111年成大測量-線性代數詳解

國立成功大學111學年度碩士班招生考試

系所:測量及空間資訊學系
科目:線性代數

解答: $$\textbf{(a)}\; AA^{-1}=I \Rightarrow \det(AA^{-1}) =\det(I) \Rightarrow \det(A)\cdot \det(A^{-1})=1 \Rightarrow \det(A^{-1})={1\over \det(A)}\\ \quad \bbox[red, 2pt]{Q.E.D.}\\ \textbf{(b)}\; \det\begin{bmatrix}2I_3 & A \\0 & CB \end{bmatrix} =\det(2I_3)\cdot \det(CB)= 8 \det(B)\det(C) =-8\cdot {\det(A)\over 2} =\bbox[red, 2pt]{-8}$$

解答: $$A=\begin{bmatrix}0 & 1 & 2 \\1 & 0 & 1 \end{bmatrix} \Rightarrow B= A^TA =\begin{bmatrix}1 & 0 & 1 \\0 & 1 & 1\\ 1& 2& 5 \end{bmatrix} \\ \det(B-\lambda I)=0\Rightarrow \text{eigenvalues: }\lambda=6,1,0 \\ \Rightarrow \text{normailized eigenvectors: } v_1 = \begin{bmatrix} 1/\sqrt{30} \\2/\sqrt{30} \\ 5/\sqrt{30}\end{bmatrix}, v_2= \begin{bmatrix} -2/\sqrt 5 \\ 1/\sqrt 5 \\0 \end{bmatrix} , v_3= \begin{bmatrix}-1/\sqrt 6 \\ -2/\sqrt 6 \\1/\sqrt 6 \end{bmatrix}\\ \Rightarrow V=[v_1\mid v_2 \mid v_3]=\begin{bmatrix}1/\sqrt{30} & -2/\sqrt 5 & -1/\sqrt 6 \\2/\sqrt{30} & 1/\sqrt 5 & -2/\sqrt 6\\ 5/\sqrt{30} & 0 & 1/\sqrt 6 \end{bmatrix} \\ \Rightarrow \text{square root of the nonzero eigenvalues: }\sigma_1=\sqrt 6, \sigma_2=1 \Rightarrow \sum= \begin{bmatrix} \sigma_1 & 0 & 0 \\0 & \sigma_2 & 0 \end{bmatrix} =\begin{bmatrix} \sqrt 6 & 0 & 0 \\0 & 1 & 0 \end{bmatrix} \\ \cases{u_1={1\over \sigma_1} A v_1 =\begin{bmatrix}2/\sqrt 5 \\1/\sqrt 5 \end{bmatrix} \\ u_2= {1\over \sigma_1} A v_2 =\begin{bmatrix}1/\sqrt 5 \\ -2/\sqrt 5 \end{bmatrix}} \Rightarrow U=\begin{bmatrix}2/\sqrt 5 & 1/\sqrt 5 \\1/\sqrt 5 & -2/\sqrt 5 \end{bmatrix}\\ \Rightarrow \bbox[red, 2pt]{A=U\sum V^T, \text{ where }U = \begin{bmatrix} 2/ \sqrt 5 & 1/\sqrt 5 \\1/\sqrt 5 & -2/\sqrt 5 \end{bmatrix}, \sum =\begin{bmatrix} \sqrt 6 & 0 & 0 \\0 & 1 & 0 \end{bmatrix}, \\V= \begin{bmatrix}1/\sqrt{30} & -2/\sqrt 5 & -1/\sqrt 6 \\2/\sqrt{30} & 1/\sqrt 5 & -2/\sqrt 6\\ 5/\sqrt{30} & 0 & 1/\sqrt 6 \end{bmatrix}}$$

解答: $$\textbf{(a)}\; A＝\begin{bmatrix}0 & 1 & 1 \\1 & 0 & 1 \\1 & 1 & 0 \end{bmatrix} \Rightarrow \det(A-\lambda I)=-\lambda^3+ 3\lambda+2 = -(\lambda+1)^2 (\lambda-2) \\ \Rightarrow \text{eigenvalues: }2,-1 \Rightarrow \bbox[red, 2pt]{\text{indefinite}} \\\textbf{(b)}\; \lambda_1=-1 \Rightarrow (A-\lambda_1 I)v =0 \Rightarrow \begin{bmatrix}1 & 1 & 1 \\1 & 1 & 1 \\1 & 1 & 1 \end{bmatrix} \begin{bmatrix}x_1 \\x_2\\x_3 \end{bmatrix}=0 \Rightarrow x_1+x_2+x_3=0\\ \lambda_2=2 \Rightarrow (A-\lambda_2 I)v =0 \Rightarrow \begin{bmatrix}-2 & 1 & 1 \\1 & -2 & 1 \\1 & 1 & -2 \end{bmatrix} \begin{bmatrix}x_1 \\x_2\\x_3 \end{bmatrix}=0 \Rightarrow \cases{x_1  -x_3  =  0 \\x_2  -x_3  =  0}\\ \Rightarrow \text{there is only one eigenvalue }-1\text{ on the subspace }M \Rightarrow \bbox[red, 2pt]{\text{possitive definite}} \\\textbf{(c)}\; f(x_1,x_2,x_3) = [x_1,x_2,x_3]\begin{bmatrix}1 & \beta &-1\\ \beta & 1 &2\\ -1 & 2 &5\end{bmatrix} \begin{bmatrix}x_1 \\x_2\\ x_3 \end{bmatrix} =\mathbf x^TA\mathbf x \\ A \text{ is positive defininte} \Rightarrow \cases{\begin{vmatrix}1 & \beta \\\beta & 1 \end{vmatrix}=1-\beta^2 \gt 0 \\ \begin{vmatrix} 1 & \beta &-1\\ \beta & 1 &2\\ -1 & 2 &5\end{vmatrix}=a(-5a-4) \gt 0} \Rightarrow \cases{-1\lt \beta \lt 1 \\ -4/5\lt \beta\lt 0} \\ \Rightarrow \bbox[red, 2pt]{-{4\over 5}\lt \beta\lt 0}$$

