115年台大碩士班- 數學B詳解

國立臺灣大學 115 學年度碩士班招生考試試題

科目: 數學(B)

解答:

$$\textbf{(a) }\cases{u=x+y\\ v=x-y} \Rightarrow \cases{x=(u+v)/2\\ y=(u-v)/2} \Rightarrow \begin{Vmatrix} x_u& x_v\\ y_u& y_v\end{Vmatrix} = \begin{Vmatrix} 1/2& 1/2\\ 1/2& -1/2\end{Vmatrix} ={1\over 2} \\\quad \Rightarrow \iint_R e^{x+y}\,dx\,dy = \int_{-1}^1 \int_{-1}^1 e^u\cdot {1\over 2}\,du\,dv = \bbox[red, 2pt]{e-e^{-1}} \\\textbf{(b) }x^2+xy+y^2 =(x+{1\over 2})^2+ {3\over 4}y^2 \Rightarrow \cases{u=x+{1\over 2}y\\ v={\sqrt 3\over 2}y} \Rightarrow \begin{Vmatrix}u_x & u_y\\ v_x& v_y \end{Vmatrix} =\begin{Vmatrix}1 & 1/2\\ 0& \sqrt 3/2 \end{Vmatrix} ={\sqrt 3\over 2} \\\quad \Rightarrow J={2\over \sqrt 3} \Rightarrow I=\int_{-\infty}^\infty \int_{-\infty}^\infty e^{-(x^2+xy+y^2)} \,dx \,dy = {2\over \sqrt 3}\int_{-\infty}^\infty \int_{-\infty}^\infty e^{-(u^2+v^2)} \,du \,dv  \\ \quad \cases{u= r\cos \theta\\ v= r\sin \theta} \Rightarrow I={2\over \sqrt 3} \int_0^{2\pi} \int_0^\infty e^{-r^2} r\,dr\,d\theta ={2\pi\over \sqrt 3} =\bbox[red, 2pt]{2\sqrt 3\pi\over 3}$$

解答:$$\textbf{(a) } \cases{u=f(x) \\ dv=dx} \Rightarrow \cases{du =f'(x)dx\\ v=x} \Rightarrow  \int_a^b f(x)\,dx = \left. \left[ xf(x) \right] \right|_a^b - \int_a^b xf'(x)\,dx \\\qquad =af(a) -bf(b)- \int_a^b xf'(x)\,dx \\\qquad \text{Let }y=f(x) \Rightarrow \cases{dy=f'(x)dx\\ x=g(y)  } \Rightarrow \int_a^b xf'(x)dx = \int_{f(a)}^{f(b)} g(y)\,dy  \\ \qquad \Rightarrow \int_a^b f(x)\,dx =af(a) -bf(b)-\int_{f(a)}^{f(b)} g(y)\,dy \quad \bbox[red, 2pt]{QED.} \\\textbf{(b) } \cases{f(x)=\ln x\\ [a,b]=[1,e]} \text{ and }y=\ln x \Rightarrow x=e^y =g(y)\Rightarrow \int_1^e \ln x\,dx =e\cdot f(e)-1\cdot f(1)-\int_{f(1)}^{f(e)} g(y) \,dy \\ \qquad =e\cdot 1-1\cdot 0-\int_{0}^1 e^y\,dy =e- \left. \left[ e^y \right] \right|_0^1 =e-e+1= \bbox[red, 2pt]1$$

解答:$$\textbf{(a) } \text{Forward Implication}(\Rightarrow )\\ \qquad P \text{ is orthogonal projection  matrix } \Rightarrow Pv=v , \text{if } v\in S\\v \in \mathbb R^n \Rightarrow Pv =w\in S \Rightarrow (v-Pv) \bot w \Rightarrow Pv-P(Pv) =0 \;(Pv\in S \Rightarrow P(Pv)=Pv) \Rightarrow P^2=P \\ \text{for all }v\in \mathbb R^n \text{ and }w\in S, \text{ we have }w^T(v-Pv) =0 \Rightarrow (Pu)^T(v-Pv)=0 \Rightarrow u^TP^T(I-P)v=0 \\ \Rightarrow P^T(I-P)=0 \Rightarrow P^T=P^TP \Rightarrow (P^T)^T=(P^TP)^T \Rightarrow P=P^TP =P^T \Rightarrow P=P^T \\ \text{Reverse Implication}(\Leftarrow) \\ w\in Im(P) \Rightarrow w=Pu \Rightarrow w^T(v-Pv)=(Pu)^T (v-Pv) =u^TP^T(v-Pv) =u^TP(v-Pv) \\=u^T(Pv-P^2v) =u^T(Pv-Pv)=0\\ \text{We have shown that }P \text{ is orthogonal projection matrix iff }P^2=P \text{ and }P^T=P\;\bbox[red, 2pt]{QED.} \\\textbf{(b) } P_A= A(A^TA)^{-1}A^T \Rightarrow P_A^T = (A(A^TA)^{-1}A^T)^T =A((A^TA)^T)^{-1}A^T= A(A^TA)^{-1} A^T=P_A\\ \quad \Rightarrow P_A^T=P_A\\ (P_A)^2 =(A(A^TA)^{-1}A^T)(A(A^TA)^{-1}A^T)=A(A^TA)^{-1}A^TA(A^TA)^{-1}A^T =A(A^TA)^{-1}IA^T \\\quad =A(A^TA)^{-1} A^T =P_A \Rightarrow P_A^2=P_A \\ \text{That is, we have shown that } P_A^T =P_A \text{ and }P_A^2=P_A \Rightarrow P_A \text{ is an orthogonal projection matrix}\\ \qquad \bbox[red, 2pt]{QED.}$$

