國立台北科技大學110學年度碩士班招生考試
系所組別: 2121電機工程系碩士班戊組
第一節 線性代數(選考)
解答:$$\textbf{(1)}\;A=\left[\begin{matrix}2 & 4 & -2 & 1\\-2 & -5 & 7 & 3\\3 & 7 & -8 & 6\end{matrix}\right] \xrightarrow{R_1/2 \to R_1} \left[\begin{matrix}1 & 2 & -1 & 1/2\\-2 & -5 & 7 & 3\\3 & 7 & -8 & 6\end{matrix}\right] \xrightarrow{ 2R_1+R_2 \to R_2, -3R_1+R_3 \to R_3}\\ \left[\begin{matrix}1 & 2 & -1 & \frac{1}{2}\\0 & -1 & 5 & 4\\0 & 1 & -5 & \frac{9}{2}\end{matrix} \right] \xrightarrow{2R_2+R_1\to R_1, R_2+R_3\to R_3} \left[\begin{matrix}1 & 0 & 9 & \frac{17}{2}\\0 & -1 & 5 & 4\\0 & 0 & 0 & \frac{17}{2}\end{matrix}\right] \xrightarrow{-R_2\to R_2, R_3-R_1\to R_1} \\ \left[\begin{matrix}1 & 0 & 9 & 0\\0 & 1 & -5 & -4\\0 & 0 & 0 & \frac{17}{2}\end{matrix}\right] \xrightarrow{2R_3/17 \to R_3} \left[\begin{matrix}1 & 0 & 9 & 0\\0 & 1 & -5 & -4\\0 & 0 & 0 & 1\end{matrix}\right] \xrightarrow{4R_3+R_2\to R_2} \left[\begin{matrix}1 & 0 & 9 & 0\\0 & 1 & -5 & 0\\0 & 0 & 0 & 1\end{matrix}\right]\\\Rightarrow B= rref(A) =\left[ \begin{matrix}1 & 0 & 9 & 0\\0 & 1 & -5 & 0\\0 & 0 & 0 & 1\end{matrix} \right] \Rightarrow Ax=0 \Rightarrow Bx=0 \Rightarrow \left[\begin{matrix}1 & 0 & 9 & 0\\0 & 1 & -5 & 0\\0 & 0 & 0 & 1 \end{matrix} \right] \left[\begin{matrix}x_1\\x_2 \\ x_3\\ x_4 \end{matrix} \right]=0 \\ \Rightarrow \cases{x_1+9x_3=0 \\ x_2=5x_3\\ x_4=0} \Rightarrow x=x_3 \begin{pmatrix} -9\\ 5\\ 1\\ 0 \end{pmatrix} =\bbox[red, 2pt]{\left\{ k\begin{pmatrix} -9\\ 5\\ 1\\ 0\end{pmatrix} \mid k\in \mathbb R\right\}} \\ \textbf{(2)}\; Au= \left[\begin{matrix}2 & 4 & -2 & 1\\-2 & -5 & 7 & 3\\3 & 7 & -8 & 6\end{matrix}\right] \left[ \begin{matrix}3\\-2 \\ -1 \\0\end{matrix} \right] = \left[ \begin{matrix}3\\-2 \\ -1 \\0\end{matrix}\right] =\left[\begin{matrix}0\\-3\\3\end{matrix}\right] \ne 0 \Rightarrow \bbox[red, 2pt]{\mathbf u \not \in Null(A)}$$
解答:$$\textbf{(1)} \Vert VUx\Vert^2 =(VUx)^H(VUx)= x^HU^HV^HVUx = x^HU^HI_mUx = x^HU^HUx \\ \qquad =x^HI_nx = x^Hx =\Vert x\Vert^2 \Rightarrow \Vert VUx\Vert =\Vert x\Vert, \bbox[red, 2pt]{Q.E.D.}\\ \textbf{(2)}\; V \text{ is a }m\times m \text{ unitary matrix} \Rightarrow V^{-1} =V^H \Rightarrow V^HV=I_m \Rightarrow \det(V^HV)=\bbox[red, 2pt] 1 \\ \textbf{(3)}\; U \text{ is a }m\times n \text{ column-wise orthonormal matrix} \Rightarrow U^HU=I_n \Rightarrow trace(U^HU)= \bbox[red, 2pt]n\\ \textbf{(4)}\; V^HV=I_m \Rightarrow trace(V^HV)= trace(I_m)=\bbox[red, 2pt]m$$
