114年政大轉學考-線性代數詳解

 國立政治大學114學年度學士班轉學生招生考試

科目:線性代數
系所別:應用數學系

解答:$$\textbf{a. }\bbox[red, 2pt]{\text{False}}:\text{Let } A = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \in \mathbb{R}^{2 \times 2} , \text{and two linearly independent vectors, } v_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix}  \text{ and }  v_2 = \begin{bmatrix} 0 \\ 1 \end{bmatrix}.\\ \text{Then } Av_1=Av_2 = \begin{bmatrix}0\\0 \end{bmatrix}. \text{ That is, }Av_1 \text{ and }Av_2 \text{ are linearly dependent.} \\\textbf{b. }\bbox[red, 2pt]{\text{True}}: \text{Assume $\{v_1, v_2\}$ is linearly dependent. Then there exist scalars $c_1$ and $c_2$, }\\ \qquad \text{not both zero, such that: $c_1v_1+c_2v_2=0$. Then we have} \\ \qquad A(c_1v_1+ c_2v_2)= A(0) \Rightarrow c_1(Av_1)+ c_2(Av_2) =0 \Rightarrow \{Av_1,Av_2\} \text{ is linearly independent} \\ \textbf{c. }\bbox[red, 2pt]{\text{False}}: A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}  \Rightarrow \det(A)=1 \ne 0 \Rightarrow A \text{ is invertible.} \\\qquad \det(A-\lambda I)=(1-\lambda)^2 =0 \Rightarrow \lambda =1 \Rightarrow (A-\lambda I) x=0 \Rightarrow \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \\ \qquad \Rightarrow x=x_1 \begin{bmatrix}1\\ 0 \end{bmatrix} \Rightarrow \text{ the geometric multiplicity (1)} \lt \text{ the algebraic multiplicity (2)}\\ \qquad \text{the matrix $A$ cannot be diagonalized.}$$

解答:$$\textbf{a. } O= \begin{bmatrix}0&0\\0 & 0 \end{bmatrix} \Rightarrow AO=OA= O \Rightarrow O\in W \\ \qquad \cases{B_1\in W \Rightarrow AB_1=B_1A\\ B_2\in W \Rightarrow AB_2= B_2A} \Rightarrow A(B_1+B_2)= AB_1+AB_2 =B_1A+B_2A= (B_1+B_2)A \\ \qquad \Rightarrow A(B_1+B_2)= (B_1+B_2)A \Rightarrow B_1+B_2 \in W \\ B\in W\Rightarrow AB=BA \Rightarrow A(cB)= c(AB)=c(BA) =(cB)A \Rightarrow cB\in W\\ \qquad \text{Therefore, $W$ is a subspace of }\mathbb R^{2\times 2}. \\ \textbf{b. }\text{Let }B= \begin{bmatrix}x& y\\ z& w \end{bmatrix} \in \mathbb R^{2\times 2}. \text{Assum }B\in W \Rightarrow AB=BA \Rightarrow  \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} x & y \\ z & w \end{bmatrix}  =\begin{bmatrix} x & y \\ z & w \end{bmatrix}  \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \\ \qquad \Rightarrow \begin{bmatrix}x+z& y+w\\z& w \end{bmatrix}= \begin{bmatrix}x& x+y\\ z& z+w \end{bmatrix} \Rightarrow \cases{x+z = x \\y+w = x+y \\z = z \\w = z+w} \Rightarrow B= \begin{bmatrix}x& y\\0& x \end{bmatrix}  = x \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + y \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \\ \qquad \Rightarrow  \bbox[red, 2pt]{\left\{ \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}, \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \right\}} \text{ is a basis for $W$}$$

