國立臺北科技大學115學年度碩士班招生考試
系所組別:電機工程 科目:線性代數
解答:$$\textbf{(a) }A = \begin{bmatrix} 1 & 3 \\ 2 & 6 \end{bmatrix} \Rightarrow \cases{c_1= \begin{bmatrix}1\\2 \end{bmatrix} \\c_2= \begin{bmatrix}3\\6 \end{bmatrix}} \Rightarrow c_1 =3c_2 \Rightarrow \bbox[red, 2pt]{\text{No, they are linearly dependent}} \\ \textbf{(b) }\text{ a basis for }C(A): \bbox[red, 2pt]{\left\{ \begin{bmatrix}1\\2 \end{bmatrix}\right\}} \\ \textbf{(c) }RREF(A) = \begin{bmatrix}1& 3\\0&0 \end{bmatrix} \Rightarrow Rank(A)=\bbox[red, 2pt]1$$
解答:$$\textbf{(a) } \cases{\mathbf x=(1,-1,2)^T\\ \mathbf y=(2,0,1)^T} \Rightarrow \mathbf x\cdot \mathbf y=1\cdot 2+(-1)\cdot 0+2\cdot 1=\bbox[red, 2pt]4 \\\textbf{(b) }\mathbf x\cdot \mathbf y=4\ne 0 \Rightarrow \bbox[red, 2pt]{\text{No, they are not orthogonal.}} \\ \textbf{(c) }\text{proj}_{\mathbf y} \mathbf x = \left( {\mathbf x \cdot \mathbf y\over \mathbf y\cdot \mathbf y} \right)\mathbf y ={4\over 5} \begin{bmatrix}2\\0\\1 \end{bmatrix} =\bbox[red, 2pt] {(8/5,0,4/5)^T}$$
解答:$$\textbf{(a) } \cases{x + y + z = 2\\2x + 3y + z = 5 \\3x + 4y + 2z = 8} \Rightarrow \begin{bmatrix}1& 1& 1\\2& 3& 1\\3& 4& 2 \end{bmatrix} \begin{bmatrix}x\\y\\x \end{bmatrix}= \begin{bmatrix}2\\5\\8 \end{bmatrix} \Rightarrow \text{ augmented matrix:} \bbox[red, 2pt]{\left[ \begin{array}{ccc\|c} 1 & 1 & 1 & 2 \\ 2 & 3 & 1 & 5 \\ 3 & 4 & 2 & 8 \end{array} \right] } \\ \textbf{(b) }\left[ \begin{array}{ccc\|c} 1 & 1 & 1 & 2 \\ 2 & 3 & 1 & 5 \\ 3 & 4 & 2 & 8 \end{array} \right] \xrightarrow{R_2-2R_1\to R_2, R_3-3R_1\to R_3} \left[ \begin{array}{ccc\|c} 1 & 1 & 1 & 2 \\ 0 & 1 & -1 & 1 \\ 0 & 1 & -1 & 2 \end{array} \right] \xrightarrow{R_3-R_2 \to R_3} \left[ \begin{array}{ccc\|c} 1 & 1 & 1 & 2 \\ 0 & 1 & -1 & 1 \\ 0 & 0 & 0 & 1 \end{array} \right] \\ \textbf{(c) } \text{Look at the last row:}\begin{bmatrix} 0 & 0 & 0 & \mid & 1 \end{bmatrix} \Rightarrow 0x+0y+0z=1 \Rightarrow 0=1 \Rightarrow \bbox[red, 2pt]{\text{no solution}}$$
解答:$$\textbf{(a) } \mathbf{A} = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 1 & 1 & 1 \end{bmatrix} \Rightarrow RREF(A) =\begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \end{bmatrix} \Rightarrow \cases{x_1-x_3=0\\x_2+2x_3=0} \Rightarrow \mathbf x= x_3 \begin{bmatrix}1\\-2\\1 \end{bmatrix} \\ \qquad \Rightarrow \text{a basis for }N(A): \bbox[red, 2pt]{\left\{ \begin{bmatrix} 1 \\ -2 \\ 1 \end{bmatrix} \right\}} \\ \textbf{(b) } \text{the dimension of the null space}=\text{ the number of vectors in its basis} = \bbox[red, 2pt]1 \\ \textbf{(c) }\cases{A \text{ is }3\times 3 \Rightarrow n=3 \\ rank(A)=\text{the number of non-zero rows in RREF}(A)=2 \\ nullity(A)=1}\\ \qquad \Rightarrow \text{rank}(\mathbf{A}) + \text{nullity}(\mathbf{A}) = 2+1=3=n \Rightarrow \text{the rank-nullity theorem for A holds}$$
