115年成功大學電機碩士班-線性代數詳解

國立成功大學115學年度碩士班招生考試試題

系所:電機工程學系 
科目:線性代數

解答:$$\textbf{a. }\bbox[red, 2pt]{False}:rank(A)=10 \Rightarrow D \text{ is }10\times 10 \ne 13\times 13 \\\textbf{b. }\bbox[red, 2pt]{True}:rank(A)=10 \Rightarrow D \text{ is }10\times 10 \Rightarrow U \text{ is }13\times 10 \\\textbf{c. }\bbox[red, 2pt]{False}:A_{13\times 20}=U_{13\times 10}D_{10\times 10}V^T \Rightarrow V^T_{10\times 20} \Rightarrow V_{20\times 10} \ne V_{10\times 20} \\\textbf{d. }\bbox[red, 2pt]{False}:A \text{ is symmetric} \Rightarrow A\text{ is diagonalizable }\Rightarrow \text{geometric multiplicity=algebraic mulicity} \\\qquad \Rightarrow \text{dim(Nul}(A)) = \text{ geometric multicity of }\lambda=0 =3\ne 2 \\\textbf{e. }\bbox[red, 2pt]{False}:Q\text{ is orthogonal} \Rightarrow Q^T=Q^{-1}\Rightarrow \cases{(Q^T)^3=(Q^{-1})^3=Q^{-3}\\ (Q^T)^{-1} =(Q^{-1})^{-1}=Q\\ (Q^{-1})^T =(Q^T)^T=Q} \\\qquad Q^2 (Q^T)^3 Q^{-1}(Q^T)^{-1}(Q^{-1})^T = Q^2Q^{-3}Q^{-1}QQ=I \ne Q \\\textbf{f. }\bbox[red, 2pt]{True}:A\text{ is orthonormal} \Rightarrow A^{-1} \text{ exists} \Rightarrow A^{-1}AX=A^{-1}B \Rightarrow \text{ unique solution }X=A^{-1}B \\\textbf{g. }\bbox[red, 2pt]{False}: x\ne 0 \text{ and }x\in W \Rightarrow Proj_W(x)=x \ne 0 \\\textbf{h. }\bbox[red, 2pt]{False}:  A=B= \begin{bmatrix}1&0\\0& 1 \end{bmatrix} \Rightarrow \cases{\det(A+B)=4\\ \det(A)+\det(B)=2} \Rightarrow 4\ne 2 \\\textbf{i. }\bbox[red, 2pt]{False}:\cases{A= \begin{bmatrix}1& 0\\0& 0 \end{bmatrix} \\B= \begin{bmatrix}0& 0\\0& 1 \end{bmatrix}} \Rightarrow AB= 0, \text{ bu t}A\ne 0, B\ne 0 \\\textbf{j. }\bbox[red, 2pt]{True}:y\in V \Rightarrow \text{Proj}_V(y)=y \Rightarrow \text{Proj}_V(y-\text{Proj}_V(y)) =\text{Proj}_V(0)=0$$