解答:$$\textbf{a. } \text{Proof of Additivity: }T(p(x) + q(x)) = (p + q)'(x + 1) + (p + q)(1)x^2 \\= (p'(x + 1) + q'(x + 1)) + (p(1) + q(1))x^2 = (p'(x + 1) + p(1)x^2) + (q'(x + 1) + q(1)x^2) \\=T(p(x))+T(q(x)) \\ \text{Proof of Scalar Multiplication: } T(cp(x)) = (cp)'(x + 1) + (cp)(1)x^2 = c \cdot p'(x + 1) + c \cdot p(1)x^2 \\=c \left( p'(x+1)+p(1)x^2 \right) =cT(p(x)) \\ \text{Since $T$ satisfies both additivity and scalar multiplication, it is a linear transformation.}\bbox[red, 2pt]{QED.} \\\textbf{b. }   p(x)=x^3 \Rightarrow \cases{p'(x)= 3x^2\\ p(1)=1} \Rightarrow p'(x+1) =3x^2+6x+3 \Rightarrow T(p(x))= p'(x+1)+p(1)x^2\\=  4x^2+6x+3 \Rightarrow  [T(x^3)]_\gamma = \begin{bmatrix} 4 \\ 6 \\ 3 \end{bmatrix}$\\ p(x)=x^2 \Rightarrow \cases{p'(x)=2x\\ p(1)=1} \Rightarrow p'(x+1)=2x+2 \Rightarrow T(p(x))= p'(x+1)+p(1)x^2 \\=x^2+2x+2 \Rightarrow   [T(x^2)]_\gamma = \begin{bmatrix} 1 \\ 2 \\ 2 \end{bmatrix}\\p(x)=x \Rightarrow \cases{p'(x)=1\\ p(1)=1} \Rightarrow p'(x+1)=1 \Rightarrow T(p(x))=p'(x+1)+p(1)x^2 \\=x^2+1 \Rightarrow   [T(x)]_\gamma = \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} \\ p(x)=1 \Rightarrow \cases{p'(x)=0\\ p(1)=1} \Rightarrow p'(x+1)=0 \Rightarrow T(p(x))=p'(x+1)+p(1)x^2 \\=x^2 \Rightarrow  [T(1)]_\gamma = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} \\ \Rightarrow   A = [T]_\beta^\gamma = \bbox[red, 2pt]{\begin{bmatrix} 4 & 1 & 1 & 1 \\ 6 & 2 & 0 & 0 \\ 3 & 2 & 1 & 0 \end{bmatrix} }\\ \textbf{c. }   \text{RREF} \left( A \right) = \begin{bmatrix} 1 & 0 & 0 & 1/4 \\ 0 & 1 & 0 & -3/4 \\ 0 & 0 & 1 & 3/4 \end{bmatrix}  \Rightarrow \cases{x+{w\over 4}=0\\ y-{3w\over 4}=0\\z+{3w\over 4}=0} \Rightarrow \mathbf v=\begin{bmatrix} x\\y\\z\\w \end{bmatrix} =w \begin{bmatrix}-1/4\\3/4\\ -3/4\\1 \end{bmatrix} \\ \qquad \text{Choosing }\mathbf v= \begin{bmatrix}1\\-3\\3\\-4 \end{bmatrix} \Rightarrow x^3-3x^2+3x-4 \Rightarrow N(T)= \bbox[red, 2pt]{\text{span}\{x^3-3x^2+3x-4\}} \\ \textbf{d. } \begin{bmatrix} 4 & 1 & 1 & 1 & \vert{} & 1 \\ 6 & 2 & 0 & 0 & \vert{} & 0 \\ 3 & 2 & 1 & 0 & \vert{} & 0 \end{bmatrix} \xrightarrow{\text{RREF}} \begin{bmatrix} 1 & 0 & 0 & 1/4 & \vert{} & 1/2 \\ 0 & 1 & 0 & -3/4 & \vert{} & -3/2 \\ 0 & 0 & 1 & 3/4 & \vert{} & 1/2 \end{bmatrix}     \\ \qquad \Rightarrow   \bbox[red, 2pt]{p(x) = (\frac{1}{2}x^3 - \frac{3}{2}x^2 + \frac{1}{2}x) + c_0(-\frac{1}{4}x^3 + \frac{3}{4}x^2 - \frac{3}{4}x + 1) }$$