解答:$$\text{Set $\mathbf{w}_1 = \mathbf{v}_1$} = \begin{bmatrix}1\\1\\0 \end{bmatrix} \Rightarrow \mathbf u_1={\mathbf w_1\over \Vert\mathbf w_1 \Vert} ={1\over \sqrt 2} \begin{bmatrix}1\\1\\0 \end{bmatrix} \\ \mathbf{w}_2 = \mathbf{v}_2 - \text{proj}_{\mathbf{w}_1}(\mathbf{v}_2) = \mathbf{v}_2 - \frac{\mathbf{v}_2 \cdot \mathbf{w}_1}{\mathbf{w}_1 \cdot \mathbf{w}_1} \mathbf{w}_1 = \mathbf{w}_2 = \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} - \frac{1}{2} \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} - \begin{bmatrix} 1/2 \\ 1/2 \\ 0 \end{bmatrix} = \begin{bmatrix} 1/2 \\ -1/2 \\ 1 \end{bmatrix} \\ \Rightarrow \mathbf{u}_2 = \frac{1}{\Vert{}\mathbf{w}_2\Vert{}} \mathbf{w}_2 = \frac{2}{\sqrt{6}} \begin{bmatrix} 1/2 \\ -1/2 \\ 1 \end{bmatrix} = \begin{bmatrix} 1/\sqrt{6} \\ -1/\sqrt{6} \\ 2/\sqrt{6} \end{bmatrix} = \frac{1}{\sqrt{6}} \begin{bmatrix} 1 \\ -1 \\ 2 \end{bmatrix} \\ \Rightarrow \text{The orthonormal basis for the subspace spanned by $\mathbf{v}_1$ and $\mathbf{v}_2$ is:}\bbox[red, 2pt]{\left\{ \begin{bmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{bmatrix}, \begin{bmatrix} 1/\sqrt{6} \\ -1/\sqrt{6} \\ 2/\sqrt{6} \end{bmatrix} \right\} }$$
解答:$$\textbf{(a) } \mathbf{A} = \begin{bmatrix} 1 & 1 \\ 1 & 2 \\ 1 & 3 \end{bmatrix}, \quad \mathbf{b} = \begin{bmatrix} 1 \\ 2 \\ 2 \end{bmatrix} \Rightarrow \mathbf{A}^T \mathbf{A} = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 1 & 2 \\ 1 & 3 \end{bmatrix} = \begin{bmatrix}3& 6\\6& 14 \end{bmatrix} \\ \qquad \Rightarrow \mathbf{A}^T \mathbf{b} = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \end{bmatrix} \begin{bmatrix} 1 \\ 2 \\ 2 \end{bmatrix} =\begin{bmatrix}5\\11 \end{bmatrix} \Rightarrow \text{the normal equations are: } \bbox[red, 2pt]{\begin{bmatrix} 3 & 6 \\ 6 & 14 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 5 \\ 11 \end{bmatrix} } \\ \textbf{(b) } (\mathbf{A}^T \mathbf{A})^{-1} = \frac{1}{6} \begin{bmatrix} 14 & -6 \\ -6 & 3 \end{bmatrix} \Rightarrow \hat {\mathbf {x}} = (\mathbf{A}^T \mathbf{A})^{-1} \mathbf{A}^T \mathbf{b} = \frac{1}{6} \begin{bmatrix} 14 & -6 \\ -6 & 3 \end{bmatrix} \begin{bmatrix}5\\11 \end{bmatrix} =\bbox[red, 2pt]{ \begin{bmatrix}2/3\\ 1/2 \end{bmatrix}} \\\textbf{(c) } \text{The least-squares solution $\hat{\mathbf{x}}$ computes a vector $\hat{\mathbf{b}} = \mathbf{A}\hat{\mathbf{x}}$ which represents the}\\ \text{orthogonal projection of $\mathbf{b}$ onto the column space of $\mathbf{A}$.