解答:$$\textbf{a. }A= \begin{bmatrix}1 & -2 & -4 & -3 \\2 & -7 & -7 & -6 \\-1 & 2 & 6 & 4 \\-4 & -1 & 9 & 8 \end{bmatrix} \stackrel {R_2-{\color{blue}2}R_1\to R_2}{ \Longrightarrow } \begin{bmatrix} 1 & -2 & -4 & -3 \\0 & -3 & 1 & 0 \\-1 & 2 & 6 & 4 \\-4 & -1 & 9 & 8\end{bmatrix} \stackrel {R_3-{\color{blue}(-1)}R_1\to R_3}{ \Longrightarrow } \begin{bmatrix}1 & -2 & -4 & -3 \\0 & -3 & 1 & 0 \\0 & 0 & 2 & 1 \\-4 & -1 & 9 & 8 \end{bmatrix} \\ \stackrel {R_4-{\color{blue}(-4)}R_1\to R_4}{ \Longrightarrow } \begin{bmatrix} 1 & -2 & -4 & -3 \\0 & -3 & 1 & 0 \\0 & 0 & 2 & 1 \\0 & -9 & -7 & -4\end{bmatrix} \stackrel {R_4-{\color{blue}(3)}R_2\to R_4}{ \Longrightarrow } \begin{bmatrix}1 & -2 & -4 & -3 \\0 & -3 & 1 & 0 \\0 & 0 & 2 & 1 \\0 & 0 & -10 & -4 \end{bmatrix} \stackrel {R_4-{\color{blue}(-5)}R_3\to R_4}{ \Longrightarrow } \\ \begin{bmatrix} 1 & -2 & -4 & -3 \\0 & -3 & 1 & 0 \\0 & 0 & 2 & 1 \\0 & 0 & 0 & 1\end{bmatrix} \Rightarrow \bbox[red, 2pt]{L= \begin{bmatrix} 1 & 0 & 0 & 0 \\2 & 1 & 0 & 0 \\-1 & 0 & 1 & 0 \\-4 & 3 & -5 & 1\end{bmatrix} , U= \begin{bmatrix} 1 & -2 & -4 & -3 \\0 & -3 & 1 & 0 \\0 & 0 & 2 & 1 \\0 & 0 & 0 & 1\end{bmatrix}} \\\textbf{b. }A= \begin{bmatrix} 7 & 4 & 16 \\2 & 5 & 8 \\-2 & -2 & -5\end{bmatrix} \Rightarrow \det(A-\lambda I) = -(\lambda-1)(\lambda-3)^2=0 \Rightarrow \cases{\lambda_1 =1\\ \lambda_2=3} \\ \lambda_1=1 \Rightarrow (A-\lambda_1 I)v=0 \Rightarrow \begin{bmatrix} 6 & 4 & 16 \\2 & 4 & 8 \\-2 & -2 & -6\end{bmatrix} \begin{bmatrix}x_1\\ x_2\\x_3 \end{bmatrix} =0 \Rightarrow \cases{x_1+ 2x_3=0\\ x_2+x_3=0} \Rightarrow v= x_3 \begin{bmatrix}-2\\-1\\1 \end{bmatrix} \\ \qquad \text{Choosing }v_1= \begin{bmatrix}-2\\ -1\\1 \end{bmatrix} \\ \lambda_2=3 \Rightarrow (A-\lambda_2 I)v=0 \Rightarrow \begin{bmatrix}4 & 4 & 16 \\2 & 2 & 8 \\-2 & -2 & -8 \end{bmatrix} \begin{bmatrix}x_1\\ x_2\\x_3 \end{bmatrix} =0 \Rightarrow x_1+x_2+4x_3=0 \\\qquad \Rightarrow v= x_2 \begin{bmatrix}-1\\1\\0 \end{bmatrix} +x_3 \begin{bmatrix}-4\\0\\1 \end{bmatrix}, \text{ choosing }v_2= \begin{bmatrix}-1 \\1\\0 \end{bmatrix}, v_3= \begin{bmatrix}-4\\ 0\\ 1 \end{bmatrix} \\ \Rightarrow P= [v_1 \; v_2\; v_3] \Rightarrow \bbox[red, 2pt]{P= \begin{bmatrix} 1 & 1 & 4 \\1 & 2 & 4 \\-1 & -1 & -3\end{bmatrix}}, D= \begin{bmatrix}\lambda_1& 0& 0\\0& \lambda_2&0\\ 0& 0& \lambda_2 \end{bmatrix} \Rightarrow \bbox[red, 2pt]{D=\begin{bmatrix} 1& 0& 0\\0& 3&0\\ 0& 0& 3 \end{bmatrix} } \\\textbf{c. } A = \begin{bmatrix} -2 & 3 \\ 5 & 7 \\ 2 & -2 \\ 4 & 6 \end{bmatrix} \Rightarrow \vec u_1= \begin{bmatrix}-2\\5\\2\\4 \end{bmatrix} , \vec u_2= \begin{bmatrix}3\\7\\-2\\6 \end{bmatrix}   \Rightarrow \vec e_1={\vec u_1\over |\vec u_1|} ={1\over 7}\begin{bmatrix}3\\7\\-2\\6 \end{bmatrix}  \Rightarrow \vec u_2\cdot \vec e_1 =7 \\ \Rightarrow \vec v_2= \vec u_2-(\vec u_2\cdot \vec e_1)\vec e_1 =\begin{bmatrix}3\\7\\-2\\6 \end{bmatrix} - \begin{bmatrix}-2\\5\\2\\4 \end{bmatrix} = \begin{bmatrix}5\\2\\-4\\2 \end{bmatrix} \Rightarrow \vec e_2={\vec v_2\over |\vec v_2|} = {1\over 7} \begin{bmatrix}5\\2\\-4\\2 \end{bmatrix} \\ \Rightarrow Q=[\vec e_1\; \vec e_2] \Rightarrow \bbox[red, 2pt]{Q={1\over 7} \begin{bmatrix}-2& 5\\5& 2\\2&-4\\4& 2 \end{bmatrix} }\Rightarrow R= \begin{bmatrix}\vec u_1\cdot \vec e_1 & \vec u_2\cdot \vec e_1\\0& \vec u_2\cdot \vec e_2 \end{bmatrix} \Rightarrow \bbox[red, 2pt]{R = \begin{bmatrix}7&7\\0&7 \end{bmatrix}} \\ \textbf{d. } A = \frac{1}{15} \begin{bmatrix} 19 & 14 & 10 \\ 8 & -2 & 20 \end{bmatrix} \Rightarrow AA^T= \begin{bmatrix} 73/25 & 36/25 \\ 36/25 & 52/25 \end{bmatrix} \Rightarrow  \det(AA^T-\lambda I) =0\\ \Rightarrow (25\lambda-100)(25\lambda-25) =0 \Rightarrow \cases{\lambda_1=4\\ \lambda_2=1} \Rightarrow \cases{\sigma_1=\sqrt{\lambda_1}=2\\ \sigma_2 =\sqrt{\lambda_2}=1} \Rightarrow \bbox[red, 2pt] {\Sigma= \begin{bmatrix}2& 0& 0\\0& 1&0 \end{bmatrix}}\\ \cases{\lambda_1=4 \Rightarrow \text{ normalized eigenvector }u_1={1\over 5} \begin{bmatrix}4\\3 \end{bmatrix}\\ \lambda_2=1 \Rightarrow \text{normailized eigenvector }u_2= {1\over 5}\begin{bmatrix}-3\\4 \end{bmatrix}} \Rightarrow \bbox[red, 2pt]{U={1\over 5} \begin{bmatrix}4&-3\\3& 4 \end{bmatrix}} \\ \cases{v_1= {1\over \sigma_1} A^T u_1 =  \frac{1}{2} \left( \frac{1}{15} \begin{bmatrix} 19 & 8 \\ 14 & -2 \\ 10 & 20 \end{bmatrix} \right) \left( \frac{1}{5} \begin{bmatrix} 4 \\ 3 \end{bmatrix} \right) =  \frac{1}{3} \begin{bmatrix} 2 \\ 1 \\ 2 \end{bmatrix} \\ v_2= {1\over \sigma_2}A^Tu_2 =  \frac{1}{1} \left( \frac{1}{15} \begin{bmatrix} 19 & 8 \\ 14 & -2 \\ 10 & 20 \end{bmatrix} \right) \left( \frac{1}{5} \begin{bmatrix} -3 \\ 4 \end{bmatrix} \right)  = \frac{1}{3} \begin{bmatrix} -1 \\ -2 \\ 2 \end{bmatrix} }  \\ \Rightarrow v_3={v_1\times v_2\over |v_1\times v_2|} = {1\over 3} \begin{bmatrix}2\\-2\\-1 \end{bmatrix} \Rightarrow V^T= [v_1\; v_2\; v_3] \Rightarrow \bbox[red, 2pt]{V^T= {1\over 3}\begin{bmatrix}2& 1& 2\\ -1&-1&2\\2&-2&-1 \end{bmatrix}}$$

