113年台綜大轉學考-線性代數詳解

臺灣綜合大學系統113學年度學士班轉學生聯合招生考試

科目名稱:線性代數
類組代碼:A07/C11 

解答: $$\Rightarrow: x\in Null(T) \Rightarrow T(x)=0 \Rightarrow x=0=(0,0,0)\text{ (since T is one-to-one)} \\\qquad \Rightarrow  Null(T) \text{ is spanned by (0,0,0)}\\ \Leftarrow: Null(T) \text{ is spanned by (0,0,0)} \Rightarrow Null(T)=k(0,0,0)=(0,0,0)  \\ \qquad \text{ Let }T(u)=T(v), u,v \in \mathbb R^3, \text{ then }T(u)-T(v)= T(u-v)=0 \Rightarrow u-v \in \text{Null}(T)\\ \qquad \Rightarrow u-v=0 \Rightarrow u=v \Rightarrow T \text{ is one-to-one}\\\bbox[red, 2pt]{QED}$$

解答: $$\cases{T(1)=2\cdot 0+\int_0^x 1\,dt=x\\ T(x) =2\cdot 1+ \int_0^x t\,dt = 2+{1\over 2}x^2\\ T(x^2) =2\cdot (2x)+ \int_0^x t^2\,dt =4x+{1\over 3}x^3} \Rightarrow T= \bbox[red, 2pt]{\begin{bmatrix}0& 2 &0\\ 1 & 0& 4\\ 0 & {1\over 2}& 0\\ 0& 0 & {1\over 3} \end{bmatrix}}$$

解答: $$\textbf{(a) }A= W_1\cap W_2 =\begin{bmatrix}a & 0\\ 0 & -a \end{bmatrix}   \Rightarrow\text{dim}(W_1\cap W_2)=  \bbox[red, 2pt]1 \\\textbf{(b) } \text{dim}(W_1+W_2) =\text{dim }W_1+ \text{dim }W_2-\text{dim}(W_1\cap W_2) =3+2-1=\bbox[red, 2pt]4$$

解答: $$\textbf{(a) }v \in \text{Col }AB  \Rightarrow v=(AB)x =A(Bx) \Rightarrow v\in \text{Col }A \Rightarrow \text{Col }AB \subseteq \text{ Col }A \\\qquad \Rightarrow \text{dim(Col }AB) \le \text{dim(Col }A) \Rightarrow \text{rank}(AB)\le \text{rank}(A). \bbox[red, 2pt]{QED}\\ \textbf{(b) }\text{By }(a), \text{ we have} \cases{\text{rank}(AP) \le \text{rank}(A)  \cdots(1)\\ \text{rank}(A)=\text{rank}(APP^{-1}) \le \text{rank}(AP) \cdots(2)} \\\qquad \text{Then rank}(AP)= \text{rank }A. \bbox[red, 2pt]{QED}$$

解答: $$A^3= \begin{bmatrix} 1& -2 \\ 3& 4\end{bmatrix} \Rightarrow (A^3)^{-1} =B=\left[ \begin{matrix}-4 & 2 \\3 & -1 \end{matrix}\right ] \Rightarrow A^3B=I \Rightarrow A(A^2B)=I \Rightarrow A^{-1} =A^2B\\ \Rightarrow A \text{ is invertible} \;\bbox[red, 2pt]{QED.}$$

