早稲田大学 創造理工学研究科 経営システム工学専攻 2021年8月実施 线性代数

标签：

• Waseda-University
• Linear-Transformation

Author

思齐塾, 祭音Myyura

Description

$R$ を実数集合 (set of real numbers) とし、

$$A = \begin{bmatrix} 1 & 1 & 2 \\ 2 & -1 & 1 \\ 3 & 1 & 3 \\ 2 & -1 & 0 \end{bmatrix}$$

とし、線形写像 (linear transformation) $f: R^3 \to R^4$ が

$$f(\mathbf{x}) = A\mathbf{x}, \quad \mathbf{x} \in R^3$$

と表せるとする。このとき、 $f(\mathbf{x}) = f(\mathbf{y})$ ならば $\mathbf{x} = \mathbf{y}$ であることを示せ。

题目描述

记实数集为 $\mathbb R$，令

$$A=\begin{bmatrix} 1&1&2\\ 2&-1&1\\ 3&1&3\\ 2&-1&0 \end{bmatrix},$$

并定义线性映射

$$f:\mathbb R^3\to\mathbb R^4,\qquad f(\mathbf x)=A\mathbf x\quad(\mathbf x\in\mathbb R^3).$$

证明：若 $f(\mathbf x)=f(\mathbf y)$，则 $\mathbf x=\mathbf y$。

Kai

We want to show that if $f(\mathbf{x}) = f(\mathbf{y})$ , then $\mathbf{x} = \mathbf{y}$ . This is equivalent to showing that $f(\mathbf{x}) = A\mathbf{x}$ is injective (one-to-one).

If $f(\mathbf{x}) = f(\mathbf{y})$ , then $A\mathbf{x} = A\mathbf{y}$ . Thus, $A\mathbf{x} - A\mathbf{y} = \mathbf{0}$ , which means $A(\mathbf{x} - \mathbf{y}) = \mathbf{0}$ . Let $\mathbf{z} = \mathbf{x} - \mathbf{y}$ . We want to show that $A\mathbf{z} = \mathbf{0}$ implies $\mathbf{z} = \mathbf{0}$ .

In other words, we want to show that the kernel (null space) of $A$ is only the zero vector. We can determine this by row reducing $A$ .

$$A = \begin{bmatrix} 1 & 1 & 2 \\ 2 & -1 & 1 \\ 3 & 1 & 3 \\ 2 & -1 & 0 \end{bmatrix} \sim \begin{bmatrix} 1 & 1 & 2 \\ 0 & -3 & -3 \\ 0 & -2 & -3 \\ 0 & -3 & -4 \end{bmatrix} \sim \begin{bmatrix} 1 & 1 & 2 \\ 0 & 1 & 1 \\ 0 & -2 & -3 \\ 0 & -3 & -4 \end{bmatrix} \sim \begin{bmatrix} 1 & 1 & 2 \\ 0 & 1 & 1 \\ 0 & 0 & -1 \\ 0 & 0 & -1 \end{bmatrix} \sim \begin{bmatrix} 1 & 1 & 2 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} \sim \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} \sim \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}$$

Since the row reduced echelon form of $A$ has rank 3, which is the number of columns, the kernel of $A$ is $\{\mathbf{0}\}$ . Thus, $A\mathbf{z} = \mathbf{0}$ implies $\mathbf{z} = \mathbf{0}$ . Since $\mathbf{z} = \mathbf{x} - \mathbf{y}$ , we have $\mathbf{x} - \mathbf{y} = \mathbf{0}$ , so $\mathbf{x} = \mathbf{y}$ .