解答:$$\textbf{(a) } \{v_1,\dots, v_n\} \subseteq V \text{ is linearly independent if} \\\qquad c_1v_1+c_2v_2+ \cdots+c_nv_n =0 \Rightarrow c_1=c_2 =\cdots=c_n=0 \\\textbf{(b) } \text{Proof by Induction} \\\quad k=1 \Rightarrow c_1v_1=0 \Rightarrow c_1=0 \;(\text{eigenvector }v_1\ne 0) \\\text{Assume that the proposition holds for any set of }k-1.\\ \qquad c_1v_1+ \cdots+ c_kv_k=0 \cdots(1) \Rightarrow T(c_1v_1+ \cdots+ c_kv_k)=0 \Rightarrow c_1\lambda_1v_1+\cdots+ c_k\lambda_k v_k=0 \cdots(2) \\ \quad \Rightarrow \lambda_k\cdot (1)-(2) =(c_1\lambda_k-c_1\lambda_1)v_1+ (c_2\lambda_k-c_2\lambda_2) v_2+ \cdots +(c_{k-1}\lambda_k-c_{k-1}\lambda_{k-1})v_{k-1}=0 \\ \Rightarrow c_1(\lambda_k-\lambda_1) =c_2(\lambda_k-\lambda_2) = \cdots =c_{k-1}(\lambda_k-\lambda_{k-1}) =0 \Rightarrow c_1=c_2= \cdots=c_{k-1}=0\\ \quad \text{Substitue }c_1=c_2=\cdots =c_{k-1} =0 \text{ back into (1)} \Rightarrow c_kv_k=0 \Rightarrow c_k=0 \;(\text{eigenvector }v_k\ne 0) \\ \quad \Rightarrow c_1=c_2= \cdots =c_k=0 \Rightarrow \{v_1, v_2, \dots,v_k\} \text{ is linearly independent. }\bbox[red, 2pt]{Q.E.D.}$$

解答:$$\textbf{(a) } \cases{f= \sum_{i=1}^n x_iy_i \\ g= \sum_{i=1}^n x_i^2-1 \\ h= \sum_{i=1}^ny_i^2-1 } \Rightarrow \cases{f_{x_i} =\lambda g_{x_i}, i=1,\dots,n\\ f_{y_1}=\sigma h_{y_i},i=1,\dots,n} \Rightarrow \cases{y_i=2 \lambda x_i \\ x_i =2\sigma y_i \\ \sum x_i^2=1\\ \sum y_i^2=1} \Rightarrow \cases{\sum y_i^2 =\sum 4\lambda^2 x_i^2 \\ \sum x_i^2 = \sum 4\sigma^2 y_i^2} \\ \quad \Rightarrow \cases{1=4\lambda^2\\ 1=4\sigma^2} \Rightarrow \cases{\lambda=\pm 1/2\\ \sigma=\pm 1/2} \\ \lambda= \sigma=1/2 \Rightarrow y_i=x_i \Rightarrow \max(f)=\sum_{i=1}^n x_ix_i = \bbox[red, 2pt]1 \\\textbf{(b) } \text{Let }\cases{\sum_{i=1}^n a_i^2 =A\\ \sum_{i=1}^n b_i^2 =B} , \text{ then we have }\cases{b_i=2\lambda a_i\\ a_i =2\sigma b_i} \Rightarrow 4\lambda^2 A= B \Rightarrow \lambda=\pm {1\over 2} \sqrt{B\over A} \\ \Rightarrow \sum_{i=1}^n a_ib_i \le \sum_{i=1}^n a_i \cdot \sqrt{B\over A}a_i =\sqrt{B\over A}\sum_{i=1}^n a_i^2=\sqrt{B\over A} \cdot A=\sqrt{A\cdot B} =\sqrt{\sum_{i=1}^n a_i^2} \cdot \sqrt{\sum_{i=1}^n b_i^2} \\ \Rightarrow \sum_{i=1}^n a_ib_i \le \sqrt{\sum_{i=1}^n a_i^2} \cdot \sqrt{\sum_{i=1}^n b_i^2} \quad \bbox[red, 2pt]{\text{QED.}}$$