解答: $$\textbf{(a) } A \text{ has eigenvalues 1, 2, and 3, then }\cases{Av_1=v_1\\ Av_2= 2v_2\\ Av_3=3v_3}, \text{ where }v_1, v_2, \text{ and }v_3 \text{ are eigenvectors}\\ \quad av_1+bv_2=0 \Rightarrow A(av_1+ bv_2)=aA(v_1) +bA(v_2) =av_1+2b_2 v_2=0 \Rightarrow b_2v_2=0 \\\quad \Rightarrow b_2=0 \Rightarrow av_1=0 \Rightarrow a=0 \Rightarrow v_1 \text{ and }v_2 \text{ are linearly independent} \\ \quad \text{By the same way, we have }v_2\text{ and }v_3 \text{ are linearly independent, and }v_1 \text{ and }v_3\text{ are linearly independent.}\\ \text{That is, }v_1, v_2, \text{ and }v_3   \text{ are linearly independent. By the Diagonalization Theorem, } \\\quad A \text{ is 3x3 and has 3 linearly independent eigenvectors, then A is diagonalizable.} \bbox[red, 2pt]{QED} \\\textbf{(b) }\cases{2=2+0= 1+1 \Rightarrow \text{ two partitions} \\ 3=3+0 = 2+1= 1+1+1 \Rightarrow \text{ three partitions}} \Rightarrow \text{numbers of JC. forms =}2\times 3=6\\ \Rightarrow J_1=\begin{bmatrix}4 &0  & 0 & 0 & 0 \\0 & 4 & 0 & 0 &0  \\0 &0  & 5 & 0 & 0 \\0 & 0 & 0 & 5 &0  \\0 &  0 & 0 & 0 & 5\end{bmatrix}, J_2= \begin{bmatrix}4 &1  & 0 & 0 & 0 \\0 & 4 & 0 & 0 &0  \\0 &0  & 5 & 0 & 0 \\0 & 0 & 0 & 5 &0  \\0 &  0 & 0 & 0 & 5\end{bmatrix}, J_3=\begin{bmatrix}4 &0  & 0 & 0 & 0 \\0 & 4 & 0 & 0 &0  \\0 &0  & 5 & 1 & 0 \\0 & 0 & 0 & 5 &0  \\0 &  0 & 0 & 0 & 5\end{bmatrix} \\ J_4=\begin{bmatrix}4 &0  & 0 & 0 & 0 \\0 & 4 & 0 & 0 &0  \\0 &0  & 5 & 1 & 0 \\0 & 0 & 0 & 5 &1  \\0 &  0 & 0 & 0 & 5\end{bmatrix}, J_5= \begin{bmatrix}4 &1  & 0 & 0 & 0 \\0 & 4 & 0 & 0 &0  \\0 &0  & 5 & 1 & 0 \\0 & 0 & 0 & 5 &0  \\0 &  0 & 0 & 0 & 5\end{bmatrix}, J_6=\begin{bmatrix}4 &1  & 0 & 0 & 0 \\0 & 4 & 0 & 0 &0  \\0 &0  & 5 & 1 & 0 \\0 & 0 & 0 & 5 &1  \\0 &  0 & 0 & 0 & 5\end{bmatrix}$$

解答: $$\textbf{(a) }\text{a basis }\{1,x,x^2\} \in P_2, \text{ by Gram-Schmidt process}:\\ \langle 1,1\rangle =\int_{-1}^1 1\cdot 1\,dx =2 \Rightarrow ||\langle 1,1 \rangle ||=\sqrt 2 \Rightarrow e_1={1\over \sqrt 2} \\ \langle x,e_1 \rangle =\int_{-1}^1 {x\over \sqrt 2}\,dx =0 \Rightarrow  x-\langle x, e_1\rangle  e_1 =x \Rightarrow e_2={x\over \sqrt{\langle x\cdot x\rangle}} ={x\over \sqrt{\int_{-1}^1 x^2\,dx} } =\sqrt{3\over 2}x\\ \cases{ \langle x^2,e_1\rangle ={1\over \sqrt 2}\int_{-1}^1 x^2 ={\sqrt 2\over 3}\\ \langle x^2, e_2 \rangle = \sqrt{3\over 2} \int_{-1}^1 x^3\,dx =0} \Rightarrow x^2- \langle x^2,e_1\rangle e_1- \langle x^2,e_2\rangle e_2 =x^2-{1\over 3} \\ \Rightarrow e_3={x^2 -{1\over 3}\over \sqrt{\langle x^2 -{1\over 3},x^2 -{1\over 3} \rangle}} ={3\over 4}\sqrt{10}(x^2-{1\over 3}) \\ \Rightarrow \text{an orthonormal basis: }\left\{ e_1,e_2,e_3\right\} = \bbox[red, 2pt]{\left\{ {\sqrt2 \over 2}, {\sqrt 6\over 2} x, {3\over 4} \sqrt{10} x^2 -{1 \over 4}\sqrt{10}\right\} }\\\textbf{(b) }f=a+bx+cx^2 \in P_2 \Rightarrow \langle h-f\rangle = \int_{-1}^1 (x^3-cx^2-bx-a)\,dx ={1\over 2}-b \\\quad \Rightarrow ||h-f||= \sqrt{{1\over 2}-b} \ge 0 \\\text{ Then, we can choose } \bbox[red, 2pt]{u={1\over 2} x} \Rightarrow ||h-u||=0 \Rightarrow ||h-f|| \ge ||h-u||$$

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解題僅供參考，轉學考歷年試題及詳解