國立中央大學115學年度碩士班考試入學
系所:數學系碩士班
科目:線性代數
解答:$$\textbf{(a) } A = \begin{bmatrix} 1 & -2 & -2 & -1 & -2 \\ 1 & -1 & -1 & -1 & -1 \\ 2 & 3 & 3 & -2 & 3 \\ -3 & 1 & 1 & 3 & 1 \\ 1 & 2 & 2 & -1 & 2 \end{bmatrix} \xrightarrow{\matrix{R_2 \gets R_2-R_1\\ R_3\gets R_3-2R_1\\ R_4\gets R_4+3R_1\\ R_5 \gets R_5-R_1}} \begin{bmatrix} 1 & -2 & -2 & -1 & -2 \\ 0 & 1 & 1 & 0 & 1 \\ 0 & 7 & 7 & 0 & 7 \\ 0 & -5 & -5 & 0 & -5 \\ 0 & 4 & 4 & 0 & 4 \end{bmatrix} \xrightarrow{ \matrix{R_1 \gets R_1+2R_2\\ R_3\gets R_3-7R_2\\ R_4 \gets R_4 +5R_2\\ R_5 \gets R_5-4R_2}} \\ \qquad \begin{bmatrix} 1 & 0 & 0 & -1 & 0 \\ 0 & 1 & 1 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix} \Rightarrow \text{RREF}(A) = \bbox[red, 2pt]{\begin{bmatrix} 1 & 0 & 0 & -1 & 0 \\ 0 & 1 & 1 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}} \\\textbf{(b) } \begin{bmatrix} 1 & 0 & 0 & -1 & 0 \\ 0 & 1 & 1 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix} \begin{bmatrix}x_1\\ x_2\\ x_3\\ x_4\\ x_5 \end{bmatrix} =0 \Rightarrow \cases{x_1=x_4\\ x_2+x_3+x_5=0} \Rightarrow \mathbf x = \begin{bmatrix} x_4 \\ -x_3 - x_5 \\ x_3 \\ x_4 \\ x_5 \end{bmatrix} \\ \qquad = x_3 \begin{bmatrix} 0 \\ -1 \\ 1 \\ 0 \\ 0 \end{bmatrix} + x_4 \begin{bmatrix} 1 \\ 0 \\ 0 \\ 1 \\ 0 \end{bmatrix} + x_5 \begin{bmatrix} 0 \\ -1 \\ 0 \\ 0 \\ 1 \end{bmatrix} \Rightarrow R(A)^\perp = N(A) = \bbox[red, 2pt]{\text{span} \left\{ \begin{bmatrix} 0 \\ -1 \\ 1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 1 \\ 0 \\ 0 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ -1 \\ 0 \\ 0 \\ 1 \end{bmatrix} \right\} } \\ \textbf{(c) } \text{Let } \mathbf{v}_1 = \begin{bmatrix} 0 \\ -1 \\ 1 \\ 0 \\ 0 \end{bmatrix}, \quad \mathbf{v}_2 = \begin{bmatrix} 1 \\ 0 \\ 0 \\ 1 \\ 0 \end{bmatrix}, \quad \mathbf{v}_3 = \begin{bmatrix} 0 \\ -1 \\ 0 \\ 0 \\ 1 \end{bmatrix}, \text{ by Gram-Schmidt processing,} \\ \qquad \mathbf{u}_1 = \mathbf{v}_1 = \begin{bmatrix} 0 \\ -1 \\ 1 \\ 0 \\ 0 \end{bmatrix} \Rightarrow \mathbf{u}_2 = \mathbf{v}_2 - \text{proj}_{\mathbf{u}_1}(\mathbf{v}_2) = \mathbf{v}_2 - \frac{\mathbf{v}_2 \cdot \mathbf{u}_1}{\mathbf{u}_1 \cdot \mathbf{u}_1} \mathbf{u}_1= \begin{bmatrix}1\\0\\0\\1 \\0 \end{bmatrix}\\ \qquad \Rightarrow \mathbf u_3= \mathbf{v}_3 - \frac{\mathbf{v}_3 \cdot \mathbf{u}_1}{\mathbf{u}_1 \cdot \mathbf{u}_1} \mathbf{u}_1 - \frac{\mathbf{v}_3 \cdot \mathbf{u}_2}{\mathbf{u}_2 \cdot \mathbf{u}_2} \mathbf{u}_2 = \begin{bmatrix}0 \\-1/2\\ -1/2\\0 \\1 \end{bmatrix} \Rightarrow \cases{\mathbf e_1= \mathbf u_1/|| \mathbf u_1|| \\\mathbf e_2= \mathbf u_2/|| \mathbf u_2|| \\\mathbf e_3= \mathbf u_3/|| \mathbf u_3|| \\\mathbf e_4= \mathbf u_4/|| \mathbf u_4|| \\\mathbf e_5= \mathbf u_5/|| \mathbf u_5|| } \\ \qquad \Rightarrow \text{orthonormal basis: }\{\mathbf e_1, \mathbf e_2, \mathbf e_3, \mathbf e_4, \mathbf e_5 \} = \bbox[red, 2pt]{ \left\{ \frac{1}{\sqrt{2}}\begin{bmatrix} 0 \\ -1 \\ 1 \\ 0 \\ 0 \end{bmatrix}, \frac{1}{\sqrt{2}}\begin{bmatrix} 1 \\ 0 \\ 0 \\ 1 \\ 0 \end{bmatrix}, \frac{1}{\sqrt{6}}\begin{bmatrix} 0 \\ -1 \\ -1 \\ 0 \\ 2 \end{bmatrix} \right\} }$$
解答:$$W_1\cap W_2 \text{ is finite } \Rightarrow \text{dim}(W_1 \cap W_2) =k \lt \infty \Rightarrow \text{ a basis of }W_1\cap W_2 =\{v_1,v_2,\dots, v_k\} \\ \text{Suppose dim}(W_1)=m, \text{ we can find }u_1,u_2, \dots, u_{m-k} \text{ such that }\\ B_1 = \{v_1, \dots, v_k, u_1, \dots, u_{m-k}\} \text{ is a bsis of }W_1\\ \text{At the same way, dim}(W_2)=n,\text{ we can find }w_1,w_2, \dots, w_{n-k} \text{ such that }\\ B_2 = \{v_1, \dots, v_k, w_1, \dots, w_{n-k}\}\text{ is a bsis of }W_2 \\ \text{Now we want to prove that } B = B_1 \cup B_2 = \{v_1, \dots, v_k, u_1, \dots, u_{m-k}, w_1, \dots, w_{n-k}\} \\ \qquad \text{ is a basis of }W_1+W_2\\ x\in W_1+W_2 \Rightarrow x=x_1+x_2 , \text{ where }\cases{x_1\in W_1 \\x_2\in W_2} \Rightarrow \cases{x_1 = \sum_{i=1}^k a_i v_i + \sum_{j=1}^{m-k} b_j u_j\\ x_2 = \sum_{i=1}^k c_i v_i + \sum_{l=1}^{n-k} d_l w_l } \\ \Rightarrow x = x_1 + x_2 = \sum_{i=1}^k (a_i + c_i)v_i + \sum_{j=1}^{m-k} b_j u_j + \sum_{l=1}^{n-k} d_l w_l = \text{span}\{B\} \\ \text{Suppose scales }\alpha_i, \beta_j, \gamma_{\ell} \in F, \text{ such that }\sum_{i=1}^k \alpha_i v_i + \sum_{j=1}^{m-k} \beta_j u_j + \sum_{\ell=1}^{n-k} \gamma_{\ell} w_{\ell} = 0 \\ \Rightarrow \sum_{\ell=1}^{n-k} \gamma_{\ell} w_{\ell} = - \left( \sum_{i=1}^k \alpha_i v_i + \sum_{j=1}^{m-k} \beta_j u_j \right) , \text{ where} \\\sum_{\ell=1}^{n-k} \gamma_{\ell} w_{\ell} \in \text{span}\{B_2\} \text{ and } \not \in \{W_1\cap W_2\}, \text{ and} - \left( \sum_{i=1}^k \alpha_i v_i + \sum_{j=1}^{m-k} \beta_j u_j \right) \in \text{ span{$B_1$}} \\ \Rightarrow