高等代数 | 向量组、方程组与线性空间方程组 | 方程组理论问题

向量组、方程组与线性空间

方程组理论问题

(山东大学,2022)已知两个向量组的秩相同,且其中一个可以被另一个线性表出,试证明这两个向量组等价.

proof
设 ${\alpha_{1},\cdots,\alpha_{s}}$ 与 ${\beta_{1},\cdots,\beta_{t}}$ 是秩均为 ${r}$ 的向量组,且 ${\alpha_{1},\cdots,\alpha_{s}}$ 可被 ${\beta_{1},\cdots,\beta_{t}}$ 线性表出.考虑 ${\beta_{1},\cdots,\beta_{t}}$ 与 ${\alpha_{1},\cdots,\alpha_{s},\beta_{1},\cdots,\beta_{t}}$ 这两个向量组,显然它们等价,所以 ${\alpha_{1},\cdots,\alpha_{s},\beta_{1},\cdots,\beta_{t}}$ 的秩也为 ${r}$ ,现设 ${\alpha_{k_{1}},\cdots,\alpha_{k_{r}}}$ 是 ${\alpha_{1},\cdots,\alpha_{s}}$ 的一个极大线性无关组,则 ${\alpha_{k_{1}},\cdots,\alpha_{k_{r}}}$ 也是 ${\alpha_{1},\cdots,\alpha_{s},\beta_{1},\cdots,\beta_{t}}$ 中的 ${r}$ 个线性无关的向量,又由 ${\alpha_{1},\cdots,\alpha_{s},\beta_{1},\cdots,\beta_{t}}$ 的秩为 ${r}$ ,所以 ${\alpha_{k_{1}},\cdots,\alpha_{k_{r}}}$ 是 ${\alpha_{1},\cdots,\alpha_{s},\beta_{1},\cdots,\beta_{t}}$ 的极大线性无关组,从而 ${\alpha_{k_{1}},\cdots,\alpha_{k_{r}}}$ 与 ${\alpha_{1},\cdots,\alpha_{s},\beta_{1},\cdots,\beta_{t}}$ 等价,由传递性知 ${\alpha_{k_{1}},\cdots,\alpha_{k_{r}}}$ 与 ${\beta_{1},\cdots,\beta_{t}}$ 等价,即 ${\alpha_{1},\cdots,\alpha_{s}}$ 与 ${\beta_{1},\cdots,\beta_{t}}$ 等价.

(西南大学,2022)设 ${A,B}$ 均为 ${n}$ 列矩阵,证明: ${n}$ 元齐次线性方程组 ${A X=0}$ 与 ${B X=0}$ 同解当且仅当 ${A,B}$ 的行向量组等价.

proof
必要性.为了方便,记 ${C=\left(\begin{array}{c}A \\ B\end{array}\right)}$ ,由于 ${A X=0}$ 与 ${B X=0}$ 同解,所以 ${A X=0,B X=0}$ 均与 ${C X=0}$ 同解,进而系数矩阵 ${A,B,C}$ 的秩相同,而明显 ${A,B}$ 的行向量组均可由 ${C}$ 的行向量组线性表出,所以结合例题 22 可知 ${A,B}$ 的行向量组均与 ${C}$ 的行向量组等价,由传递性, ${A,B}$ 的行向量组也等价.

充分性.已知 ${A,B}$ 的行向量组等价,所以存在矩阵 ${P,Q}$ ,使得 ${P A=B,Q B=A}$ ,那么若 ${X}$ 满足 ${A X=0}$ ,则有 ${B X=P A X=0}$ ,若 ${X}$ 满足 ${B X=0}$ ,则 ${A X=Q B X=0}$ ,这说明方程组 ${A X=0}$ 与 ${B X=0}$ 同解.

