2024秋 MA103 期中答案

校对备注

如有不确定请参考原始PDF, 字太差, 生成的md比较混乱

1-1

Answer: D

解析:

$A\in {\mathbb{R}}^{m\times n},\mathrm{\forall }b\in {\mathbb{R}}^n,A{x}^{2}b$ 有解 $\Rightarrow R(A)=m$

$$R(A,b)=R(A)$$

$A:$ 如 $A=[L\mid {\alpha }_1{\alpha }_I]\alpha =\alpha$

$B:$ $A^{\mathrm{\top }}\in {\mathbb{R}}^{m\times m}$

$\mathrm{\forall }d{A}^{T}x=d$ 有解 $\Rightarrow R(A)=n$

$C:$

$$A=\left(\begin{array}{lll}{\alpha }_1& \cdots & {\alpha }_n\end{array}\right)\left(\begin{array}{l}{x}_i\\ ⋮\\ x_n\end{array}\right)=\sum _{i=1}^n{\alpha }_i{x}_i$$

$D:$ $A\in \mid {\mathbb{R}}^{m\times n}$

if $B\in \mid {\mathbb{R}}^{m\times n}$, $AB=I\Rightarrow B$ 为 $A$ 的右逆

$A_{m\times n}$ 有右逆 $\Rightarrow R(A)=m$

$A=\left(\begin{array}{lll}{\alpha }_1& \cdots & {\alpha }_n\end{array}\right)\times (\frac{I_m}{0})=A_{m\times n}^{{}^{\prime }}$ (可逆)

$\Rightarrow A$ 的右逆为 $(\frac{I_m}{0})(A_{m\times n}^{{}^{\prime }}{)}^{-1}$

${\alpha }_{1}to{\alpha }_n$线性无关

备注: 这里肯定有我没看懂的地方

1-2

Answer: A

解析:

A: $Ax=0\Rightarrow EAx=0=Bx$

B: let A =

$$\left[\begin{array}{lll}1& 0& 0\\ 0& 1& 0\\ 0& 0& 0\end{array}\right]$$

B=

$$\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ -1& 0& 0\end{array}\right]$$

备注: 也许B[1][1]是个-1, 这里写的真模糊

C:

$A^{\mathrm{\top }}x=0\iff B^{\mathrm{\top }}x=0(z)A^{\mathrm{\top }}{E}^{\mathrm{\top }}x=0$

so $x\ne E^{\mathrm{\top }}x$

D:

$$\begin{array}{c}D:B^{\mathrm{\top }}B=A^{\mathrm{\top }}{E}^{\mathrm{\top }}EA=A^{\mathrm{\top }}A?x\end{array}$$

1-3

Answer: B

解析:

A: Not symmetic

B: $Ax=0\iff A^{\mathrm{\top }}Ax=0$

C : 不能随意行变换

D: $n>m$

1-4

Answer: $C$

Hint: $C=(AB{)}^{-1}=B^{-1}{A}^{-1}$

1-5

Answer: C

解析:

$$\begin{array}{rl}& A=A^2\Rightarrow 0=A(A-I)\\ & \therefore R(A)+R(A-I)\le n\\ & \text{and}R(A)+R(B)⩾R(A+B)\\ & \text{ 取 }B=I-A\\ & R(A)+R(I-A)⩾R(I)=n\end{array}$$

2

2-(1)

$$\left[\begin{array}{ccc}1& -\frac{1}{2}& -\frac{1}{6}\\ 0& \frac{1}{2}& -\frac{1}{6}\\ 0& 0& \frac{1}{3}\end{array}\right]$$

(2)

$$\left[\begin{array}{ccc}1& 0& 0\\ 2& 1& 0\\ 3& 1& 1\end{array}\right]$$

(3)

$$\frac{1}{14}\left[\begin{array}{lll}4& 6& 2\\ 6& 9& 3\\ 2& 3& 1\end{array}\right]$$

投影到 $V,V$ besis: $V_1\dots ..V_m$

$A:[v_1\dots v_m]$

$$P_v=A(A^{\mathrm{\top }}A{)}^{-1}{A}^{\mathrm{\top }}$$

(4)

10

$A\in {\mathbb{R}}^{m\times n}$

$N(A)=n-R(A)$

$C(A)=R(A)$

$C(A^{\mathrm{\top }})=R(A)$

$N(A{)}^{\mathrm{\top }}=m-R(A)$

$m=2024,n=2025$

3

a

$$I-A=\left[\begin{array}{ccc}1& -1& 0\\ -1& 1& 1\\ 0& -1& 1\end{array}\right]$$$$I-A^2=\left[\begin{array}{ccc}0& 0& 1\\ 0& 1& 0\\ -1& 0& 2\end{array}\right]$$

b

$X-X{A}^2-AX+AX{A}^2=I$

$(I-A)X(I-A^2)=I$

$X=(I-A{)}^{-1}(I-A^2{)}^{-1}$

$={[(I-A^2)(I-A))]}^{-1}=$

$$\left[\begin{array}{lll}1\text{(or maybe 3, pdf上太模糊了)}& 1& -2\\ 1& 1& -1\\ 2& 1& -1\end{array}\right]$$

4

4-a

$$\left[\begin{array}{cccc}1& 0& 2& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\\ 0& 0& 0& 0\end{array}\right]$$

Pivots 上方均为0!

