9/22 作业

Homework 4

1. (1) $3$ , $1$ , $2$ .

   (2) $2$ , $2$ , 相同。

   (3) $3$ .

2. $\mathbf 1_4$ , $\begin{pmatrix}1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\\ 0 & 0 & 0\\ 0 & 0 & 0\end{pmatrix}$ .

3. (a) $\operatorname{diag}( \frac{2}{5}, 0, -\frac{1}{9} )$ .

   (b) $\begin{pmatrix}1 & 1 & \frac{2}{3}\\ 0 & 0 & 0\\ \frac{1}{2} & \frac{1}{2} & \frac{1}{3}\end{pmatrix}$ .

   (c) $\begin{pmatrix}\frac{8}{5} & 1 & 1\\ 0 & 3 & 1\\ 0 & 0 & -\frac{1}{9} \end{pmatrix}$ .

   (d) $\begin{pmatrix}1 & 1 & \frac{2}{3}\\ 1 & 1 & \frac{8}{9}\\ 1 & 1 & \frac{7}{9}\end{pmatrix}$ .

4. (a) $\mathbf 1_3$ because rank of the coefficient matrix is $3$ .

   (b) Unique because $A$ is invertible.

   (c) $0$ or $1$ .

   $\operatorname{rank}(A) = \operatorname{rank}(A | \vec b)=3$ since $A \vec x = \vec b$ has unique solution.

   $\operatorname{rank}(A | \vec c) = 3$ or $4$ , then $\operatorname{rank}(A) \le \operatorname{rank}(A | \vec c)$ , and the system has either one or no solution.

5. (a) Infinitely many.

   Because $\operatorname{rank}(A)=n$ , $A$ has right inverse.

   Therefore $AA^+ = \mathbf 1_n$ .

   Lemma 1 $\forall X : \R^{m \times n}$ , $A (A^+ + (\mathbf 1_m - A^+A)X) = \mathbf 1_n$ .

   Proof

   $$\begin{align*} A (A^+ + (\mathbf 1_m - A^+ A) X) \\ & = A A^+ + A (\mathbf 1_m - A^+ A) X \\ & = \mathbf 1_n + (A - A A^+ A) X \\ & = \mathbf 1_n + (A - A) X \\ & = \mathbf 1_n \end{align*}$$

   $\square$ .

   By lemma 1, obviously there are infinitely many right inverse of $A$ .

   (b) Only $A^{-1}$ .

Optional

1. a. $\vec x = \mathbf 0$ is a solution.

   b. $\operatorname{rank}(A) = \operatorname{rank}(A|\vec 0) \lt \text{number of variables}$ .

   c. $A (\vec x_1 + \vec x_2) = A \vec x_1 + A \vec x_2 = \vec 0$ .

   d. $A(k \vec x) = k(A \vec x) = \vec 0$ .

2. a. $A (\vec x_1 + \vec x_h) = A \vec x_1 + A \vec x_h = \vec b + \vec 0 = \vec b$ .

   b. $A (\vec x_2 - \vec x_1) = A \vec x_2 - A \vec x_1 = \vec b - \vec b = \vec 0$ .

   c. Plot.

   

3. (1)(2) 证明

   不妨令 $A : \R ^{n \times n}$ 是上三角矩阵。

   $(A^k)_{i,i} = (A_{i,i})^k$ , $A^k = \mathbf 0 \implies A_{i,i} = 0 \space \forall i$ .

   展开 $A^n$

   $$\begin{equation} (A^n)_{i, j} = \sum_{i_1, i_2, \cdots, i_{n-1}} A_{i, i_1} A_{i_1, i_2} \cdots A_{i_{n-1}, j} \end{equation}$$

   如果 $A$ 的主对角线均为 $0$ , 由于 $A$ 是上三角矩阵

   $$A_{i, j} \ne 0 \implies i < j$$

   由于 $i,j : \N^* \le n$ , 不可能找到连续上升的 $n + 1$ 个下标 $i, i_1, i_2, \cdots, j$ . 故至少一项 $A_{i_k, i_{k+1}} = 0$ . 因此求和 (1) 的每一项均为 $0$ , $(A^n)_{i, j} = 0$ , $A^n = \mathbf 0_{n \times n}$ $\square$ .

   (3) 证明 $(A^k)^n = \mathbf 0_{n \times n}$ $\square$ .