112年政大轉學考-線性代數詳解

國立政治大學112學年度轉學生招生考試

科目:線性代數

系所別:應用數學系三年級

解答:$$\ell \text{ is a subspace of }\mathbb R^2 \Rightarrow (0,0)\in \ell \Rightarrow 0=m\cdot 0+b \Rightarrow b=0\\ b=0 \Rightarrow \ell =\{(x,y) \in \mathbb R^2:y=mx\}\\\qquad \text{Suppose }\cases{w_1=(x_1,y_1) \in \ell \\w_2=(x_2,y_2) \in \ell} \Rightarrow \cases{y_1=mx_1 \\y_2=mx_2} \Rightarrow ay_1+by_2=(ax_1+bx_2)m\\ \qquad \Rightarrow (ax_1+bx_2, ay_1+by_2) =aw_1+bw_2 \in \ell \Rightarrow \ell \text{ is a subspace of }\mathbb R^2\\ \bbox[red, 2pt]{QED}$$

解答:$$A= \begin{bmatrix} 2 & -1 & -1 \\-1 & 2 & -1 \\-1 & -1 & 2\end{bmatrix} \Rightarrow \det(A-\lambda I) =-\lambda(\lambda-3)^2=0 \Rightarrow \lambda=0,3\\ \lambda_1=0 \Rightarrow (A-\lambda_1 I)v=0 \Rightarrow \begin{bmatrix} 2 & -1 & -1 \\-1 & 2 & -1 \\-1 & -1 & 2\end{bmatrix} \begin{bmatrix} x_1 \\x_2 \\x_3 \end{bmatrix} =0 \Rightarrow \cases{x_1=x_3\\ x_2=x_3} \Rightarrow v= x_3 \begin{bmatrix} 1 \\1 \\1 \end{bmatrix}\\\qquad \Rightarrow \text{ we choose }v_1= \begin{bmatrix} 1 \\1 \\1 \end{bmatrix} \\ \lambda_2=3 \Rightarrow  (A-\lambda_3 I)v=0 \Rightarrow \begin{bmatrix} -1 & -1 & -1 \\-1 & -1 & -1 \\-1 & -1 & -1 \end{bmatrix} \begin{bmatrix} x_1 \\x_2 \\x_3\end{bmatrix} =0 \Rightarrow x_1+x_2+ x_3=0 \\ \qquad \Rightarrow v= \begin{bmatrix} -x_2-x_3 \\x_2 \\x_3 \end{bmatrix} =x_2 \begin{bmatrix} -1 \\1 \\0 \end{bmatrix} +x_3\begin{bmatrix} -1 \\0 \\1 \end{bmatrix} \Rightarrow  \text{ we choose }v_2= \begin{bmatrix} -1 \\1 \\0 \end{bmatrix}, v_3= \begin{bmatrix} -1 \\0 \\1 \end{bmatrix} \\ \Rightarrow P=[v_1 v_2 v_3], D=\begin{bmatrix} \lambda_1 & 0 & 0\\0 & \lambda_2 & 0\\0 & 0 & \lambda_3 \end{bmatrix} \Rightarrow \bbox[red, 2pt] {A=\begin{bmatrix}  1 & -1 & -1 \\1 & 1 & 0 \\1 & 0 & 1\end{bmatrix} \begin{bmatrix} 0 & 0 & 0 \\0 & 3 & 0 \\ 0 & 0 & 3 \end{bmatrix} \begin{bmatrix} \frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\\frac{-1}{3} & \frac{2}{3} & \frac{-1}{3} \\\frac{-1}{3} & \frac{-1}{3} & \frac{2}{3} \end{bmatrix}}$$