4-b

$$\begin{array}{rl}& c(A)=k_1\left[\begin{array}{c}1\\ 1\\ 1\\ 3\end{array}\right]+k_2\left[\begin{array}{c}0\\ 1\\ 0\\ 2\end{array}\right]+k_3\left[\begin{array}{c}0\\ 0\\ 1\\ 0\end{array}\right]\\ & c(A^{\mathrm{\top }})=k_1\left[\begin{array}{c}1\\ 0\\ 2\\ 0\end{array}\right]+k_2\left[\begin{array}{c}0\\ 1\\ 1\\ 0\end{array}\right]+k_3\left[\begin{array}{c}0\\ 0\\ 0\\ 1\end{array}\right]\end{array}$$$$N(A^{\mathrm{\top }})=k_1\left[\begin{array}{c}-1\\ -2\\ 0\\ 1\end{array}\right]$$

4-c

$$[A\mid b]\to \left[\begin{array}{ccccc}1& 0& 2& 0& 3\\ 0& 1& 1& 0& 0\\ 0& 0& 0& 1& 0\\ 0& 0& 0& 0& 6\end{array}\right]$$

特解:

$$x_r=\left[\begin{array}{c}1\\ -1\\ 1\\ 0\end{array}\right]$$

通解:

$$x_p=k\left[\begin{array}{c}-2\\ -1\\ 1\\ 0\end{array}\right]$$$$\Rightarrow x=x_1+x_p=\left[\begin{array}{c}1\\ -1\\ 1\\ 0\end{array}\right]+k\left[\begin{array}{c}-2\\ -1\\ 1\\ 0\end{array}\right]$$

5

5-(a)

(1) $v_1,v_2$ 线性无关

(2) $V_1,V_2\in \mathrm{V}$, 代入

$\mathrm{span}(v_1,v_2)\ge V$

解

$$x_1+x_2+k_3=v\to k_1\left[\begin{array}{c}1\\ 0\\ -1\end{array}\right]+k_2\left[\begin{array}{c}1\\ -1\\ 0\end{array}\right]$$

5-(b)

$T(v_1,v_2)=(e_1,e_2,e_3)A_{3\times 2}$

$$T(v_1)=\left[\begin{array}{c}-3\\ 1\\ -1\end{array}\right]=\left[\begin{array}{lll}{e}_1& e_2& e_3\end{array}\right]\left[\begin{array}{c}-3\\ 1\\ -1\end{array}\right]$$$$T(v_2)=\left[\begin{array}{c}-2\\ 1\\ -1\end{array}\right]=\left[\begin{array}{lll}{e}_1& e_2& e_3\end{array}\right]\left[\begin{array}{c}-2\\ 1\\ -1\end{array}\right]$$$$A=\left[\begin{array}{cc}-3& -2\\ 1& 1\\ -1& -1\end{array}\right]$$

线性变换: $7:V^n\to W^m$ (回顾)

$$\begin{array}{r}{v}_1,\dots ,v_n\to w_1,\dots ,w_m\\ T_{v_1}=?T_{v_2}=?\\ & T(v_1,v_2\dots v_n)=(T_{v_1},T_2\dots T_{v_n})\\ & =[{\omega }_1\cdots {\omega }_n]A_{m\times n}\end{array}$$

5-c

$$\begin{array}{rl}& \text{(c) }{T}_V=\left[\begin{array}{c}3\\ 2\\ 1\end{array}\right]\Rightarrow T(v_1,v_2)=\underset{3\times 3}{\underbrace{\left[\begin{array}{lll}{e}_1& e_2& e_3\end{array}\right]}}\left[\begin{array}{cc}-3& -2\\ 1& 1\\ -1& -1\end{array}\right]\cdot \left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]=\left[\begin{array}{c}3\\ 2\\ 1\end{array}\right]\end{array}$$$$\Rightarrow \left[\begin{array}{cc}-3& -2\\ 1& 1\\ -1& -1\end{array}\right]\left[\begin{array}{l}{x}_1\\ x_2\end{array}\right]=\left[\begin{array}{l}3\\ 2\\ 1\end{array}\right]$$$$\left[\begin{array}{ccc}-3& -2& 3\\ 1& 1& 2\\ -1& -1& 1\end{array}\right]⟶\cdots \left[\begin{array}{ccc}-3& -2& 3\\ 0& \frac{1}{3}& 3\\ 0& -\frac{1}{3}& 0\end{array}\right]\to \text{无解}$$