(电子科技大学,2022)若非齐次线性方程组 ${A X=\beta(\beta \neq 0)}$ 有解,齐次线性方程组 ${A X=0}$ 有 ${k(k < n)}$ 个线性无关解,证明: ${A X =\beta}$ 有 ${k+1}$ 个线性无关解,不存在 ${k+2}$ 个线性无关解.

proof
首先设 ${\eta_{0}}$ 为方程组 ${A X=\beta}$ 的一个特解, ${\eta_{1},\eta_{2},\cdots,\eta_{k}}$ 为 ${A X=0}$ 的 ${k}$ 个线性无关解,则

$\eta_{0},\eta_{0}+\eta_{1},\eta_{0}+\eta_{2},\cdots,\eta_{0}+\eta_{k} \quad (1)$

为 ${A X=\beta}$ 的 ${k+1}$ 个解,若 ${l_{0} \eta_{0}+l_{1}\left(\eta_{0}+\eta_{1}\right)+l_{2}\left(\eta_{0}+\eta_{2}\right)+\cdots+l_{k}\left(\eta_{0}+\eta_{k}\right)=0}$ ,即
$\left(l_{0}+l_{1}+\cdots+l_{k}\right) \eta_{0}+l_{1} \eta_{1}+l_{2} \eta_{2}+\cdots+l_{k} \eta_{k}=0 .\quad (2)$
上述等式两边同时被 ${A}$ 作用可得
${ 0=\left(l_{0}+l_{1}+\cdots+l_{k}\right) A \eta_{0}+l_{1} A \eta_{1}+l_{2} A \eta_{2}+\cdots+l_{k} A \eta_{k}=\left(l_{0}+l_{1}+\cdots+l_{k}\right) \beta . }$
由于 ${\beta \neq 0}$ ,所以
${ l_{0}+l_{1}+\cdots+l_{k}=0 . }$
将其代入到(2)式,结合 ${\eta_{1},\eta_{2},\cdots,\eta_{k}}$ 线性无关可得 ${l_{1}=l_{2}=\cdots=l_{k}=0}$ ,进而也有 ${l_{0}=0}$ ,这说明向量组(1)是方程组 ${A X=\beta}$ 的 ${k+1}$ 个线性无关的解向量.另外,若 ${A X=\beta}$ 存在 ${k+2}$ 个线性无关解 ${\alpha_{1},\alpha_{2},\cdots,\alpha_{k+2}}$ ,那么
$\alpha_{1}-\alpha_{k+2},\alpha_{2}-\alpha_{k+2},\cdots,\alpha_{k+1}-\alpha_{k+2} \quad (3)$
均是导出组 ${A X=0}$ 的解,同时若 ${t_{1}\left(\alpha_{1}-\alpha_{k+2}\right)+t_{2}\left(\alpha_{2}-\alpha_{k+2}\right)+\cdots+t_{k+1}\left(\alpha_{k+1}-\alpha_{k+2}\right)=0}$ ,即
${ t_{1} \alpha_{1}+t_{2} \alpha_{2}+\cdots+t_{k+1} \alpha_{k+1}-\left(t_{1}+t_{2}+\cdots+t_{k+1}\right) \alpha_{k+2}=0 . }$
根据 ${\alpha_{1},\alpha_{2},\cdots,\alpha_{k+2}}$ 线性无关可得 ${t_{1}=t_{2}=\cdots=t_{k+1}=0}$ ,即向量组(3)是 ${A X=0}$ 的 ${k+1}$ 个线性无关解,这与已知矛盾.即 ${A X=\beta}$ 不存在 ${k+2}$ 个线性无关解.

(重庆大学,2022)设 ${A=\left(a_{i j}\right)_{n \times n}}$ 是实矩阵,证明: 线性方程组 ${A X=B}$ 有解的充要条件是向量 ${B}$ 与齐次线性方程组 ${A^{T} X=0}$ 的解空间正交.

proof
必要性.由于方程组 ${A X=B}$ 有解,设解为 ${X_{0}}$ ,即 ${B=A X_{0}}$ ,那么对任意满足 ${A^{T} X=0}$ 的 ${X}$ ,有 ${B^{T} X=\left(A X_{0}\right)^{T} X=X_{0}^{T}\left(A^{T} X\right)=0},$ 即 ${B}$ 与齐次线性方程组 ${A^{T} X=0}$ 的解空间正交.

充分性.由于向量 ${B}$ 与齐次线性方程组 ${A^{T} X=0}$ 的解空间正交,即若 ${A^{T} X=0}$ ,则有 ${B^{T} X=0}$ ,这说明方程组 ${A^{T} X=0}$ 与 ${\left(\begin{array}{c}A^{T} \\ B^{T}\end{array}\right) X=0}$ 同解,进而系数矩阵的秩相同,那么取转置就有 ${r(A)=r(A,B)}$ ,所以线性方程组 ${A X=B}$ 有解.

