國立臺北科技大學113學年度碩士班招生考試
系所組別: 電機工程系碩士班丁組
第一節 線性代數
解答:$$A=\begin{bmatrix}-1 & 2 \\2 & -3 \\-1 & 3\end{bmatrix} \Rightarrow \cases{A^TA= \begin{bmatrix}6 & -11 \\-11 & 22 \end{bmatrix} \\[1ex]A^T\mathbf b =\begin{bmatrix}-4 \\11 \end{bmatrix}}\\ \Rightarrow rref([A^TA \mid A^Tb ]) =rref\left( \left[\begin{array}{rr|r} 6 & -11 &-4\\-11 & 22 &11 \end{array} \right] \right) =\left[\begin{array}{rr|r} 1 & 0 &3\\0 & 0 &2 \end{array} \right] \\\Rightarrow \text{the least-squares solution: } \bbox[red, 2pt]{\begin{bmatrix} 3 \\2 \end{bmatrix}}$$
解答:$$\textbf{(1)}\; \text{Let }E_a(A) =\{\mathbf x\mid A\mathbf x=a\mathbf x\}\text{ be the eigenspace of }A\text{ corresponding to }a.\\\qquad \text{(a)}\; A\vec o=a\vec 0=\vec 0 \Rightarrow \vec 0 \in E_a(A)\\ \qquad\text{(b)}\; \mathbf x \in E_a(A) \Rightarrow A(t \mathbf x)= t A(\mathbf x)=t a\mathbf x= a t \mathbf x \Rightarrow A(t\mathbf x)=at\mathbf x \Rightarrow t\mathbf x\in E_a(A)\\ \qquad \text{(c)}\; \cases{\mathbf x_1\in E_a(A)\\ \mathbf x_2\in E_a(A)} \Rightarrow \cases{A\mathbf x_1= a\mathbf x_1 \\A\mathbf x_2= a\mathbf x_2} \Rightarrow A(\mathbf x_1+ \mathbf x_2)= A(\mathbf x_1)+A(\mathbf x_2) =a\mathbf x_1+ a\mathbf x_2\\ \qquad = a(\mathbf x_1+ \mathbf x_2) \Rightarrow \mathbf x_1+ \mathbf x_2 \in E_a(A)\\\quad \text{By (a),(b),(c), we can concloude that } E_a(A) \text{ is a subspace.} \bbox[red, 2pt]{QED} \\\textbf{(2)}\; t_1\mathbf x+t_2\mathbf y=0 \Rightarrow t_1a \mathbf x+t_2 a\mathbf y=0 \cdots(1)\\ t_1\mathbf x+t_2\mathbf y=0 \Rightarrow A(t_1\mathbf x+t_2\mathbf y) =A(t_1\mathbf x)+A(t_2\mathbf y)= t_1a\mathbf x+ t_2b\mathbf y =0 \cdots(2)\\ (1)-(2) \Rightarrow t_2(a-b)\mathbf y=0 \Rightarrow t_2=0 \Rightarrow t_1\mathbf x=0 \Rightarrow t_1=0 \Rightarrow \mathbf x,\mathbf y\text{ are linearly independent}\\ \text{By the same way,} t_1\mathbf x+t_2\mathbf y+ t_3\mathbf z=0 \Rightarrow t_1c\mathbf x+t_2c\mathbf y+ t_3c\mathbf z=0 \cdots(3)\\ t_1\mathbf x+t_2\mathbf y+ t_3\mathbf z=0 \Rightarrow A(t_1\mathbf x+t_2\mathbf y+ t_3\mathbf z)= t_1a\mathbf x+t_2b\mathbf y+ t_3c\mathbf z=0 \cdots(4)\\ (3)-(4) \Rightarrow t_1(c-a)\mathbf x+t_2(c-b)\mathbf y=0 \Rightarrow t_1(c-a)= t_2(c-b)=0 \;(\because \mathbf x,\mathbf y \text{ independent}) \\ \Rightarrow t_1=t_2=0 \Rightarrow t_3z=0 \Rightarrow t_3=0 \Rightarrow \mathbf x,\mathbf y, \mathbf z \text{ are linearly independent} \\\text{Continuing on the same way, }t_1\mathbf x+t_2\mathbf y+ t_3\mathbf z+ t_4\mathbf w=0 \Rightarrow t_1d\mathbf x+t_2d\mathbf y+ t_3d \mathbf z+t_4 d\mathbf w=0 \cdots(5)\\ t_1\mathbf x+t_2\mathbf