6

6-(a)

不始设 $B_1,B_2\in U$

$$\begin{array}{c}A{B}_1=B_{1}A,A{B}_2=B_{2}A\\ \Rightarrow A(B_1+B_2)=A{B}_1+A{B}_2\text{ 验证加法, 数乘封闭 }\\ A(k{B}_1)=k(A{B}_1)\end{array}$$

6-(b)

$$B=\left[\begin{array}{lll}{b}_{11}& b_{21}& b_{13}\\ b_{21}& b_{22}& b_{23}\\ b_{31}& b_{32}& b_{33}\end{array}\right]$$$$AB=\left[\begin{array}{ccc}{b}_{11}+b_{21}+b_{31}& b_{12}+b_{22}+b_{32}& b_{13}+b_{12}+b_{33}\\ b_{31}& -b_{32}& b_{33}\\ -b_{31}& -b_{32}& -b_{33}\end{array}\right]$$$$BA=\left[\begin{array}{lll}{b}_{11}& b_{11}& b_{11}+b_{12}-b_{13}\\ b_{21}& b_{21}& b_{21}+b_{22}-b_{23}\\ b_{31}& b_{31}& b_{33}+b_{32}-b_{33}\end{array}\right]$$$$AB=BA\Rightarrow (AB{)}_{ij}=(BA{)}_{ij}$$$$\begin{gathered} \Rightarrow b_{31}=0 \\ {b}_{32}=0 \\ {b}_{21}=0 \\ {b}_{22}-b_{23}=b_{33} \\ {b}_{11}=b_{12}+b_{22} \\ {b}_{11}+b_{12}-b_{13}=b_{13}+b_{23}+b_{33} \\ \Rightarrow b_{12}=b_{13} \end{gathered}$$$$B=\left[\begin{array}{ccc}{b}_{12}+b_{22}& b_{12}& b_{12}\\ 0& b_{22}& b_{23}\\ 0& 0& b_{22}-b_{23}\end{array}\right]$$$$b_{213}:\{\left[\begin{array}{lll}1& 1& 1\\ 0& 0& 0\\ 0& 0& 0\end{array}\right],\left[\begin{array}{lll}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right],\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 1\\ 0& 0& -1\end{array}\right]\}$$

6-(c)

$\dim (u)=3$

7

7-(a)

$C(A^{\mathrm{\top }}):v_1,\dots ,v_s$

$N(A):w_1,\dots ,w_t$

$$\begin{array}{rl}& t=R(N(A))=n\cdot R(A)=n-\dim (C(A^{\mathrm{\top }}))=n-S\\ & \Rightarrow s+t=n\\ & k_1{v}_1+\dots +k_s{v}_s+\cdots +k_n{w}_t=0\iff k_i=0,k_1,\dots ,n.\\ & A=\left[\begin{array}{lll}{\alpha }_1& \alpha \dots & {\alpha }_m\end{array}\right]A^{\mathrm{\top }}=\left[\frac{{\alpha }_1^{\mathrm{\top }}}{{\alpha }_m^{\mathrm{\top }}}\right]\end{array}$$$$\begin{array}{rl}& C(A^{\mathrm{\top }})=\mathrm{span}({\alpha }_1,\cdots {\alpha }_m)\\ & N(A)=\{x:Ax=0\}\\ & =\{x:{\alpha }_i^{\mathrm{\top }}\times x=0,i=1,2,\dots ,m\}\\ & \iff V_i^{\mathrm{\top }}{w}_j=0,i=1,\cdots ,s,j=1,\dots ,t\\ & N(A)\perp C(A^{\mathrm{\top }})\end{array}$$$$\begin{array}{rl}& \alpha =k_{1}v+\cdots +k_1{v}_s+\cdots +k_n{w}_t=0\\ & \Rightarrow v+w=0\\ & v^{\mathrm{\top }}(v+w)=v^{\mathrm{\top }}v=0\to v=0\\ & \Rightarrow k_1,\dots ,k_s=0\\ & \text{ 同理 }{k}_{s+1},\cdots k_n=0.\\ & \Rightarrow \text{ 线性无关 }\to \text{ 为一组基. }\end{array}$$

7-(b)

$A\in {\mathbb{R}}^{m\times n}$

$$\begin{array}{rl}& R(A)=m\Rightarrow R(A^{\mathrm{\top }})=m\\ & R(A{A}^{\mathrm{\top }})=m-N(A{A}^{\mathrm{\top }})=m-N(A^{\mathrm{\top }})=m-(m-R(A))=R(A)\end{array}$$