(复旦大学,2022)设 ${\mathbb{K}}$ 为一数域, ${A_{1},A_{2} \in \mathbb{K}^{n \times n},b_{1},b_{2} \in \mathbb{K}^{n \times 1}}$ ,若线性方程组 ${A_{1} X=b_{1}}$ 和 ${A_{2} X=b_{2}}$ 的解集相同,证明: 存在可逆的 ${n}$ 阶方阵 ${P}$ ,使得 ${P A_{1}=A_{2}}$ ,且 ${P b_{1}=b_{2}}$ .

proof
由于 ${A_{1} X=b_{1}}$ 和 ${A_{2} X=b_{2}}$ 的解集相同,所以 ${A_{1} X=b_{1},A_{2} X=b_{2},\left(\begin{array}{l}A_{1} \\ A_{2}\end{array}\right) X=\left(\begin{array}{l}b_{1} \\ b_{2}\end{array}\right)}$ 的解集也相同,进而增广矩阵的秩相同,即
${ r\left(A_{1},b_{1}\right)=r\left(A_{2},b_{2}\right)=r\left(\begin{array}{cc} A_{1} & b_{1} \\ A_{2} & b_{2} \end{array}\right) . }$
而明显 ${\left(A_{1},b_{1}\right),\left(A_{2},b_{2}\right)}$ 的行向量均为 ${\left(\begin{array}{ll}A_{1} & b_{1} \\ A_{2} & b_{2}\end{array}\right)}$ 的行向量,所以前两者的行向量组均与后者行向量组等价,根据传递性便知 ${\left(A_{1},b_{1}\right)}$ 与 ${\left(A_{2},b_{2}\right)}$ 的行向量组等价,现在设 ${\alpha_{1},\alpha_{2},\cdots,\alpha_{r}}$ 与 ${\beta_{1},\beta_{2},\cdots,\beta_{r}}$ 分别为 ${\left(A_{1},b_{1}\right)}$ 与 ${\left(A_{2},b_{2}\right)}$ 行向量组的极大线性无关,则它们等价,即存在 ${r}$ 阶可逆矩阵 ${Q_{1}}$ ,使得
${ Q_{1}\left(\begin{array}{c} \alpha_{1} \\ \vdots \\ \alpha_{r} \end{array}\right)=\left(\begin{array}{c} \beta_{1} \\ \vdots \\ \beta_{r} \end{array}\right) }$
进而
${ \left(\begin{array}{cc} Q_{r} & O \\ O & E_{n-r} \end{array}\right)\left(\begin{array}{c} \alpha_{1} \\ \vdots \\ \alpha_{r} \\ 0 \\ \vdots \\ 0 \end{array}\right)=\left(\begin{array}{c} \beta_{1} \\ \vdots \\ \beta_{r} \\ 0 \\ \vdots \\ 0 \end{array}\right) . }$
另外,根据矩阵的初等变换,还存在 ${n}$ 阶可逆矩阵 ${S}$ 和 ${T}$ ,使得
${ S\left(A_{1},b_{1}\right)=\left(\begin{array}{c} \alpha_{1} \\ \vdots \\ \alpha_{r} \\ 0 \\ \vdots \\ 0 \end{array}\right),T\left(A_{2},b_{2}\right)=\left(\begin{array}{c} \beta_{1} \\ \vdots \\ \beta_{r} \\ 0 \\ \vdots \\ 0 \end{array}\right) . }$
所以
${ T\left(A_{2},b_{2}\right)=\left(\begin{array}{c} \beta_{1} \\ \vdots \\ \beta_{r} \\ 0 \\ \vdots \\ 0 \end{array}\right)=\left(\begin{array}{cc} Q_{r} & O \\ O & E_{n-r} \end{array}\right)\left(\begin{array}{c} \alpha_{1} \\ \vdots \\ \alpha_{r} \\ 0 \\ \vdots \\ 0 \end{array}\right)=\left(\begin{array}{cc} Q_{r} & O \\ O & E_{n-r} \end{array}\right) S\left(A_{1},b_{1}\right) . }$

