111年台大碩士班-線性代數C詳解

 國立臺灣大學111學年度碩士班招生考試

科目:線性代數(C)

解答: $$(a)\bbox[red, 2pt]{\text{false}}:A=\begin{bmatrix}0& 0\\1& 0 \end{bmatrix} \Rightarrow A^T=A \Rightarrow A \text{ is symmetric, but it cannot be diagonalizable} \\(b)\bbox[red, 2pt]{\text{false}}:A=\begin{bmatrix}1& 1\\1& 1 \end{bmatrix} \Rightarrow A^T=A \Rightarrow A \text{ is symmetric, but }\det(A)=0 \Rightarrow A^{-1} \text{does not exist} \\(c)\bbox[red, 2pt]{\text{true}}:\cases{A^T=A\\ Ax_1=\lambda_1 x_1 \\ Ax_2 =\lambda_2 x_2} \Rightarrow \lambda_1 \langle x_1, x_2\rangle  =\langle \lambda_1 x_1, x_2\rangle  =\langle Ax_1, x_2\rangle =\langle x_1, A^Tx_2\rangle =\langle x_1, Ax_2\rangle \\\qquad =\langle x_1, \lambda_ 2x_2\rangle =\lambda_2 \langle x_2, x_2 \rangle  \Rightarrow \lambda_1\langle x_1, x_2\rangle =\lambda_2\langle x_1, x_2\rangle \Rightarrow (\lambda_1-\lambda_2) \langle x_1, x_2\rangle =0 \Rightarrow x_1\bot x_2 (\lambda_1\ne \lambda_2)\\ (d)\bbox[red, 2pt]{\text{true}}: \text{from (c), they are othogonal and then linearly independent}$$

解答: $$A=\begin{bmatrix}1 &2  &2  &0  &1  \\0 &2  & 1 & 1 &0  \\0 & 0 &0  &0  & 1 \\0 & 0 & 0 &0  &0 \end{bmatrix} \Rightarrow B=rref(A)= \begin{bmatrix}1 &0  &1  &-1  &0  \\0 &1  & 1/2 & 1/2 &0  \\0 & 0 &0  &0  & 1 \\0 & 0 & 0 &0  &0 \end{bmatrix}\\Bx=0 \Rightarrow \cases{x_1+x_3-x_4=0\\ x_2+x_3/2+x_3/2=0\\ x_5=0} \Rightarrow x= \begin{bmatrix} -x_3+x_4\\ -(x_3+x_4)/2\\x_3\\x_4 \\ 0\end{bmatrix} =x_3\begin{bmatrix} -1\\-1/2\\1\\0 \\0\end{bmatrix} +x_4 \begin{bmatrix} 1\\-1/2\\0\\1 \\0 \end{bmatrix}  \\ \Rightarrow \bbox[red, 2pt]{\text{null space =}\left\{ a\begin{bmatrix} -1\\-1/2 \\1\\0 \\0 \end{bmatrix} +b \begin{bmatrix} 1\\-1/2\\0\\1 \\0 \end{bmatrix} \mid a,b \in \mathbb R\right\}}\\\text{From the nonzero rows of the reduced matrix, we have } \bbox[red, 2pt]{\text{ row space =Span}\left\{ \begin{bmatrix} 1\\2\\2\\0 \\1\end{bmatrix},  \begin{bmatrix} 0\\2\\1\\1\\ 0 \end{bmatrix},  \begin{bmatrix} 0\\0\\0\\0 \\1\end{bmatrix}\right\}} \\\text{From the pivot columns of the reduced matrix, we have} \bbox[red, 2pt]{\text{ column space =Span }\left\{ \begin{bmatrix} 1\\0\\0 \\0 \end{bmatrix},  \begin{bmatrix} 2\\2\\0\\0 \end{bmatrix},  \begin{bmatrix} 1\\0\\1\\0 \end{bmatrix}\right\}} \\ \text{Finally, from rref(A), we have }\bbox[red, 2pt]{rank(A)=3}$$