z=\sum_{\ell=1}^{n-k} \gamma_{\ell} w_{\ell} \in W_1\cap W_2 \Rightarrow z = \sum_{i=1}^k \delta_i v_i \Rightarrow \sum_{i=1}^k \delta_i v_i - \sum_{l=1}^{n-k} \gamma_l w_l = 0 \\ \text{Since }\{v_1, \dots, v_k, w_1, \dots, w_{n-k}\}=B_2 \text{ is linearly independent}, \gamma_{\ell}=0 (\ell=1,\dots, n-k) \\ \Rightarrow \sum_{i=1}^k \alpha_i v_i + \sum_{j=1}^{m-k} \beta_j u_j = 0 \Rightarrow \alpha_i = 0 (i = 1, \dots, k) ,\beta_j = 0 (j = 1, \dots, m-k)\\ \Rightarrow B \text{ is linearly independent. That is, $B$ is a basis of $W_1+W_2$} \\ \Rightarrow \dim(W_1 + W_2) = k + (m - k) + (n - k) = m + n - k = \dim(W_1) + \dim(W_2) - \dim(W_1 \cap W_2) \\ \bbox[red, 2pt]{QED.}$$
解答:$$\textbf{(a) }\cases{a_0=0\\ a_1 =0\\ a_2 =0} \Rightarrow f(x)= a_0 + a_1 \cos(2\pi x) + a_2 \sin(2\pi x) =0 \Rightarrow 0 \in C^1(\mathbb R) \\ \cases{f(x) \in W \Rightarrow f(x)= a_0+a_1 \cos(2\pi x)+ a_2 \sin(2\pi x) \\ g(x)\in W \Rightarrow g(x)=b_0+ b_1\cos(2\pi x)+ b_2\sin(2\pi x)} \\\Rightarrow (f+g)(x) = (a_0+b_0)+ (a_1+b_1)\cos(2\pi x) + (a_2+b_2) \sin(2\pi x) \in W \;(\because a_i+bi\in \mathbb R) \\ \cases{f\in W\\ c\in \mathbb R} \Rightarrow (cf)(x) =c(a_0+a_1 \cos(2\pi x) +a_2 \sin(2\pi x)) =(ca_0)+ (ca_1) \cos(2\pi x) +ca_2\sin(2\pi x) \\ \text{Since }ca_i \in \mathbb R, cf \in W. \text{ In a additon, $\cos (2\pi x)$ and $\sin(2\pi x)$ are continuously differentiable. } \\ \text{Therefore, $W$ is a subspace of }C^1(\mathbb R) \\[2em] \textbf{(b) } W = \text{span}\{1, \cos(2\pi x), \sin(2\pi x)\}, \text{ then we choose }\beta=\{1, \cos(2\pi x), \sin(2\pi x)\}\\ \text{Now we want to prove that $\beta$ is a basis of $W$.}\\ \text{Suppose } c_1(1) + c_2 \cos(2\pi x) + c_3 \sin(2\pi x) = 0 , \text{ for all }x\in \mathbb R \\ \Rightarrow \cases{x=0 \Rightarrow c_1+c_2=0\\ x=1/4 \Rightarrow c_1+c_3=0\\ x=1/2 \Rightarrow c_1= c_2} \Rightarrow c_1=c_2=c_3=0 \Rightarrow \beta \text{ is linearly independent} \\ \Rightarrow \text{$\beta$ is a basis of $W$ and contains 3 elements. That is, dim}(W)=3. \quad \bbox[red, 2pt]{QED.}\\[2em] \textbf{(c) } f\in W \Rightarrow f(x) = a_0 + a_1 \cos(2\pi x) + a_2 \sin(2\pi x) \Rightarrow f'(x) = -2\pi a_1 \sin(2\pi x) + 2\pi a_2 \cos(2\pi x) \\\qquad =0\cdot 1+(2\pi a_2)\cos(2\pi x)+(-2\pi a_1)\sin(2\pi x) \Rightarrow f'\in W \Rightarrow D(f) \in W\\ \text{Additionally, }\cases{f,g\in W\\ c\in \mathbb R} \Rightarrow \cases{D(f+g) = (f+g)'=f'+g'=D(f)+D(g) \\ D(cf)=(cf)'=cf'=cD(f)} \\\qquad \Rightarrow D \text{ is a linear transformation.