解答:$$\textbf{a. }\det(A-\lambda I)=  -\lambda^3 - 3\lambda^2 + 6\lambda + 8 =-(\lambda+1)(\lambda+4)(\lambda-2)=0 \Rightarrow \bbox[red, 2pt]{\text{eigenvalues: -1,-4,2}} \\ \lambda_1=-1 \Rightarrow (A- \lambda_1 I)v=0 \Rightarrow   \begin{bmatrix} 1 & 0 & 2 \\ 0 & -1 & -2 \\ 2 & -2 & 0 \end{bmatrix} \begin{bmatrix}x\\ y\\ z \end{bmatrix}=0 \Rightarrow \cases{x_1+2x_3=0\\ x_2+2x_3=0} \\ \qquad \Rightarrow v= x_2 \begin{bmatrix}-2\\-2\\1 \end{bmatrix}  \Rightarrow \text{choosing }v_1= \begin{bmatrix} -2\\-2\\ 1 \end{bmatrix} \\ \lambda_2=-4 \Rightarrow (A-\lambda_2 I) v=0 \Rightarrow   \begin{bmatrix} 4 & 0 & 2 \\ 0 & 2 & -2 \\ 2 & -2 & 3 \end{bmatrix} \begin{bmatrix}x\\ y\\ z \end{bmatrix}=0 \Rightarrow \cases{x_2=x_3\\ 2x_1-2x_2+3x_3=0} \\ \qquad \Rightarrow v= -x_1 \begin{bmatrix}1\\ 2\\ 2 \end{bmatrix} \Rightarrow \text{choosing }v_2= \begin{bmatrix}1\\2\\2 \end{bmatrix} \\ \lambda_3= 2 \Rightarrow (A-\lambda_3 I )v=0 \Rightarrow  \begin{bmatrix} -2 & 0 & 2 \\ 0 & -4 & -2 \\ 2 & -2 & -3 \end{bmatrix} \begin{bmatrix}x\\ y\\ z \end{bmatrix} =0 \Rightarrow \cases{2x_2+x_3=0\\ x_1=x_3} \\ \qquad \Rightarrow v= x_1 \begin{bmatrix}1\\-1/2\\ 1 \end{bmatrix}\Rightarrow \text{choosing }v_3= \begin{bmatrix}2\\-1\\2 \end{bmatrix} \\ \Rightarrow \bbox[red, 2pt]{\text{eigenvectors: } \begin{bmatrix} -2\\-2\\ 1 \end{bmatrix}, \begin{bmatrix}1\\2\\2 \end{bmatrix},\begin{bmatrix}2\\-1\\2 \end{bmatrix}} \\\textbf{b. } \text{eigenvalues of }A:-1,-4,2 \Rightarrow \text{eigenvalues of }(A^3+I):(-1)^3+1,(-4)^3+1,2^3+1\\ \text{That is, }0,-63,9 \Rightarrow \det(A^3+I)= 0\cdot (-63)\cdot 9=\bbox[red, 2pt]0$$

解答:$$\cases{tr(A)=tr(B) \\ \det(A)=\det(B)} \Rightarrow \cases{a+3=b-1\\ 2a-4=-2b} \Rightarrow \bbox[red, 2pt]{\cases{a=-1\\ b=3}}$$

解答:$$\text{We want to show that }T \text{ is 1-1 } \Leftrightarrow N(T)=\{0\} \\(\Rightarrow ) \text{$T$ is linear }\Rightarrow T(0)=0. \text{ Let }v\in N(T) \Rightarrow T(v)=0=T(0)\\ \text{Since $T$ is 1-1 and $T(v)=T(0)$, then }v=0 \Rightarrow N(T)=\{0\} \\ (\Leftarrow) \text{Suppose $u,v \in V$ satisfy $T(u)=T(v)$. Since $T$ is linear, }T(u)-T(v)=0 \Rightarrow T(u-v)=0 \\ \qquad \Rightarrow u-v \in N(T) =\{0\} \Rightarrow u-v=0 \Rightarrow u=v \Rightarrow T\text{ is 1-1}\\ \text{Both direction have been prove. }\bbox[red, 2pt]{QED.}$$

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