解答:$$\textbf{(a) } \det(A-\lambda I) =-(\lambda-1)(\lambda-3)^2=0 \Rightarrow \lambda=1,3 \\ \lambda_1=1 \Rightarrow (A-\lambda_1 I) v=0 \Rightarrow \begin{bmatrix} 1 & 1 & 0 \\1 & 1 & 0 \\0 & 0 & 2\end{bmatrix} \begin{bmatrix}x_1\\ x_2\\ x_3 \end{bmatrix} =0 \Rightarrow \cases{x_1+x_2=0 \\ x_3=0 } \\ \qquad \Rightarrow v=x_2 \begin{bmatrix}-1\\1\\0 \end{bmatrix}, \text{ choosing }v_1=\begin{bmatrix}-1\\1\\0 \end{bmatrix} \\ \lambda_2=3 \Rightarrow (A- \lambda_2 I)v=0 \Rightarrow  \begin{bmatrix}-1 & 1 & 0 \\1 & -1 & 0 \\0 & 0 & 0 \end{bmatrix} \begin{bmatrix}x_1\\ x_2\\ x_3 \end{bmatrix}=0 \Rightarrow x_1=x_2 \\ \qquad \Rightarrow v= x_2 \begin{bmatrix}1\\1\\0 \end{bmatrix}+x_3 \begin{bmatrix}0\\ 0\\1 \end{bmatrix}, \text{ choosing }v_2= \begin{bmatrix}1\\1\\0 \end{bmatrix} , v_3= \begin{bmatrix}0\\ 0\\1 \end{bmatrix} \\ \Rightarrow P=[v_1\; v_2\; v_3] = \begin{bmatrix}-1&1& 0\\ 1& 1& 0\\ 0& 0& 3\end{bmatrix}, D= \begin{bmatrix} \lambda_1 & 0& 0\\ 0& \lambda_2 & 0\\ 0& 0& \lambda_2 \end{bmatrix} = \begin{bmatrix} 1& 0& 0\\ 0& 3 & 0\\ 0& 0& 3 \end{bmatrix} \\ \Rightarrow \bbox[red, 2pt]{A= \begin{bmatrix}-1&1& 0\\ 1& 1& 0\\ 0& 0& 3\end{bmatrix} \begin{bmatrix}1& 0& 0\\ 0& 3 & 0\\ 0& 0& 3 \end{bmatrix} \begin{bmatrix}-1&1& 0\\ 1& 1& 0\\ 0& 0& 3\end{bmatrix}^{-1}} \\\textbf{(b) }B^2=A \Rightarrow B =\sqrt A  \Rightarrow B= \bbox[red, 2pt]{P \begin{bmatrix} \pm 1 & 0& 0\\ 0& \pm \sqrt 3& 0\\ 0& 0&\pm \sqrt 3\end{bmatrix}P^{-1}}$$

解答:$$\textbf{positivity: } (A^TA)_{jj} = \sum_{i=1}^n (A^T)_{ji}a_{ij} = \sum_{i=1}^n a_{ij}a_{ij}= \sum_{i=1}^n a_{ij}^2 \Rightarrow tr(A^TA) =\sum_{j=1}^n \sum_{i=1}^n a_{ij}^2 \ge 0 \\ \qquad \cases{tr(A^TA) =0 \Rightarrow a_{ij}^2=0 \Rightarrow a_{ij}=0 \Rightarrow A=0 \\ A\ne 0 \Rightarrow  \text{ for some }a_{ij}\ne 0 \Rightarrow tr(A^TA) \gt 0} \\ \textbf{symmetry: } tr(M)=tr(M^T) \Rightarrow tr(A^TB)= tr(A^TB)^T= tr(B^TA) \Rightarrow tr(A^TB)=tr(B^TA) \\\textbf{additivity: }tr((A+C)^TB) =tr((A^T+C^T)B) =tr(A^TB+C^TB)= tr(A^TB)+ tr(C^TB) \\\textbf{homogeneity: }tr((kA)^TB) =tr(kA^TB) =k\cdot tr(A^TB) \\ \Rightarrow \langle A, B\rangle=tr(A^TB) \text{ is an inner product }\; \bbox[red, 2pt]{QED.}$$

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解題僅供參考，碩士班歷年試題及詳解