y+ t_3\mathbf z+ t_4\mathbf w=0 \Rightarrow A(t_1\mathbf x+t_2\mathbf y+ t_3\mathbf z+ t_4\mathbf w)= t_1a\mathbf x+t_2 b\mathbf y+ t_3c\mathbf z+ t_4d\mathbf w=0 \cdots(6)\\ (5)-(6) \Rightarrow t_1(a-d)\mathbf x+t_2 (b-d)\mathbf y+ t_3 (c-d)\mathbf z=0 \\ \mathbf x,\mathbf y,\mathbf z \text{ are linealy independent } \Rightarrow \cases{t_1(a-d)=0\\ t_2(b-d)=90\\ t_3(c-d)=0} \Rightarrow \cases{t_1=0\\ t_2=0 \\ t_3=0} \Rightarrow t_4 \mathbf z=0 \Rightarrow t_4=0\\\text{Finally, we have }\mathbf x,\mathbf y, \mathbf z, \mathbf w \text{ are linealy independent.} \bbox[red, 2pt]{QED}$$
解答:$$A\text{ is full rank} \Rightarrow rank(A)=N \Rightarrow nullity(A)=N-rank(A)=0 \Rightarrow nullity(A)=0\\ \Rightarrow Ax=0 \text{ has only the trivial solution}. \bbox[red, 2pt]{QED}$$
解答:$$\textbf{(1)}\; A=PBP^{-1} \Rightarrow A^2=PBP^{-1}PBP^{-1} =PBBP^{-1}=PB^2P^{-1} \\ \quad \Rightarrow A^2\text{ is similar to }B^2\;\bbox[red, 2pt]{QED} \\ \textbf{(2)}\; A=PBP^{-1}=QDQ^{-1}, \text{where }Q\text{ is diagonal} \\ \quad \Rightarrow BP^{-1} =P^{-1}QDQ^{-1} \Rightarrow B=P^{-1}QDQ^{-1}P = (P^{-1}Q) D(P^{-1}Q)^{-1} \\\quad \Rightarrow B\text{ is diagonalizable }\bbox[red, 2pt]{QED}$$
解答:$$\textbf{(1)}\; \det(H-\lambda I)=0 \Rightarrow (\lambda-3)(\lambda-4)^3=0 、\\\qquad \Rightarrow \text{ eigenvalues: }\bbox[red, 2pt]{3,4}, \text{ and their multiplicities are }\bbox[red, 2pt]{1，3 } \\\textbf{(2)}\; \det(UHU^T) = \det(U)\det(H)\det(U^T) = \det(H)\det(U) \det(U^T) = \det(H) \det(UU^T) \\\qquad =\det(H)\det(I)= \det(H)= 4^3\cdot 3= \bbox[red, 2pt]{192} \\ \textbf{(3)} \; \text{trace}(UHU^T) =\text{trace}(HUU^T) =\text{trace}(HI) = \text{trace}(H) = 4\times 3+3= \bbox[red, 2pt]{15} \\\textbf{(4)}\; Hx= \begin{bmatrix}4 & 3 & 2& 1\\0 & 4 & 3& 2 \\0 & 0 & 4& 3\\ 0 & 0 & 0 & 3 \end{bmatrix} \begin{bmatrix}a \\ b \\c \\ d \end{bmatrix} = \begin{bmatrix} 4a+3b+2c+d \\ 4b+3c+2d \\4c+3d \\3 d \end{bmatrix}\\ ||U(Hx)||^2=(U(Hx))^T(U(Hx))= (Hx)^T U^T U(Hx)= (Hx)^T(Hx)= ||Hx||^2 \\\quad \Rightarrow ||U(Hx)|| =\bbox[red, 2pt]{\sqrt{(4a+3b+2c+d)^2+ (4b+3c+2d)^2+ (4c+3d)^2+ (3d)^2}}$$
解答:$$\cases{\mathbf X=\begin{bmatrix} 1& 0 \\ 0& -1\end{bmatrix} \\[1ex] \mathbf Y= \begin{bmatrix} -1& 0 \\ 0& 0\end{bmatrix}} \Rightarrow \mathbf X+\mathbf Y=\begin{bmatrix} 0& 0 \\ 0& 0\end{bmatrix} \Rightarrow \cases{\det(\mathbf X)= \det(\mathbf Y) =-1\\ \det(\mathbf X+\mathbf Y) =0} \\\Rightarrow \det(\mathbf X)+\det(\mathbf Y)=-2 \ne \det(\mathbf X+\mathbf Y)=0 \Rightarrow \bbox[red, 2pt]{\text{disprove}}$$
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