111年北科大電機碩士班戊組-線性代數詳解

 

 國立臺北科技大學 111學年度碩士班 招 生考試

系所組別 :2151電機工程系碩士班戊組
第一節 線性代數 試題 (選考)

解答：$$A \text{ and }B \text{ are similar} \Rightarrow B=P^{-1}AP \Rightarrow A=PBP^{-1}\\ \text{If } Av =\lambda v, \text{ then }PBP^{-1}v=\lambda v \Rightarrow B(P^{-1}v) = \lambda (P^{-1}v) \\ \text{That is, }\lambda \text{ is  a eigenvalue of }A, \text{ then }\lambda \text{ is  a eigenvalue of }B. \bbox[red, 2pt]{QED.}$$

解答：$$A^TA=\left( \begin{matrix}1 & 1 & 1 & 1 \\-6 & -2 & 1 & 7\end{matrix} \right) \left(\begin{matrix} 1 & -6 \\1 & -2 \\1 & 1 \\1 & 7 \end{matrix} \right) =\left( \begin{matrix}4 & 0 \\
0 & 90 \end{matrix}\right)\\ A^Tb=\left( \begin{matrix}1 & 1 & 1 & 1 \\-6 & -2 & 1 & 7\end{matrix} \right) \left( \begin{matrix}-1 \\2 \\1 \\6 \end{matrix} \right)=\left( \begin{matrix}8 \\45\end{matrix} \right) \\ \Rightarrow \left( \begin{matrix}4 & 0 \\0 & 90 \end{matrix}\right)\left( \begin{matrix}x_1 \\
x_2\end{matrix} \right) =\left( \begin{matrix}8 \\45\end{matrix} \right) \Rightarrow \cases{4x_1=8 \\90x_2=45} \Rightarrow \cases{x_1=2 \\x_2= 1/2} \\ \Rightarrow \textbf{x}=\bbox[red, 2pt]{ \left( \begin{matrix}2 \\1/2\end{matrix} \right)}$$

解答：$$\textbf{(1)}\; \bbox[red, 2pt]{Prove: }A \text{ is }M\times N  \Rightarrow  A^H \text{ is }N\times M  \Rightarrow A^HA \text{ is }N\times N\\ A \text{ is column orthonormal} \Rightarrow A=[\vec a_1\; \vec a_2\; \cdots \vec a_N] \Rightarrow A^H=\begin{bmatrix} \vec a_1^*\\ \vec a_2^* \\ \vdots \\\vec a_N^*\\\end{bmatrix} \\ \Rightarrow A^HA =\begin{bmatrix} \vec a_1 \cdot \vec a_1^* & \vec a_2\cdot \vec a_1^* & \cdots & \vec a_N\cdot \vec a_1^*\\  \vec a_1 \cdot \vec a_2^* & \vec a_2\cdot \vec a_2^* & \cdots & \vec a_N\cdot \vec a_2^*\\ \cdots \\ \vec a_1\cdot \vec a_N^* & \vec a_2\cdot \vec a_N^* & \cdots & \vec a_N\cdot \vec a_N^*  \end{bmatrix} =[I]_{N\times N}\\ \textbf{(2)}\; \bbox[red, 2pt]{Disprove: }A=\begin{bmatrix}1 & 0 \\0 & 1 \\ 0 & 0\\ 0 & 0\end{bmatrix} \Rightarrow A^H=\begin{bmatrix}1 & 0 & 0 & 0 \\0 & 1 & 0 & 0\end{bmatrix} \Rightarrow AA^H =\begin{bmatrix}1 & 0 & 0 & 0 \\0 & 1  & 0 & 0 \\0 & 0 & 0 & 0  \\0 & 0 & 0 & 0 \end{bmatrix} \ne I$$

解答：$$\textbf{(1)}\; \text{If one diagonal element of }B \text{ is zero, then} \det(H) =\det(ABC)= \det(A)\det(B) \det(C) =0 \\ \quad \Rightarrow \det(H)=0 \Rightarrow H\text{ is invertible}, \bbox[red, 2pt]{QED} \\\textbf{(2)}分解過程需計算H^TH的特徵值及相對應的特徵向量, 若特徵值排序不同,\\ 相對應的特徵向量也不同, 分解結果隨之不同\\ \textbf{(3)}\; \Vert ACx\Vert ^2= (ACx)^T ACx =x^TC^TA^T ACx =x^TC^TICx =x^TIx = x^Tx= \Vert x\Vert^2\\ \quad \Rightarrow \Vert ACx \Vert=\Vert x\Vert ,\bbox[red, 2pt]{QED}\\ \textbf{(4)}\; A^{-1}=A^T =A^H \Rightarrow A^{-1}=A^H, \bbox[red, 2pt]{QED}$$

解答：$$ 8\cdot \vec 0+0\cdot \vec v=0, \text{but }8\ne 0 \Rightarrow \vec 0 \text{ and } \vec v \text{ are linearly dependent }, \bbox[red, 2pt]{QED}$$

解答：$$假設A為一m\times n的矩陣,則A的n個\text{column vector: }\vec a_1, \vec a_2,\dots, \vec a_n所形成的線性組合空間,\\ 稱為矩陣A的\text{ column space =}Col(A). \\A\textbf x=0所有解的集合稱之為\text{ null space=}N(A)=\{\textbf x \in \mathbb R^n \mid A\textbf x=0\}$$

解答：$$T(a_0+a_1t +a_2t^2)= 4a_0+(5a_0-2a_1) t+(4a_1+a_2)t^2\\ \Rightarrow A \begin{bmatrix} a_0   \\a_1\\ a_2 \end{bmatrix} = \begin{bmatrix} 4a_0   \\5a_0-2a_1\\ 4a_1+a_2 \end{bmatrix} \Rightarrow A= \bbox[red, 2pt] {\begin{bmatrix}4 & 0& 0 \\5& -2& 0 \\ 0& 4& 1 \end{bmatrix}}$$

解答：$$A=QR,其中R為上三角矩陣,Q為N\times N 且\text{orthogonal},即Q^TQ=I_N,\\因此 A\textbf{x}=\textbf b  \Rightarrow QR\textbf x=\textbf b \Rightarrow R\textbf x= Q^T\textbf b為三角形,可用\text{ back substitution}求解$$

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解題僅供參考,其他歷年試題及詳解