} \\[2em] \text{(d) } \cases{f_1=1\\ f_2=\cos(2\pi x) \\ f_3 =\sin(2\pi x)} \Rightarrow \cases{D(f_1) =0 =0\cdot 1+0\cdot \cos(2\pi x) + 0\cdot \sin(2\pi x)\Rightarrow v_1 \begin{bmatrix}0\\0\\0 \end{bmatrix} \\ D(f_2)=-2\pi\sin(2\pi x) \Rightarrow v_2= \begin{bmatrix}0\\0\\ -2\pi \end{bmatrix} \\ D(f_3)=2\pi \cos(2\pi x) \Rightarrow v_3= \begin{bmatrix}0\\2\pi \\0 \end{bmatrix}} \\ \qquad \Rightarrow [D]_{\beta} =[v_1\; v_2\; v_3] = \bbox[red, 2pt] { \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 2\pi \\ 0 & -2\pi & 0 \end{bmatrix} } $$
解答:$$\textbf{(a) }A \text{ is diagonalizable} \Rightarrow A= QDQ^{-1}, \text{ where $D$ is a diagonal matrix and the diagonal elmenets: }\\ \qquad \lambda_1, \lambda_2, \dots, \lambda_n\text{ are the eigenvalues of }A. \text{ Then we have } \\\qquad P(x)= \det(A-x I)= c(x - \lambda_1)(x - \lambda_2)\cdots(x - \lambda_n), c=(-1)^n \\ \Rightarrow P(A)= c(A - \lambda_1 I)(A - \lambda_2 I)\cdots(A - \lambda_n I) \\\qquad = c \left( Q(D - \lambda_1 I)Q^{-1} \right) \left( Q(D - \lambda_2 I)Q^{-1} \right) \cdots \left( Q(D - \lambda_n I)Q^{-1} \right) \\\qquad = c \, Q \left( (D - \lambda_1 I)(D - \lambda_2 I)\cdots(D - \lambda_n I) \right) Q^{-1} \\ \text{Let } M = (D - \lambda_1 I)(D - \lambda_2 I)\cdots(D - \lambda_n I), \text{ then } [M]_{k,k} = (\lambda_k - \lambda_1)(\lambda_k - \lambda_2)\cdots(\lambda_k - \lambda_n) =0 \\ \Rightarrow P(A)=cQ\cdot 0\cdot Q^{-1}=0 \quad \bbox[red, 2pt]{QED.} \\[2em] \textbf{(b) }P(x) =\det(A-xI) = c_0 + c_1 x + c_2 x^2 + \dots + c_n x^n \Rightarrow c_0=P(0) =\det(A-0\cdot I)= \det(A)\\ \quad 0 \text{ is not an eigenvalue of $A$} \Rightarrow c_0=P(0)=\det(A) \ne 0 \Rightarrow \text{ A is invertible} \\ P(A)=0 \Rightarrow c_0 I + c_1 A + c_2 A^2 + \dots + c_n A^n = 0 \Rightarrow c_0 I = - (c_1 A + c_2 A^2 + \dots + c_n A^n) \\ \Rightarrow I = A \left( -\frac{c_1}{c_0} I - \frac{c_2}{c_0} A - \dots - \frac{c_n}{c_0} A^{n-1} \right) \Rightarrow \bbox[red, 2pt]{ A^{-1} = -\frac{1}{c_0} (c_1 I + c_2 A + \dots + c_n A^{n-1}) }\\[2em] \textbf{(c) } P(x) = (-1)^n x^n + c_{n-1} x^{n-1} + \dots + c_1 x + c_0 = (-1)^n (x - \lambda_1)(x - \lambda_2)\cdots(x - \lambda_n) \\ P(0)= \det(A-0\cdot I)= \det(A)=c_0 \\ \text{coefficient of }x^{n-1}:(-1)^n(-\lambda_1-\lambda_2-\cdots-\lambda_n)= (-1)^{n+1}(\lambda_1+ \lambda_2+ \cdots +\lambda_n)\\\qquad =(-1)^{n+1} \text{tr}(A) \Rightarrow c_{n-1} = (-1)^{n+1} \text{tr}(A) \\ \text{Therefore, }\bbox[red, 2pt]{\det(A) =c_0, \text{tr}(A)=(-1)^{n-1}c_{n-1}}$$
解答:$$\textbf{(a) }\textbf{Additivity: }\langle A+C, B\rangle_F = \text{tr}(B^*(A+C)) =\text{tr}(B^*A) +\text{tr}(B^*C) =\langle A, B\rangle_F +\langle C, B\rangle_F \\ \qquad \Rightarrow \langle A+C, B\rangle_F = \langle A, B\rangle_F +\langle C, B\rangle_F \\ \textbf{Homogeneity: } \langle cA,B \rangle_F= \text{tr}(B^*(cA)) = \text{tr}(c(B^*A)) = c\cdot \text{tr}(B^*A) =c \langle A,B \rangle_F \\ \textbf{Conjugate Symmetry: } \overline{\langle A, B \rangle_F} = \overline{ \text{tr}(B^*A)} = \text{tr}(\overline{B^*A}) = \text{tr}((B^*A)^*) = \textbf{tr}(A^*B) = \langle B,A \rangle_F \\ \textbf{Positive Definiteness: }\langle A, A \rangle_F = \text{tr}(A^*A) = \sum_{j=1}^n \sum_{i=1}^m \vert{}a_{ij}\vert{}^2 \ge 0 \\ \text{Additionally, } \sum_{j=1}^n \sum_{i=1}^m \vert{}a_{ij}\vert{}^2 = 0 \Rightarrow \vert{}a_{ij}\vert{}^2 = 0 \Rightarrow a_{ij}=0 \Rightarrow A=0 \\ \text{All of four properties are satisfied, $\langle A, B \rangle_F := \text{tr}(B^* A)$ defines in inner product on $M_{m \times n}(\mathbb{C})$ } \\\textbf{(b) }A = U \Sigma V^* \Rightarrow \Vert{}A\Vert{}_F^2 = \text{tr}(A^*A) = \text{tr}((U\Sigma V^*)^*(U\Sigma V^*)) = \text{tr}(V\Sigma^*U^*U\Sigma V^*) = \text{tr}(V\Sigma^*\Sigma V^*) \\ = \text{tr}(V^*V\Sigma^*\Sigma) = \text{tr}(\Sigma^* \Sigma) = \sum_{i=1}^k \sigma_i^2(A) \Rightarrow \Vert{}A\Vert{}_F = \sqrt{ \sum_{i=1}^k \sigma_i^2(A)} \\ \Vert{}A\Vert{}_*^2 = \left( \sum_{i=1}^k \sigma_i \right)^2 = \sum_{i=1}^k \sigma_i^2 + \sum_{i \neq j} \sigma_i \sigma_j \Rightarrow \sum_{i=1}^k \sigma_i^2 \le \left( \sum_{i=1}^k \sigma_i \right)^2 \Rightarrow \Vert{}A\Vert{}_F^2 \le \Vert{}A\Vert{}_*^2 \Rightarrow \Vert{}A\Vert{}_F \le \Vert{}A\Vert{}_* \\ \Rightarrow \Vert{}A\Vert{}_F \le \sqrt{k} \Vert{}A\Vert{}_* \quad \bbox[red, 2pt]{QED.}$$
