}=0 ; -6 x_{1}+2 x_{2}+3 x_{5}=-12 . }$
于是方程组
${ (I I)\left\{\begin{array}{l} 2 x_{2}+3 x_{3}=3 \\ -3 x_{1}-x_{2}+3 x_{4}=0 \\ -6 x_{1}+2 x_{2}+3 x_{5}=-12 . \end{array}\right. }$
的导出组为 ${(I)}$ ,同时以 ${\alpha_{0}=(1,-3,3,0,0)^{\prime}}$ 为特解.

(合肥工业大学,2020)设 ${W=L\left(\alpha_{1},\alpha_{2},\alpha_{3}\right)}$ ,其中
${ \alpha_{1}=(1,2,-1,0,4)^{\prime},\alpha_{2}=(-1,3,2,4,1)^{\prime},\alpha_{3}=(2,9,-1,4,13)^{\prime} }$

求以 ${W}$ 作为其解空间的齐次线性方程组;

求以 ${W_{1}=\{\eta+\alpha | \alpha \in W\}}$ 为解集的非齐次线性方程组,其中 ${\eta=(1,2,1,2,1)^{\prime}}$ .

solution

首先记 ${A=\left(\alpha_{1},\alpha_{2},\alpha_{3}\right)}$ ,则
${ A^{\prime}=\left(\begin{array}{c} \alpha_{1}^{\prime} \\ \alpha_{2}^{\prime} \\ \alpha_{3}^{\prime} \end{array}\right)=\left(\begin{array}{ccccc} 1 & 2 & -1 & 0 & 4 \\ -1 & 3 & 2 &4 & 1 \\ 2 & 9 & -1 & 4 & 13 \end{array}\right) . }$
对 ${A^{\prime}}$ 进行初等行变换化为阶梯形有
${ A^{\prime} \longrightarrow\left(\begin{array}{ccccc} 1 & 2 & -1 & 0 & 4 \\ 0 &5 & 1 & 4 & 5 \\ 0 &0 &0 & 0 & 0 \end{array}\right) }$
因此 ${r\left(A^{\prime}\right)=2}$ ,结合 ${\alpha_{1},\alpha_{2}}$ 线性无关可知 ${\alpha_{1},\alpha_{2}}$ 是 ${\alpha_{1},\alpha_{2},\alpha_{3}}$ 的一个极大线性无关组,从而 ${W=L\left(\alpha_{1},\alpha_{2}\right)}$ .同时根据阶梯形,可得方程组 ${A^{\prime} X=0}$ 的一组基础解系为
${ \eta_{1}=(7,-1,5,0,0)^{\prime},\eta_{2}=(8,-4,0,5,0)^{\prime},\eta_{3}=(-2,-1,0,0,1)^{\prime} . }$
此时显然有 ${A^{\prime} \eta_{i}=0(i=1,2,3)}$ ,记 ${B=\left(\eta_{1},\eta_{2},\eta_{3}\right)}$ ,即 ${A^{\prime} B=O}$ ,于是取转置可知 ${B^{\prime} A=O}$ ,这说明 ${A}$ 的列向量(即 ${\alpha_{1},\alpha_{2},\alpha_{3}}$ ) 都是方程组 ${B^{\prime} X=0}$ 的解,而显然 ${r\left(B^{\prime}\right)=3}$ ,从而 ${B^{\prime} X=0}$ 的基础解系中含有 ${5-r\left(B^{\prime}\right)=2}$ 个向量,且 ${\alpha_{1},\alpha_{2}}$ 已经是 ${B^{\prime} X=0}$ 的两个线性无关的解向量,从而 ${\alpha_{1},\alpha_{2}}$ 就是 ${B^{\prime} X=0}$ 的基础解系,这说明 ${W}$ 就是 ${B^{\prime} X=0}$ 的解空间.故 ${B^{\prime} X=0}$ 即
${ \left\{\begin{array}{l} 7 x_{1}-x_{2}+5 x_{3}=0 \\ 8 x_{1}-4 x_{2}+5 x_{4}=0 \\ -2 x_{1}-x_{2}+x_{5}=0 \end{array}\right. }$
是满足条件的一个方程组.
当 ${x_{1}=1,x_{2}=2,x_{3}=1,x_{4}=2,x_{5}=1}$ 时,有
${ 7 x_{1}-x_{2}+5 x_{3}=10,8 x_{1}-4 x_{2}+5 x_{4}=10,-2 x_{1}-x_{2}+x_{5}=-3 }$
从而 ${\eta=(1,2,1,2,1)^{\prime}}$ 就是方程组
${ \left\{\begin{array}{l} 7 x_{1}-x_{2}+5 x_{3}=10 \\ 8 x_{1}-4 x_{2}+5 x_{4}=10 \\ -2 x_{1}-x_{2}+x_{5}=-3 . \end{array}\right. }$
的一个特解,而由第一问可知此方程组导出组的解空间为 ${W}$ ,从而根据非齐次线性方程组解的性质可知 ${W_{1}=\{\eta+\alpha | \alpha \in W\}}$ 就是上述方程组的解集.