解答: $$\cases{x_1+x_3=q\\ x_2+2x_4=0\\ x_1+2x_3+3x_4=0\\ 2x_2+3x_3+ px_4 =3} \Rightarrow \begin{bmatrix} 1& 0& 1& 0\\ 0& 1& 0& 2\\ 1& 0& 2& 3\\ 0& 2& 3& p \end{bmatrix}  \begin{bmatrix} x_1\\ x_2\\ x_3\\ x_4\end{bmatrix}= \begin{bmatrix} q\\ 0\\ 0\\ 3\end{bmatrix} \\ \textbf{(a) }A=\begin{bmatrix} 1& 0& 1& 0\\ 0& 1& 0& 2\\ 1& 0& 2& 3\\ 0& 2& 3& p \end{bmatrix} \Rightarrow \det(A)=p-13 \Rightarrow \bbox[red, 2pt]{p\ne13} \\\textbf{(b) }p=13 \Rightarrow \cases{x_1+x_3=q\\ x_2+2x_4=0\\ x_1+2x_3+3x_4=0\\ 2x_2+3x_3+ 13x_4 =3} \Rightarrow \cases{x_3+3x_4=-q\\ 3x_3+9x_4=3} \Rightarrow \text{no solution: } \bbox[red, 2pt]{p=13, q\ne-1} \\\textbf{(c) }\text{ infinity solutions }\Rightarrow {1\over 3}={3\over 9}={-q\over 3} \Rightarrow q=-1 \Rightarrow \bbox[red, 2pt]{p-13,q=-1}$$

解答: $$\text{Let } \cases{x_1=\begin{bmatrix} 1\\  2\\  3 \end{bmatrix}  \\[1ex]x_2 =\begin{bmatrix} 3\\ 1\\ 4 \end{bmatrix}  \\[1ex]b= \begin{bmatrix} 1\\0\\ -1 \end{bmatrix}} \Rightarrow \cases{Ax_1=b \\ Ax_2=b} \Rightarrow Ax_1+Ax_2=A(x_1+x_2) =2b \Rightarrow A({x_1+x_2\over 2})=b\\ \Rightarrow \text{another solution }x_3={x_1+x_2\over 2} = \bbox[red, 2pt] {\begin{bmatrix} 2\\  3/2\\  7/2 \end{bmatrix} }$$

解答: $$\lambda \text{ is an eigenvalue of }A \Rightarrow Ax=\lambda x \Rightarrow A^2x= \lambda Ax= \lambda^2 x\Rightarrow A^2x = \lambda^2 x\\ \Rightarrow \lambda^2 \text{ is an eigenvalue of }A^2 \quad \bbox[red, 2pt]{QED}$$

解答: $$A=\begin{bmatrix}-2& 12\\ -1&5 \end{bmatrix} =\begin{bmatrix}4& 3\\1& 1 \end{bmatrix} \begin{bmatrix}1& 0\\0& 2 \end{bmatrix} \begin{bmatrix}1&-3\\-1&4 \end{bmatrix}  \Rightarrow A^{10} = \begin{bmatrix}4& 3\\1& 1 \end{bmatrix} \begin{bmatrix}1^{10}& 0\\0& 2^{10} \end{bmatrix} \begin{bmatrix}1&-3\\-1&4 \end{bmatrix}\\= \begin{bmatrix}4-3\cdot 2^{10}& -12(1-2^{10})\\1-2^{10}&  -3+4\cdot 2^{10} \end{bmatrix} = \bbox[red, 2pt]{\begin{bmatrix}-3068& 12276\\-1023& 4093 \end{bmatrix}}$$

解答: $$2x_1-x_3+x_4=0 \Rightarrow (x_1, x_2,x_3, x_4) =(x_1,x_2,2x_1+x_4,x_4)\\ =x_1(1,0,2,0)+ x_2(0,1,0,0)+ x_4(0,0,1,1) \Rightarrow \{(1,0,2,0), (0,1,0,0), (0,0,1,1)\} \text{ is a basis for }W\\ \text{Applying Gram-Schmidt process, we have the orthonormal basis: }\\ \bbox[red, 2pt]{\{({\sqrt 5\over 5}, 0, {2\sqrt 5\over 5},0), (0,1,0,0), (-{\sqrt{30} \over 15},0, {\sqrt{30} \over 30}, {\sqrt{30}\over 6})\}}$$