解答:$$\textbf{a. } \cases{p_0(t)=1\\ p_1(t)=t\\ p_2(t)=t^2} \Rightarrow \cases{v_0=(1,1,1,1) \\ v_1=(-3,-1,1,3) \\v_2=(9,1,1,9)} \Rightarrow \cases{\langle p_2,p_0 \rangle =9+1+1+9=20\\ \langle p_0,p_0\rangle =1^2+1^2+1^2+1^2=4\\ \langle p_2,p_1 \rangle=-27-1+1+27=0\\ \langle p_1,p_1\rangle=9+1+1+9=20} \\ \Rightarrow \text{proj}_W(p_2)= {\langle p_2,p_0 \rangle\over \langle p_0,p_0 \rangle}p_0 +{\langle p_2,p_1 \rangle\over \langle p_1,p_1 \rangle}p_1 ={20\over 4}p_0+{0\over 20}p_1=5p_0= \bbox[red, 2pt]5 \\ \textbf{b. }\text{Using the GramSchmidt process:} \\\qquad \hat q(t)=p_2(t)-\text{proj}_W(p_2)=t^2-5 \Rightarrow \cases{\hat q(-3)=9-5=4 \\ \hat q(-1)=-4\\\hat q(1) =-4\\ \hat q(3) =4} \Rightarrow (4,-4,-4,4) =4(1,-1,-1,1) \\ \Rightarrow q(t)={1\over 4}\hat q(t)={1\over 4}(t^2-5) \Rightarrow \bbox[red, 2pt]{q(t)={1\over 4}t^2-{5\over 4}}$$

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解題僅供參考，碩士班歷年試題及詳解