112年北科大電機碩士班-線性代數詳解(未)

 國立臺北科技大學112學年度碩士班招生考試

系所組別:電機工程系碩士班丁組
科目:線性代數

解答: $$\textbf{(1) }\cases{A= \begin{bmatrix}-1 & 2\\ 2& -3\\ -1 & 3 \end{bmatrix} \\[1ex]\mathbf b= \begin{bmatrix} 4\\ 1\\2\end{bmatrix}} \Rightarrow \cases{ A^TA =\begin{bmatrix} 6 & -11\\-11 & 22\end{bmatrix} \\[1ex] A^Tb= \begin{bmatrix}-4\\11 \end{bmatrix}} \Rightarrow [A^TA \mid A^T b] = \left[\begin{array} {rr|r} 6& -11&-4\\ -11&22& 11\end{array} \right]  \\ \quad \Rightarrow rref([A^TA \mid A^T b])=  \left[ \begin{array} {rr|r} 1& 0& 3\\ 0& 1 & 2\end{array} \right] \Rightarrow \text{an LS solution: } \bbox[red, 2pt]{\hat x = \begin{bmatrix}3 \\2 \end{bmatrix} }\\ \textbf{(2) } \text{Since }Ax=b \text{ has no solution, we need a "best" solution }\hat x \text{ such that }\\\quad dist(b,A\hat x) \le dist(b,Ax)$$

解答: $$\text{Let }\mathbf y_i=(\mathbf x_i-\mathbf x_j), \text{ then }\mathbf y_i \text{ is a non-zero }N\times 1 \text{ vector}. \\\text{We have }\cases{ ||H(\mathbf y_i)||_2^2 =(H\mathbf y_i)^H(H\mathbf y_i) =(\mathbf y_i^H H^H)(H\mathbf y_i) \\ ||\mathbf y_i||_2^2 = \mathbf y_i^H \mathbf y_i}. \text{ Then} \\ \lambda_{max}^2 \mathbf x^H \mathbf x \ge \mathbf x^H \mathbf H^HH \mathbf x \ge \lambda_{min}^2 \mathbf x^H \mathbf x \Rightarrow \mathbf y_i^H H^H H\mathbf y_i \ge \lambda_{min}^2 \mathbf y_i^H \mathbf y_i \Rightarrow ||H(\mathbf y_i)||_2^2 \ge \lambda_{min}^2||\mathbf y_i||_2^2 \\ \Rightarrow \min||H(\mathbf y_i)||_2^2 \ge \lambda_{min}^2 \min||\mathbf y_i||_2^2 \Rightarrow \min_{\mathbf x_i\ne \mathbf x_j}||H(\mathbf x_i-\mathbf x_j)||_2^2 \ge \lambda_{min}^2 \min_{\mathbf x_i\ne \mathbf x_j}||\mathbf (\mathbf x_i-\mathbf x_j)||_2^2 \qquad \bbox[red, 2pt]{QED.}$$

解答: $$\textbf{(1) } A: \text{N x N matrix of the orthonormal eigenvectors of } HH^T\\\quad B:\text{diagonal matrix with r elements equal to the root of the positive eigenvalues of } AAᵀ \\C: \text{transpose of an N x N matrix containing the orthonormal eigenvectors of }AA^T \\\textbf{(2) }\text{If  singular values are repeated, there will be more than one SVD. In this case, SVD is not unique. }$$

解答:

解答: $$\textbf{(1) } Null(A): \{\mathbf x\mid A\mathbf x=0,\mathbf x\in \mathbb R^n\} \\\textbf{(2) }A\mathbf 0 =\mathbf 0 \Rightarrow \mathbf 0 \in Null(A)\\ \text{Suppose }u,v \in Null(A), \text{ then }A(u+v)= A(u)+A(v)=0+0=0 \Rightarrow u+v \in Null(A)\\ \text{Suppose }u \in Null(A), A(cu) =cA(u) =c\cdot 0=0 \Rightarrow cu\in Null(A), c \text{ is a scalar}\\ \text{Therefore, }Null(A) \text{ is a subspace} \\\textbf{(3) }rank(A)+ nullity(A)=N \\\textbf{(4) }A \text{ is not of full-rank } \Rightarrow rank(A)\lt N \Rightarrow dim( Col(A)) \lt N \\\quad \Rightarrow \text{ the columns of A are linearly dependent.}$$

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解題僅供參考，碩士班歷年試題及詳解