解答: $$\textbf{(a) }\hat y_i=C+D\cdot 2^{x_i} \Rightarrow E=\sum(\hat y_i-y_i)^2 =\sum(C+D\cdot 2^{x_i}-y_i)^2 \\\qquad =\sum \left(C^2+D^2\cdot 2^{2x_i}+y_i^2+ 2CD\cdot 2^{x_i}-2Dy_i2^{x_i} -2Cy_i\right) \\ \qquad  {\partial E\over \partial C} =0 \Rightarrow \sum(C+ D\cdot 2^{x_i}-y_i) =0 \Rightarrow nC+D\sum 2^{x_i} =\sum y_i \\ {\partial E\over \partial D} =0 \Rightarrow \sum(D\cdot 2^{2x_i} + C\cdot 2^{x_i}- y_i 2^{x_i}) =0 \Rightarrow C\sum 2^{x_i}+D \sum 2^{2x_i} =\sum y_i2^{x_i} \\ \Rightarrow \begin{bmatrix}n&  \sum 2^{x_i} \\ \sum 2^{x_i} & \sum 2^{2x_i} \end{bmatrix} \begin{bmatrix} C\\ D\end{bmatrix}= \begin{bmatrix} \sum y_i\\ \sum y_i2^{x_i} \end{bmatrix} \Rightarrow \begin{bmatrix}3&  7\\ 7 & 21\end{bmatrix} \begin{bmatrix} C\\ D \end{bmatrix} = \begin{bmatrix} 10\\ 14\end{bmatrix}  \\\qquad \Rightarrow \begin{bmatrix} C\\ D \end{bmatrix} = \begin{bmatrix}3&  7\\ 7 & 21\end{bmatrix}^{-1} \begin{bmatrix} 10\\ 14 \end{bmatrix} = \begin{bmatrix}\frac{3}{2} & \frac{-1}{2} \\\frac{-1}{2} & \frac{3}{14} \end{bmatrix}  \begin{bmatrix} 10\\ 14\end{bmatrix} =\begin{bmatrix}8\\ -2\end{bmatrix} \Rightarrow \bbox[red, 2pt]{\cases{C=8\\ D=-2}}\\ \textbf{(b) } y=8-2\cdot 2^{x} \Rightarrow \bbox[red, 2pt]{\cases{x=0 \Rightarrow   y=6\\ x=1 \Rightarrow  y=4\\ x=2 \Rightarrow   y= 0}} \Rightarrow \text{ The best curve  is }y=8-2\cdot 2^x \text{ not }y=0$$

解答: $$\textbf{(a) }A= \bbox[red, 2pt]{\begin{bmatrix} 4/5& 1/10\\ 1/5& 9/10 \end{bmatrix}} \\ \textbf{(b) }A= \begin{bmatrix} -1 & \frac{1}{2} \\1 & 1 \end{bmatrix} \begin{bmatrix}\frac{7}{10} & 0 \\0 & 1 \end{bmatrix} \begin{bmatrix} \frac{-2}{3} & \frac{1}{3} \\\frac{2}{3} & \frac{2}{3} \end{bmatrix} \Rightarrow A^\infty  =\begin{bmatrix} -1 & \frac{1}{2} \\1 & 1 \end{bmatrix} \begin{bmatrix} (\frac{7}{10})^\infty & 0 \\0 & 1^\infty \end{bmatrix} \begin{bmatrix} \frac{-2}{3} & \frac{1}{3} \\\frac{2}{3} & \frac{2}{3} \end{bmatrix} \\=\begin{bmatrix} -1 & \frac{1}{2} \\1 & 1 \end{bmatrix} \begin{bmatrix} 0 & 0 \\0 & 1  \end{bmatrix} \begin{bmatrix} \frac{-2}{3} & \frac{1}{3} \\\frac{2}{3} & \frac{2}{3} \end{bmatrix} =  \begin{bmatrix} \frac{1}{3} & \frac{1}{3} \\\frac{2}{3} & \frac{2}{3} \end{bmatrix}  \Rightarrow \lim_{k\to \infty} A^k \begin{bmatrix} 1\\0\end{bmatrix} = \begin{bmatrix} \frac{1}{3} & \frac{1}{3} \\\frac{2}{3} & \frac{2}{3} \end{bmatrix} \begin{bmatrix} 1\\0\end{bmatrix} = \bbox[red, 2pt]{\begin{bmatrix} {1 \over 3}\\{2\over 3}\end{bmatrix}}$$

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