求线性空间的和与交

(浙江工商大学,2020)已知向量组
${ \begin{array}{c} \alpha_{1}=(1,2,-1,-2),\alpha_{2}=(3,1,1,1),\alpha_{3}=(-1,0,1,-1) \\ \beta_{1}=(2,5,-6,-5),\beta_{2}=(-1,2,-7,3) \end{array} }$
记子空间 ${V_{1}=L\left(\alpha_{1},\alpha_{2},\alpha_{3}\right)}$ 及 ${V_{2}=L\left(\beta_{1},\beta_{2}\right)}$ .求 ${V_{1}+V_{2}}$ 及 ${V_{1} \cap V_{2}}$ 的维数和基.

solution
首先记矩阵 ${A=\left(\alpha_{1}^{\prime},\alpha_{2}^{\prime},\alpha_{3}^{\prime},\beta_{1}^{\prime},\beta_{2}^{\prime}\right)}$ ,对 ${A}$ 进行初等行变换,化为阶梯形,有
$\begin{aligned} A &=\left(\begin{array}{ccccc} 1 & 3 & -1 & 2 & -1 \\ 2 & 1 & 0 & 5 & 2 \\ -1 & 1 & 1 & -6 & -7 \\ -2 & 1 & -1 & -5 & 3 \end{array}\right) \rightarrow\left(\begin{array}{ccccc} 1 & 3 & -1 & 2 & -1 \\ 0 & -5 & 2 & 1 & 4 \\ 0 & 4 & 0 & -4 & -8 \\ 0 & 7 & -3 & -1 & 1 \end{array}\right) \\ & \rightarrow\left(\begin{array}{ccccc} 1 & 3 & -1 & 2 & -1 \\ 0 & 1 & 0 & -1 & -2 \\ 0 & 0 & 2 & -4 & -6 \\ 0 & 0 & -3 & 6 & 15 \end{array}\right) \rightarrow\left(\begin{array}{ccccc} 1 & 3 & -1 & 2 & -1 \\ 0 & 1 & 0 & -1 & -2 \\ 0 & 0 & 1 & -2 & -3 \\ 0 & 0 & 0 & 0 & 6 \end{array}\right) \end{aligned}$
由此可知 ${\alpha_{1},\alpha_{2},\alpha_{3},\beta_{2}}$ 为向量组 ${\alpha_{1},\alpha_{2},\alpha_{3},\beta_{1},\beta_{2}}$ 的极大线性无关组,而明显 ${V_{1}+V_{2}=L\left(\alpha_{1},\alpha_{2},\alpha_{3},\beta_{1},\beta_{2}\right)}$ ,所以 ${\alpha_{1},\alpha_{2},\alpha_{3},\beta_{2}}$ 为 ${V_{1}+V_{2}}$ 的一组基,且 ${\operatorname{dim}\left(V_{1}+V_{2}\right)=3}$ .
另外,对任意的 ${\alpha \in V_{1} \cap V_{2}}$ ,不妨设
${ \alpha=k_{1} \alpha_{1}+k_{2} \alpha_{2}+k_{3} \alpha_{3}=l_{1} \beta_{1}+l_{2} \beta_{2} . }$
则有
${ k_{1} \alpha_{1}^{\prime}+k_{2} \alpha_{2}^{\prime}+k_{3} \alpha_{3}^{\prime}-l_{1} \beta_{1}^{\prime}-l_{2} \beta_{2}^{\prime}=0 . }$
将上式看作关于 ${k_{1},k_{2},k_{3},-l_{1},-l_{2}}$ 的线性方程组,则其系数矩阵为 ${A}$ ,而根据上述的阶梯形可知方程组 ${A X=0}$ 的通解为
${ \left(k_{1},k_{2},k_{3},-l_{1},-l_{2}\right)^{\prime}=k(-3,1,2,1,0)^{\prime} . }$
其中 ${k}$ 为任意常数,由此可知 ${l_{1}=-k,l_{2}=0}$ ,所以 ${\alpha=-k \beta_{1}}$ .这说明 ${\beta_{1}}$ 即为 ${V_{1} \cap V_{2}}$ 的基,且 ${\operatorname{dim}\left(V_{1} \cap V_{2}\right)=1}$ .

