【线性代数基础进阶】向量-补充+练习 & 线性方程组-练习

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概念和定理

向量

$\alpha_{1},\alpha_{2},\cdots,\alpha_{r}$及$\alpha_{1},\alpha_{2},\cdots,\alpha_{r},\cdots,\alpha_{s}(其中s\geq r)$，称$\alpha_{1},\alpha_{2},\cdots,\alpha_{r}$是$\alpha_{1},\alpha_{2},\cdots,\alpha_{s}$的部分组，$\alpha_{1},\alpha_{2},\cdots,\alpha_{s}$是整体组

 

向量组$\alpha_{1}=(a_{11},a_{21},\cdots,a_{r1})^{T},\alpha_{2}=(a_{12},a_{22},\cdots,a_{r2})^{T},\cdots,\alpha_{m1}=(a_{1m},a_{2m},\cdots,a_{rm})^{T}$及$\widetilde{\alpha_{1}}=(a_{11},a_{21},\cdots,a_{r1},\cdots,a_{s1})^{T},\widetilde{\alpha_{2}}=(a_{12},a_{22},\cdots,a_{r2},\cdots,a_{s2})^{T},\cdots,\widetilde{\alpha_{m}}=(a_{1m},a_{2m},\cdots,a_{rm},\cdots,a_{sm})^{T}$，则称$\widetilde{\alpha_{1}},\widetilde{\alpha_{2}},\cdots,\widetilde{\alpha_{m}}$为向量组$\alpha_{1},\alpha_{2},\cdots,\alpha_{m}$的延伸组（或称$\alpha_{1},\alpha_{2},\cdots,\alpha_{m}$是$\widetilde{\alpha_{1}},\widetilde{\alpha_{2}},\cdots,\widetilde{\alpha_{m}}$的缩短组）

 

定理：

任何部分组$\alpha_{1},\alpha_{2},\cdots,\alpha_{r}$相关$\Rightarrow$整体组

$\alpha_{1},\alpha_{2},\cdots,\alpha_{s}$相关

整体组$\alpha_{1},\alpha_{2},\cdots,\alpha_{s}$无关

$\Rightarrow$任何部分组

$\alpha_{1},\alpha_{2},\cdots,\alpha_{r}$无关，反之都不成立

 

定理：

$\alpha_{1},\alpha_{2},\cdots,\alpha_{m}$线性无关$\Rightarrow$延伸组$\widetilde{\alpha_{1}},\widetilde{\alpha_{2}},\cdots,\widetilde{\alpha_{m}}$线性无关

$\widetilde{\alpha_{1}},\widetilde{\alpha_{2}},\cdots,\widetilde{\alpha_{m}}$线性相关$\Rightarrow \alpha_{1},\alpha_{2},\cdots,\alpha_{m}$线性相关

 

向量组的秩

向量组的极大线性无关组的向量个数称为向量组的秩，记为$r(\alpha_{1},\alpha_{2},\cdots,\alpha_{s})$

 

线性相关例题进一步说明

例：

已知$\alpha_{1},\alpha_{2},\alpha_{3}$线性无关，证明$\alpha_{1}+\alpha_{2},\alpha_{2}+\alpha_{3},\alpha_{3}+\alpha_{1}$线性无关

 

已知某向量组$\alpha_{1},\alpha_{2},\cdots,\alpha_{n}$线性无关，推新的向量组$\beta_{1},\beta_{2},\cdots,\beta_{s}$线性无关大致思路，设

$k_{1}\beta_{1}+k_{2}\beta_{2}+\cdots+k_{s}\beta_{s}=0$

向$\alpha_{1},\alpha_{2},\cdots,\alpha_{n}$化简，即类似

$m_{1}\alpha_{1}+m_{2}\alpha_{2}+\cdots+m_{n}\alpha_{n}=0$

其中$m_{1},\cdots,m_{n}$由$k_{1},\cdots,k_{s}$表示，通过条件

$b_{1}\alpha_{1}+b_{2}\alpha_{2}+\cdots+b_{n}\alpha_{n}=0$

只有当$b_{1}=b_{2}=\cdots=b_{n}=0$时上式成立，证明

$k_{1}=k_{2}=\cdots=k_{s}=0$

 

设

因为$\alpha_{1},\alpha_{2},\alpha_{3}$线性无关

由

齐次方程组$(1)$只有$0$解，即必有$k_{1}=0,k_{2}=0,k_{3}=0$，因此$\alpha_{1}+\alpha_{2},\alpha_{2}+\alpha_{3},\alpha_{3}+\alpha_{1}$线性无关

 

已知$\alpha_{1},\alpha_{2},\alpha_{3}$线性无关，无法证明$\alpha_{1}-\alpha_{2},\alpha_{2}-\alpha_{3},\alpha_{3}-\alpha_{1}$线性无关

 

设

因为$\alpha_{1},\alpha_{2},\alpha_{3}$线性无关

由

齐次方程组$(1)$有非零解，无法证明$k_{1}=0,k_{2}=0,k_{3}=0$，因此无法证明$\alpha_{1}-\alpha_{2},\alpha_{2}-\alpha_{3},\alpha_{3}-\alpha_{1}$线性无关

 

秩

例：向量组$(1):\alpha_{1},\alpha_{2},\alpha_{3};(2):\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4};(3):\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{5}$，若秩$r(1)=3,r(2)=3,r(3)=4$，则$r(\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4}+\alpha_{5})=()$

 

由$r(1)=3$知，$\alpha_{1},\alpha_{2},\alpha_{3}$线性无关

由$r(2)=4$知，$\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4}$线性相关

故$\alpha_{4}$可以由

 

以下可以用分析或计算

 

$\alpha_{1},\alpha_{2},\alpha_{3}$线性表出，那么向量组$\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{5}$与$\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4}+\alpha_{5}$可以相互线性表示，即向量组等价

所以$r(\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4}+\alpha_{5})=r(3)=4$

或用计算

设$\alpha_{4}=k_{1}\alpha_{1}+k_{2}\alpha_{2}+k_{3}\alpha_{3}$，则

由于矩阵$\begin{pmatrix}1 & 0 & 0 & k_{1} \\ 0 & 1 & 0 & k_{2} \\ 0 & 0 & 1 & k_{3} \\ 0 & 0 & 0 & 1\end{pmatrix}$可逆，故$r(\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4}+\alpha_{5})=r(3)=4$

 

矩阵的秩

例：设$A,B$都是$n$阶非零矩阵，且$AB=O$，证明$A$和$B$的秩都小于$n$

 

不会0_o，以后会补上的

正交规范化、正交矩阵在特征值和特征向量里

 

$Ax=b$解的性质

例：设$\alpha_{1},\alpha_{2},\alpha_{3}$是四元非齐次线性方程组$Ax=b$的三个就行了，且秩$r(A)=3$，$\alpha_{1}=(1,2,3,4)^{T},\alpha_{2}+\alpha_{3}=(0,1,2,3)^{T}$，则方程组$Ax=b$的通解是（）

 

由于$n-r(A)=4-3=1$，所以方程组通解形式为$\alpha+k \eta$，现在特解已知可取$\alpha_{1}$，下面就是应找出导出组$Ax=0$的一个非零解

因为$A a_{i}=b,(i=1,2,3)$，有$A[2\alpha_{1}-(\alpha_{2}+\alpha_{3})]=0$，即

是$Ax=0$的一个非零解，于是方程组通解为$(1,2,3,4)^{T}+k(2,3,4,5)^{T}$，$k$是任意常数