解答: $$T(x)=0 \Rightarrow \cases{a_1-2a_2=0\\ 3a_3=0} \Rightarrow x= \left(\begin{array}{c}2a_2\\ a_2\\ 0\end{array}\right) \Rightarrow \bbox[red, 2pt]{Null(T)=\left\{ k\left(\begin{array}{c}2 \\ 1\\ 0\end{array}\right) \mid k\in \mathbb R\right\}} \\ \Rightarrow \bbox[red, 2pt]{\text{Nullity(A)}=1}, Rank(A)=3-Nullity(A)=2 \Rightarrow \bbox[red, 2pt]{Rank(A)=2}$$

解答: $$\textbf{(a)}\; T(\mathbf u)=\begin{bmatrix}1 & -3 \\3 & 5\\ -1 & 7 \end{bmatrix} \begin{bmatrix}2 \\-1 \end{bmatrix} = \bbox[red, 2pt]{\begin{bmatrix}5 \\ 1 \\ -9 \end{bmatrix}} \\\textbf{(b)}\; T(\mathbf x)=b \Rightarrow B=[A\mid b]= \left[\begin{matrix}1 & -3 & 3\\3 & 5 & 2\\-1 & 7 & -5\end{matrix} \right] \Rightarrow rref(B)= \left[\begin{matrix}1 & 0 & \frac{3}{2}\\0 & 1 & - \frac{1}{2}\\0 & 0 & 0\end{matrix}\right]\\\qquad  \Rightarrow \bbox[red, 2pt]{\mathbf  x=\begin{bmatrix} 3/2 \\-1/2 \end{bmatrix}}\\ \textbf{(c)}\; rref(A) = \begin{bmatrix}1` & 0 \\0 & 1\\ 0 & 0 \end{bmatrix} \Rightarrow rank(A)=2 \Rightarrow  \bbox[red, 2pt]{NO} \\ \textbf{(d)}\; T(\mathbf x)=\mathbf c \Rightarrow B=[A\mid c] = \left[ \begin{matrix}1 & -3 & 3\\3 & 5 & 2\\-1 & 7 & 5\end{matrix}\right] \Rightarrow rref(B)=\left[\begin{matrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{matrix}\right] \Rightarrow\bbox[red, 2pt]{ \mathbf c \not \in T(\mathbf x)}$$

解答: $$\textbf{(a)}\; Az=b \equiv \begin{bmatrix}1 & 0 \\1 & 1 \\1 & 2 \\ 1& 3\end{bmatrix} \begin{bmatrix}\alpha \\ \beta \end{bmatrix}＝\begin{bmatrix}0 \\2 \\ 4\\ 6 \end{bmatrix} \\ \textbf{(b)}\; \cases{A^TA= \left[\begin{matrix}1 & 1 & 1 & 1\\0 & 1 & 2 & 3\end{matrix}\right] \left[\begin{matrix}1 & 0\\1 & 1\\1 & 2\\1 & 3\end{matrix} \right] = \left[\begin{matrix}4 & 6\\6 & 14\end{matrix}\right] \\ A^Tb = \left[\begin{matrix}1 & 1 & 1 & 1\\0 & 1 & 2 & 3\end{matrix}\right] \left[\begin{matrix}0\\2\\4\\6\end{matrix} \right] =\left[ \begin{matrix} 12\\28\end{matrix} \right]} \\\Rightarrow \mathbf z= (A^TA)^{-1}(A^Tb)   =\left[\begin{matrix} \frac{7}{10} & - \frac{3}{10}\\- \frac{3}{10} & \frac{1}{5} \end{matrix} \right] \left[ \begin{matrix} 12\\28 \end{matrix}\right] = \begin{bmatrix}0 \\2 \end{bmatrix} \Rightarrow \cases{\alpha=0\\ \beta=2}\\ f=2x \Rightarrow \cases{f(0)=0\\ f(1)=2\\ f(2)=4\\ f(3)=6} \Rightarrow \text{square error }=0+0+0+0=\bbox[red, 2pt]0$$
 

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