解答:$$A \text{ is invertible and }A^n=A \Rightarrow A^{n-1}=I \Rightarrow \lambda^{n-1}=1 \Rightarrow |\lambda|=1 \Rightarrow \lambda =e^{i\theta}\\ \lambda \text{ is an eigenvalue of $A$ }\Rightarrow \text{ eigenvalue of }A^*:\bar \lambda= e^{-i\theta}\\ \Rightarrow \text{eigenvalue of }A+A^*: \mu=e^{i\theta}+e^{-i\theta} = 2\cos \theta \Rightarrow \text{eigenvalue of }(A+A^*)^2 :\mu^2 =4\cos^2\theta \ge 0\\ \text{Suppose }\cos \theta=0 \Rightarrow \theta={\pi \over 2}+k\pi \;(k\in \mathbb Z) \Rightarrow \lambda=e^{i\theta} = \pm i\\ \Rightarrow \cases{\theta=i \Rightarrow \lambda^{n-1}=1 \Rightarrow i^{n-1}=1 \Rightarrow n-1 \equiv 0 \text{ (mod 4)} \\ \theta=-i \Rightarrow (-i)^{n-1}=1 \Rightarrow n-1 \equiv 0 \text{ (mod 4)}} \Rightarrow \text{ contradiction} \Rightarrow \cos\theta \ne 0\\ \Rightarrow \mu^2 \gt 0 \Rightarrow (A+A^*)^2 \text{ is positive-definite} \quad \bbox[red, 2pt]{QED.}$$
解答:$$\textbf{(a) }A = \begin{pmatrix} 5&-3& -3\\3&-1& -3\\ 3&-3& -1 \end{pmatrix} \Rightarrow \det(A-\lambda I) = -(\lambda-2)^2(\lambda-1)=0 \Rightarrow \lambda=2,1\\\quad \lambda_1 =2 \Rightarrow (A-\lambda_1 I)v=0 \Rightarrow \begin{pmatrix} 3 & -3 & -3 \\ 3 & -3 & -3 \\ 3 & -3 & -3 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \Rightarrow x_1=x_2+x_3 \\\qquad \Rightarrow v= \begin{pmatrix} x_2+x_3\\ x_2\\ x_3 \end{pmatrix}= x_2 \begin{pmatrix} 1\\1\\0 \end{pmatrix}+x_3 \begin{pmatrix} 1\\0\\1 \end{pmatrix} ,\text{ choosing } v_1 = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} , v_2 = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} \\ \lambda_2=1 \Rightarrow (A-\lambda_2 I)v=0 \Rightarrow \begin{pmatrix} 4 & -3 & -3 \\ 3 & -2 & -3 \\ 3 & -3 & -2 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \Rightarrow x_1=x_2=x_3 \\\qquad \Rightarrow v=x_1 \begin{pmatrix} 1\\1\\1 \end{pmatrix}, \text{ choosing }v_3 = \begin{pmatrix} 1\\1\\ 1 \end{pmatrix}\\ \bbox[red, 2pt]{\text{eigenvalues: 2,1 and the corresponding eigenvectors:} \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix},\begin{pmatrix} 1\\1\\1 \end{pmatrix}} \\[2em] \textbf{(b) }P = \begin{pmatrix} v_1\; v_2; v_3 \end{pmatrix} \Rightarrow \bbox[red, 2pt]{P= \begin{pmatrix} 1& 1& 1\\1& 0& 1\\0& 1& 1 \end{pmatrix}} \text{ and }P^{-1}AP = \begin{pmatrix} \lambda_1& 0& 0\\0& \lambda_1& 0\\0& 0& \lambda_2 \end{pmatrix} \text{ is a diagonal matrix}$$
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解題僅供參考，碩士班歷年試題及詳解
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