解答:$$A=\begin{bmatrix}1 & 0 & 0   \\0 & 2 &  0 \\0 & 0 & 3  \end{bmatrix} \Rightarrow \det(A-\lambda I)= - (\lambda-1)(\lambda-2) (\lambda-3)=0\\ \lambda_1=3 \Rightarrow (A-\lambda_1 I)v=0 \Rightarrow \begin{bmatrix}-2 &   0 & 0 \\0 & -1& 0 \\0 &   0&0   \end{bmatrix} \begin{bmatrix}x_1 \\x_2 \\x_3 \end{bmatrix} =0 \Rightarrow  x_1=x_2=0  \\\qquad \Rightarrow v=x_3\begin{bmatrix}0 \\0 \\1 \end{bmatrix}  , \text{we choose }v_1=\begin{bmatrix} 0 \\0 \\1 \end{bmatrix}\\ \lambda_2=2 \Rightarrow (A-\lambda_2 I)v=0 \Rightarrow  \begin{bmatrix}-1 &   0 & 0 \\0 & 0 &   0 \\0 & 0 & 1  \end{bmatrix} \begin{bmatrix}x_1 \\x_2 \\x_3 \\x_4 \end{bmatrix} =0 \Rightarrow x_1=x_3=0 \\\qquad \Rightarrow v=x_2 \begin{bmatrix} 0 \\1 \\0 \end{bmatrix}, \text{we choose }v_2=\begin{bmatrix} 0 \\1 \\0 \end{bmatrix}\\ \lambda_3=1 \Rightarrow (A-\lambda_3 I)v=0  \Rightarrow \begin{bmatrix}0 &   0 & 0 \\0 & 1& 0 \\0 &   0&2   \end{bmatrix} \begin{bmatrix} x_1 \\x_2 \\x_3 \end{bmatrix} =0 \Rightarrow x_2=x_3=0 \\ \qquad \Rightarrow v= x_1\begin{bmatrix} 1 \\0 \\0 \end{bmatrix}, \text{we choose }v_3=\begin{bmatrix} 1 \\0 \\0 \end{bmatrix}\\ \Rightarrow P=[v_1 v_2 v_2] \Rightarrow \bbox[red, 2pt]{\text{Yes, they are similar. }P=\begin{bmatrix} 0& 0& 1 \\0 & 1& 0\\1& 0& 0 \end{bmatrix} }$$

解答:$$\mathbf{(a) }T:\mathbb R^n \to \mathbb R \Rightarrow [T] \text{ is 1 by }n \Rightarrow  rank([T])=1 \Rightarrow dim(N(T))= n-rank([T])=n-1\; \bbox[red, 2pt]{QED} \\\mathbf{(b) } \text{Suppose }av_i+bw=0,i=1-(n-1), \text{ then }T(av_i+bw)=0\\ \qquad \Rightarrow aT(v_i)+bT(w)=0 \Rightarrow bT(w)=0 \Rightarrow b=0 ( w \not \in N(T) \Rightarrow T(w)\ne 0) \\ \qquad \Rightarrow av_i+bw=av_i=0 \Rightarrow a=0 \Rightarrow v_i \text{ and }w \text{ are linearly independent }\\\qquad \Rightarrow B' \text{ is a basis of }\mathbb R^n  \bbox[red, 2pt]{QED} \\\mathbf{(c) } \text{From (b), we have }u \in \mathbb R^n \Rightarrow  u \in \text{span}(B') \Rightarrow u= a_1v_1+ a_2v_2+\cdots +a_{n-1}v_{n-1} + bw\\ \text{ We can choose }v=a_1v_1+ a_2v_2+\cdots +a_{n-1}v_{n-1} \in \text{span}(B) \Rightarrow v\in N(T) \\\quad \text{ ,and choose }b={T(u)\over T(w)}, \text{ where }T(w)\ne 0. \text{ Then }u=v+{T(u)\over T(w)}w\quad \bbox[red, 2pt]{QED}$$