}\\ \text{This means $\mathbf{A}\hat{\mathbf{x}}$ is the closest possible point in the column space of $\mathbf{A}$ to the vector $\mathbf{b}$,}\\ \text{ thereby minimizing the Euclidean distance $\Vert{}\mathbf{b} - \mathbf{Ax}\Vert{}$.}$$
解答:$$\textbf{(a) } \mathbf{A} = \begin{bmatrix} 4 & 1 \\ 2 & 3 \end{bmatrix} \Rightarrow \det(A-\lambda I) =\lambda^2-7\lambda+10=0 \Rightarrow (\lambda-2)(\lambda-5)=0 \\ \qquad \Rightarrow \text{The eigenvalues are: }\bbox[red, 2pt]{2,5} \\ \textbf{(b) }\lambda_1=2 \Rightarrow (A-\lambda_1 I)v =0 \Rightarrow \begin{bmatrix} 2 & 1 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \Rightarrow x_2=-2x_1 \\ \qquad \Rightarrow v= x_1 \begin{bmatrix} 1\\-2 \end{bmatrix}, \text{ choosing }v_1= \bbox[red, 2pt]{\begin{bmatrix}1\\-2 \end{bmatrix} }\\ \lambda_2=5 \Rightarrow (A-\lambda_2 I)v =0 \Rightarrow \begin{bmatrix} -1 & 1 \\ 2 & -2 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \Rightarrow x_1=x_2 \\\qquad \Rightarrow v =x_1 \begin{bmatrix}1\\1 \end{bmatrix} , \text{ choosing }v_2= \bbox[red, 2pt]{\begin{bmatrix}1\\1 \end{bmatrix}} \\\textbf{(c) } A= \begin{bmatrix}1& 1\\-2& 1 \end{bmatrix} \begin{bmatrix}2& 0\\0& 5 \end{bmatrix} \begin{bmatrix}1& 1\\-2& 1 \end{bmatrix}^{-1} \Rightarrow \bbox[red, 2pt]{\text{Yes, $A$ is diagonalizable.}}$$
解答:$$\textbf{(a) } \text{If $A$ is an $n\times n$ real symmetric matrix, then $A$ has the following properties:} \\\quad \text{1. $A$ has exactly $n$ real eigenvalues}\\\quad \text{2. The dimension of the eigenspace for each eigenvalue equals the algebraic}\\\qquad \text{multiplicity of that eigenvalue.}\\ \quad \text{3. Eigenspaces for distinct eigenvalues are mutually orthogonal.} \\\quad \text{4. $ A $ is orthogonally diagonalizable.} \\ \textbf{(b) }\cases{Av_1= \lambda_1v_1\\ Av_2= \lambda_2 v_2} , \lambda_1\ne \lambda_2 \Rightarrow \cases{(Av_1)^Tv_2 =(\lambda_1v_1)^Tv_2 = \lambda_1 v_1^Tv_2 \\(Av_1)^Tv_2 = v_1^TA^Tv_2 =v_1^TAv_2=\lambda_2v_1^Tv_2} \Rightarrow \lambda_1 v_1^Tv_2= \lambda_2v_1^Tv_2 \\\qquad \Rightarrow (\lambda_1-\lambda_2)v_1^Tv_2=0 \Rightarrow v_1^Tv_2=0 \Rightarrow v_1 \text{ and }v_2 \text{ are orthogonal} \quad \bbox[red, 2pt]{QED.}$$
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解題僅供參考，碩士班歷年試題及詳解
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