线性空间的交与和

(北京科技大学,2022)设 ${W_{1}=L\left(f_{1}(x),f_{2}(x),f_{3}(x)\right.}$ ), ${W_{2}=L\left(f_{4}(x),f_{5}(x)\right.}$ ),其中
${ \begin{array}{c} f_{1}(x)=1-3 x-x^{2}+2 x^{3},f_{2}(x)=-1+x+2 x^{2}-x^{3},f_{3}(x)=-1+5 x-3 x^{2} ; \\ f_{4}(x)=-1-2 x+4 x^{2},f_{5}(x)=-14 x+9 x^{2}+5 x^{3} . \end{array} }$

\item 求 ${W_{1} \cap W_{2}}$ 的基与维数;
\item 将 ${W_{1}}$ 的基扩为 ${P[x]_{4}}$ 的基.

solution

\item 首先取 ${P[x]_{4}}$ 的基 ${1,x,x^{2},x^{3}}$ ,显然
${ \left(f_{1}(x),f_{2}(x),f_{3}(x),f_{4}(x),f_{5}(x)\right)=\left(1,x,x^{2},x^{3}\right) A . }$
其中
$A=\left(\begin{array}{ccccc} 1 & -1 & -1 & -1 & 0 \\ -3 & 1 & 5 & -2 & -14 \\ -1 & 2 & -3 & 4 & 9 \\ 2 & -1 & 0 & 0 & 5 \end{array}\right)$
将矩阵 ${A}$ 进行初等行变换化为阶梯形,有
$A \rightarrow\left(\begin{array}{ccccc} 1 & -1 & -1 & -1 & 0 \\ 0 & -2 & 2 & -5 & -14 \\ 0 & 1 & -4 & 3 & 9 \\ 0 & 1 & 2 & 2 & 5 \end{array}\right) \rightarrow\left(\begin{array}{ccccc} 1 & -1 & -1 & -1 & 0 \\ 0 & 1 & 2 & 2 & 5 \\ 0 & 0 & 6 & -1 & -4 \\ 0 & 0 & -6 & 1 & 4 \end{array}\right) \rightarrow\left(\begin{array}{ccccc} 1 & -1 & -1 & -1 & 0 \\ 0 & 1 & 2 & 2 & 5 \\ 0 & 0 & 6 & -1 & -4 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right)$
根据主元位置可知 ${f_{1}(x),f_{2}(x),f_{3}(x)}$ 为 ${f_{1}(x),f_{2}(x),f_{3}(x),f_{4}(x),f_{5}(x)}$ 的极大线性无关组,即 ${f_{4}(x),f_{5}(x)}$ 均可由 ${f_{1}(x),f_{2}(x),f_{3}(x)}$ 线性表出,所以 ${f_{4}(x),f_{5}(x) \in W_{1}}$ ,也就是 ${W_{2} \subseteq W_{1}}$ ,于是 ${W_{1} \cap W_{2}=W_{2}=L\left(f_{4}(x),f_{5}(x)\right)}$ 而明显 ${f_{4}(x),f_{5}(x)}$ 线性无关,所以 ${f_{4}(x),f_{5}(x)}$ 就是 ${W_{1} \cap W_{2}}$ 的一组基, ${\operatorname{dim}\left(W_{1} \cap W_{2}\right)=2}$ .
\item 取 ${g(x)=x^{3}}$ ,则
${ \left(f_{1}(x),f_{2}(x),f_{3}(x),g(x)\right)=\left(1,x,x^{2},x^{3}\right) B . }$
其中
${ B=\left(\begin{array}{cccc} 1 & -1 & -1 & 0 \\ -3 & 1 & 5 & 0 \\ -1 & 2 & -3 & 0 \\ 2 & -1 & 0 & 1 \end{array}\right) }$
将矩阵 ${B}$ 进行初等行变换化为阶梯形,有
$B \rightarrow\left(\begin{array}{cccc} 1 & -1 & -1 & 0 \\ 0 & -2 & 2 & 0 \\ 0 & 1 & -4 & 0 \\ 0 & 1 & 2 & 1 \end{array}\right) \rightarrow\left(\begin{array}{cccc} 1 & -1 & -1 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & -3 & 0 \\ 0 & 0 & 3 & 1 \end{array}\right) \rightarrow\left(\begin{array}{cccc} 1 & -1 & -1 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & -3 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right)$
由此可知矩阵 ${B}$ 可逆,那么 ${f_{1}(x),f_{2}(x),f_{3}(x),g(x)}$ 便构成 ${P[x]_{4}}$ 的基.

