# 一文带你直观理解线性变换

### 一、线性变换的定义

In mathematics, and more specifically in linear algebra, a linear map (also called a linear mapping, linear transformation, vector space homomorphism, or in some contexts linear function) is a mapping)$$\displaystyle V\to W$$between two vector spaces that preserves the operations of vector addition and scalar multiplication.

### 二、如何判断是否为线性变换

$$\begin{array}{l} \mathbf{T}(\mathbf{x}+\mathbf{y})=\mathbf{T}(\mathbf{x})+\mathbf{T}(\mathbf{y}) \\ \mathbf{T}(a \mathbf{x})=a \mathbf{T}(\mathbf{x}) \end{array}\tag{1}$$

$$\mathbf{f}(x, y, z)=(3 x-y, 3 z, 0, z-2 x)\tag{2}$$

$$\mathbf{g}(x, y, z)=(3 x-y, 3 z+2,0, z-2 x) \\ \mathbf{h}(x, y, z)=(3 x-y, 3 x z, 0, z-2 x) \tag{3}$$

### 三、线性变换的直观解释

1. 变换前是直线，变换后依然是直线
2. 原点保持固定不动

#### 1、一点不太直观的东西

$$\mathbf{f}(x, y)=(3 x-y, 2 x) \tag{4}$$

$$f\left(\left[\begin{array}{l} x \\ y \end{array}\right]\right)=\left[\begin{array}{cc} 3 & -1\\ 2 & 0 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]\tag{5}$$

#### 2、该矩阵的本质

$$\vec{A}=\left[\begin{array}{l}3 \\2\end{array}\right]$$

$$\vec{i}=\left[\begin{array}{l}1 \\0\end{array}\right]$$

$$\vec{j}=\left[\begin{array}{l}0 \\1\end{array}\right]$$

$$\vec{A}$$可以写成：

$$\vec{A}=3\left[\begin{array}{l} 1 \\ 0 \end{array}\right]+2\left[\begin{array}{l} 0 \\ 1 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]\left[\begin{array}{l} 3 \\ 2 \end{array}\right]\tag{6}$$

$$\overrightarrow{A^{\prime}}=3\overrightarrow{i^{\prime}}+2 \overrightarrow{j^{\prime}}$$

$$\overrightarrow{i^{\prime}}=\left[\begin{array}{l}1 \\-2\end{array}\right]$$

$$\overrightarrow{j^{\prime}}=\left[\begin{array}{l}3 \\0\end{array}\right]$$

$$\overrightarrow{A^{\prime}}$$可以表示为：

$$\overrightarrow{A^{\prime}}=3\left[\begin{array}{l} 1 \\ -2 \end{array}\right]+2\left[\begin{array}{l} 3 \\ 0 \end{array}\right]=\left[\begin{array}{ll} 1 & 3 \\ -2 & 0 \end{array}\right]\left[\begin{array}{l} 3 \\ 2 \end{array}\right]\tag{7}$$

\begin{aligned} \left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] &=\left[\begin{array}{ll} \vec{i} & \vec{j} \end{array}\right] \\ \left[\begin{array}{ll} 1 & 3 \\ -2 & 0 \end{array}\right] &=\left[\begin{array}{ll} \overrightarrow{i^{\prime}} & \overrightarrow{j^{\prime}} \end{array}\right] \end{aligned} \tag{8}

$$\overrightarrow{A^{\prime}}=\left[\begin{array}{ll} 1 & 3 \\ -2 & 0 \end{array}\right]\left[\begin{array}{l} 3 \\ 2 \end{array}\right]=\left[\begin{array}{ll} 1 & 3 \\ -2 & 0 \end{array}\right]\vec{A}\tag{9}$$

$$\vec{A}=\left[\begin{array}{l}3 \\2\end{array}\right]$$

$$\left[\begin{array}{ll} 1 & 3 \\ -2 & 0 \end{array}\right]=\left[\begin{array}{ll} \overrightarrow{i^{\prime}} & \overrightarrow{j^{\prime}} \end{array}\right]$$

#### 3. 推广

$$\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]$$

$$(a, c)$$就看作是变换后第一个基向量的落脚点，$$(b, d)$$看作是变换后第二个基向量的落脚点。当你想知道任意一个向量

$$\vec{B}=\left[\begin{array}{l}x \\ y\end{array}\right]$$

$$\overrightarrow{B^{\prime}}=\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]\tag{10}$$

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