解答:$$\textbf{(1)}\; \bbox[red, 2pt]{Prove}A\text{ is diagonalizable and invertible }\Rightarrow A=PDP^{-1} \text{ where D is a diagonal matrix}\\\qquad i.e., d_{i,j}=0, \forall i\ne j \Rightarrow D^{-1}=E, e_{i,j}=1/d_{i,j} \Rightarrow D^{-1} =E\text{ is also diagonal} \\\qquad A=PDP^{-1} \Rightarrow A^{-1} =(PDP^{-1})^{-1} =PD^{-1}P^{-1} =PEP^{-1}\\\qquad \Rightarrow A^{-1} \text{ is also diagonalizable and invertible} \\\textbf{(2)}\; \bbox[red, 2pt]{Prove}:\text{If }a=0, \det(A-a I)=\det(A)=0 \Rightarrow A \text{ is NOT invertible.}\\\qquad \text{ That is } a\ne 0 \Rightarrow Av=av \Rightarrow A^{-1}Av =A^{-1}av \Rightarrow v=aA^{-1}v\\\qquad \Rightarrow A^{-1}v ={1\over a}v \Rightarrow a^{-1} \text{ is an eigenvalue of }A^{-1} \\\textbf{(3)}\; \bbox[red, 2pt]{Prove}: A=QR \Rightarrow Q^{-1}A=R \Rightarrow Q^{-1}AQ=RQ \Rightarrow A \text{ is similar to }RQ \\ \textbf{(4)}\; \bbox[red, 2pt]{Prove}: \text{Let }\lambda \text{ be a eigenvalue of }A, \text{ and v be the corresponding eigenvector.} \\ \qquad Av=\lambda v \Rightarrow AAv =A\lambda v \Rightarrow A^2v =\lambda Av \Rightarrow 0=\lambda \lambda v=\lambda^2v \Rightarrow \lambda =0\\ \textbf{(5)} \; \bbox[red, 2pt]{Prove}: 1\cdot 0+0\cdot v_1+ 0\cdot v_2+ 0\cdot v_3=0 \Rightarrow 0,v_1,v_2,v_3\text{ are linear dependent} \\\textbf{(6)}\; \bbox[red, 2pt]{Disprove}: \text{Let }\cases{v_1=(1,2)\\ v_2=(3,4)} \Rightarrow av_1+bv_2=0 \Rightarrow \cases{a+3b=0\\ 2a+4b=0} \Rightarrow a=b=0\\ \qquad \Rightarrow v_1\text{ and }v_2 \text{ are independent} \Rightarrow v_1\cdot v_2=3+8=11\ne 0 \Rightarrow \text{ Not orthogoanl} \\\textbf{(7)}\; \bbox[red, 2pt]{Disprove}: v_1=0 \Rightarrow v_1 \text{ and }v_1 \text{ are orthogonal but dependent}$$
解答:$$略$$
解答:$$A=\left[ \begin{matrix}4 & 0\\0 & 2\\1 & 1\end{matrix} \right] \Rightarrow \cases{ A^T A=\left[\begin{matrix}17 & 1\\1 & 5\end{matrix} \right] \\[1ex] A^Tb= \left[ \begin{matrix} 19 \\ 11 \end{matrix} \right]} \Rightarrow (A^TA)x=A^Tb \Rightarrow x=(A^TA)^{-1}(A^Tb) = \bbox[red, 2pt]{\left[ \begin{matrix} 1 \\ 2 \end{matrix} \right]}$$
解答:$$\textbf{(1)} \;H=\left[\begin{matrix}3 & -1 & 2 & -5\\0 & 5 & -3 & -6\\-6 & 7 & -7 & 4\\-5 & -8 & 0 & 9\end{matrix}\right] \\\Rightarrow \det(H)=3\left| \begin{matrix}5 & -3 & -6\\7 & -7 & 4\\-8 & 0 & 9\end{matrix}\right| +\left|\begin{matrix}0 & -3 & -6\\-6 & -7 & 4\\-5 & 0 & 9\end{matrix} \right| +2\left| \begin{matrix}0 & 5 & -6\\-6 & 7 & 4\\-5 & -8 & 9\end{matrix}\right| +5 \left| \begin{matrix}0 & 5 & -3\\-6 & 7 & -7\\-5 & -8 & 0\end{matrix}\right| \\=3\times 306+108+2\times (-328)+5\times (-74)= \bbox[red, 2pt]0 \\ \textbf{(2)} \; \det(H)=0 \Rightarrow H\text{ is }\bbox[red, 2pt] {\text{not invertible}}\\ \bbox[cyan,2pt]註:A應該是H$$
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