(同济大学,2022)设向量
${ \alpha_{1}=(1,-1,1,0){\prime},\alpha_{2}=(1,1,0,2){\prime},\alpha_{3}=(-2,1,1,3){\prime},\alpha_{4}=(2,0,1,2){\prime},\alpha_{5}=(1,2,-2,1){\prime} . }$ ${W_{1}=L\left(\alpha_{1},\alpha_{2}\right),W_{2}=L\left(\alpha_{3},\alpha_{4},\alpha_{5}\right)}$ ,求 ${W_{1}+W_{2}}$ 和 ${W_{1} \cap W_{2}}$ 的维数和一组基.

solution
记矩阵 ${A=\left(\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4},\alpha_{5}\right)}$ ,将 ${A}$ 进行初等行变换化为阶梯形,有
$\begin{aligned} A &=\left(\begin{array}{ccccc} 1 & 1 & -2 & 2 & 1 \\ -1 & 1 & 1 & 0 & 2 \\ 1 & 0 & 1 & 1 & -2 \\ 0 & 2 & 3 & 2 & 1 \end{array}\right) \rightarrow\left(\begin{array}{ccccc} 1 & 1 & -2 & 2 & 1 \\ 0 & 2 & -1 & 2 & 3 \\ 0 & -1 & 3 & -1 & -3 \\ 0 & 2 & 3 & 2 & 1 \end{array}\right) \rightarrow\left(\begin{array}{ccccc} 1 & 1 & -2 & 2 & 1 \\ 0 & -1 & 3 & -1 & -3 \\ 0 & 0 & 5 & 0 & -3 \\ 0 & 0 & 9 & 0 & -5 \end{array}\right) \\ & \rightarrow\left(\begin{array}{ccccc} 1 & 0 & 1 & 1 & -2 \\ 0 & -1 & 3 & -1 & -3 \\ 0 & 0 & 5 & 0 & -3 \\ 0 & 0 & 0 & 0 & \frac{2}{5} \end{array}\right) \rightarrow\left(\begin{array}{ccccc} 1 & 0 & 1 & 1 & 0 \\ 0 & -1 & 3 & -1 & 0 \\ 0 & 0 & 5 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{array}\right) \rightarrow\left(\begin{array}{ccccc} 1 & 0 & 0 & 1 & 0 \\ 0 & -1 & 0 & -1 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{array}\right) \end{aligned}$
由此可知 ${\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{5}}$ 是 ${\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4},\alpha_{5}}$ 的极大线性无关组,所以也是 ${W_{1}+W_{2}=L\left(\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4},\alpha_{5}\right)}$ 的一组基,进而 ${\operatorname{dim}\left(W_{1}+W_{2}\right)=4}$ .

另外,对任意的 ${\alpha \in W_{1} \cap W_{2}}$ ,可设
${ \alpha=x_{1} \alpha_{1}+x_{2} \alpha_{2}=x_{3} \alpha_{3}+x_{4} \alpha_{4}+x_{5} \alpha_{5} }$
那么 ${x_{1} \alpha_{1}+x_{2} \alpha_{2}-x_{3} \alpha_{3}-x_{4} \alpha_{4}-x_{5} \alpha_{5}=0}$ ,将此看作关于 ${x_{1},x_{2},-x_{3},-x_{4},-x_{5}}$ 的齐次线性方程组,根据上述阶梯形可知方程组的通解为
${ x_{1}=x_{2}=x_{4},x_{3}=x_{5}=0 . }$
其中 ${x_{4}}$ 为自由末知量,所以 ${\alpha=x_{4} \alpha_{4}}$ ,这说明 ${W_{1} \cap W_{2}=L\left(\alpha_{4}\right)}$ ,进而 ${\operatorname{dim}\left(W_{1} \cap W_{2}\right)=1}$ .