解答:$$\textbf{(a) }[A \mid I] = \left[ \begin{array}{rrr|rrr}1& 2 & 1& 1& 0 & 0\\ 2 & 5& 4& 0 & 1& 0\\ 1& 1& 0 & 0 & 0& 1 \end{array} \right] \xrightarrow{R_2-2R_1\to R_2, R_3-R_1\to R_3} \left[ \begin{array}{rrr|rrr}1 & 2 & 1 & 1 & 0 & 0\\0 & 1 & 2 & -2 & 1 & 0\\0 & -1 & -1 & -1 & 0 & 1\end{array} \right] \\ \xrightarrow{R_1- 2R_2 \to R_2, R_3+R_2\to R_3} \left[ \begin{array}{rrr|rrr}1 & 0 & -3 & 5 & -2 & 0\\0 & 1 & 2 & -2 & 1 & 0\\0 & 0 & 1 & -3 & 1 & 1\end{array} \right] \xrightarrow{R_1+3R_3\to R_1,R_2-2R_3\to R_2} \\ \left[ \begin{array}{rrr|rrr}1 & 0 & 0 & -4 & 1 & 3\\0 & 1 & 0 & 4 & -1 & -2\\0 & 0 & 1 & -3 & 1 & 1\end{array} \right] \Rightarrow \bbox[red, 2pt] {A^{-1} = \left[\begin{matrix}-4 & 1 & 3\\4 & -1 & -2\\-3 & 1 & 1\end{matrix}\right]} \\\textbf{(b) } rref(A)=I_3 \Rightarrow \bbox[red, 2pt]{\text{Yes, they are independent.}} \\\textbf{(c) } A^{-1}b= \begin{bmatrix} -3\\ 3\\-2\end{bmatrix} \Rightarrow \bbox[red, 2pt]{b= -3A_1+3A_2 -2A_3}$$

解答:$$A=\begin{bmatrix}2 & -1 & 0 & 1 \\0 & 3 & -1 & 0 \\0 & 1 & 1 & 0 \\0 & -1 & 0 & 3 \end{bmatrix} \Rightarrow \det(A-\lambda I)= (\lambda-2)^3(\lambda-3)=0\\ \lambda_1=2 \Rightarrow (A-\lambda_1 I)v=0 \Rightarrow \begin{bmatrix}0 & -1 & 0 & 1 \\0 & 1 & -1 & 0 \\0 & 1 & -1 & 0 \\0 & -1 & 0 & 1 \end{bmatrix} \begin{bmatrix} x_1 \\x_2 \\x_3 \\x_4 \end{bmatrix} =0 \Rightarrow \cases{x_2=x_4\\ x_2=x_3} \\\qquad \Rightarrow v=a\begin{bmatrix}1 \\0 \\0 \\0 \end{bmatrix}+ b\begin{bmatrix}0 \\1 \\1 \\1\end{bmatrix} , \text{we choose }v_1=\begin{bmatrix}1 \\0 \\0 \\0 \end{bmatrix},v_2=\begin{bmatrix}0 \\1 \\1 \\1\end{bmatrix}\\ \lambda_2=3 \Rightarrow (A-\lambda_2 I)v=0 \Rightarrow  \begin{bmatrix}-1 & -1 & 0 & 1 \\0 & 0 & -1 & 0 \\0 & 1 & -2 & 0 \\0 & -1 & 0 & 0 \end{bmatrix} \begin{bmatrix} x_1 \\x_2 \\x_3 \\x_4 \end{bmatrix} =0 \Rightarrow \cases{x_1=x_4\\ x_2=x_3=0} \\\qquad \Rightarrow v=a\begin{bmatrix}1 \\0 \\0 \\1 \end{bmatrix}, \text{we choose }v_3=\begin{bmatrix}1 \\0 \\0 \\1 \end{bmatrix}\\ A \text{ is 4 by 4 and has 3 eigenvectors, then }\bbox[red, 2pt]{J= \begin{bmatrix} 2 & 1 & 0 & 0\\ 0 & 2& 0 & 0\\ 0 & 0 & 2& 0\\ 0 & 0 & 0 & 3\end{bmatrix}}$$

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