(武汉理工大学,2021)设向量组
${ \begin{array}{c} A: \alpha_{1}=(1,0,2),\alpha_{2}=(1,1,3),\alpha_{3}=(1,-1,a+2) ; \\ B: \beta_{1}=(1,2,a+3),\beta_{2}=(2,1,a+6),\beta_{3}=(2,1,a+4) . \end{array} }$

求 ${a}$ 满足什么条件时,向量组 ${A,B}$ 等价;

求 ${a}$ 满足什么条件时,使得向量组 ${A,B}$ 不等价;

记向量组 ${A,B}$ 生成的子空间分别为 ${W_{1},W_{2}}$ ,当 ${A,B}$ 不等价时,求 ${W_{1}+W_{2}}$ 的基与维数.

solution
首先记 ${P=\left(\alpha_{1}^{\prime},\alpha_{2}^{\prime},\alpha_{3}^{\prime}\right),Q=\left(\beta_{1}^{\prime},\beta_{2}^{\prime},\beta_{3}^{\prime}\right)}$ ,对 ${P,Q}$ 进行初等行变换,化为阶梯形,有
$\begin{array}{c} P=\left(\begin{array}{ccc} 1 & 1 & 1 \\ 0 & 1 & -1 \\ 2 & 3 & a+2 \end{array}\right) \rightarrow\left(\begin{array}{ccc} 1 & 1 & 1 \\ 0 & 1 & -1 \\ 0 & 1 & a \end{array}\right) \rightarrow\left(\begin{array}{ccc} 1 & 1 & 1 \\ 0 & 1 & -1 \\ 0 & 0 & a+1 \end{array}\right) \\ Q=\left(\begin{array}{ccc} 1 & 2 & 2 \\ 2 & 1 & 1 \\ a+3 & a+6 & a+4 \end{array}\right) \rightarrow\left(\begin{array}{ccc} 1 & 2 & 2 \\ 0 & -3 & -3 \\ 0 & -a & -a-2 \end{array}\right) \rightarrow\left(\begin{array}{ccc} 1 & 2 & 2 \\ 0 & 1 & 1 \\ 0 & 0 & 2 \end{array}\right) . \end{array}$
由此可知 ${\beta_{1},\beta_{2},\beta_{3}}$ 线性无关.

当 ${a \neq-1}$ 时,显然 ${P,Q}$ 可逆,从而向量组 ${A,B}$ 等价.
当 ${a=-1}$ 时,由于 ${r(P)=2,r(Q)=3}$ ,所以此时向量组 ${A,B}$ 不等价.
由 (2) 可知 ${a=-1}$ ,此时对 ${(P,Q)}$ 进行初等行变换,化为阶梯形,有
$(P,Q)=\left(\begin{array}{cccccc} 1 & 1 & 1 & 1 & 2 & 2 \\ 0 & 1 & -1 & 2 & 1 & 1 \\ 2 & 3 & 1 & 2 & 5 & 3 \end{array}\right) \rightarrow\left(\begin{array}{cccccc} 1 & 1 & 1 & 1 & 2 & 2 \\ 0 & 1 & -1 & 2 & 1 & 1 \\ 0 & 1 & -1 & 0 & 1 & -1 \end{array}\right) \rightarrow\left(\begin{array}{cccccc} 1 & 1 & 1 & 1 & 2 & 2 \\ 0 & 1 & -1 & 2 & 1 & 1 \\ 0 & 0 & 0 & -2 & 0 & -2 \end{array}\right)$
由此可知 ${\alpha_{1},\alpha_{2},\beta_{1}}$ 是 ${\alpha_{1},\alpha_{2},\alpha_{3},\beta_{1},\beta_{2},\beta_{3}}$ 的极大线性无关组,进而也是 ${V_{1}+V_{2}=L\left(\alpha_{1},\alpha_{2},\alpha_{3},\beta_{1},\beta_{2},\beta_{3}\right)}$ 的一组基,且 ${\operatorname{dim}\left(V_{1}+V_{2}\right)=3}$ .

高等代数 | 向量组、方程组与线性空间 | 由基础解系反解方程组 | 线性空间的和与交