matrix

定义

矩阵：$m\times n$个数$a_{ij}(i=1,2,...,m;j=1,2,...,n)$排成一个m行n列的数表，称为一个$m\times n$矩阵。
矩阵的行数和列数不一定相等。
符号:$()或[]$

同型矩阵：矩阵A与B是同型矩阵$\iff$ A与B的行数相等，列数也相等。
例如
$A=\left\{\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\end{array}\right\}$与$B=\left(\begin{array}{ccc}0& -1& 2\\ 3& 1& 4\end{array}\right)$是同型矩阵

矩阵相等：矩阵A与B相等$\iff$A=B$\iff \{\begin{array}{l}A\text{ 与 }B\text{ 是同型矩阵;}\\ \text{对应位置的元素相等 }.\end{array}$

矩阵的加法

两个同型的矩阵相加，就是对应位置的元素相加
即

$$A={\left(a_{ij}\right)}_{m\times n},B={\left(b_{ij}\right)}_{m\times n},$$

则

$$A+B=(a_{ij}+b_{ij}{)}_{m\times n}.$$ $$设A=\left(\begin{array}{ccc}3& 0& 1\\ -2& -1& 4\end{array}\right),B=\left(\begin{array}{ccc}1& 2& 3\\ 0& 5& 1\end{array}\right),\text{则 }A+B=\left(\begin{array}{ccc}4& 2& 4\\ -2& 4& 5\end{array}\right)$$

相减同理。

矩阵的数乘

数k乘以矩阵A,就是用数k乘以矩阵A的每一个元素.
$A={\left(a_{ij}\right)}_{m\times n},\text{则 }kA={\left(k{a}_{ij}\right)}_{m\times n}.$
$\text{设 A=}\left(\begin{array}{cccc}1& 2& 3& 4\\ 0& -1& 5& 6\end{array}\right)\text{,则 3A=}\left(\begin{array}{cccc}3& 6& 9& 12\\ 0& -3& 15& 18\end{array}\right).$

矩阵的乘法

只有当左边矩阵的列数等于右边矩阵的行数，两个矩阵才可以相乘
乘积矩阵AB的行数 = 左边矩阵A的行数。
乘机矩阵AB的列数 = 右边矩阵B的列数

如

$$\begin{array}{rl}& A_{m\times n},B_{n\times s},\text{ 则 }AB\text{ 为}m\times s\text{ 矩阵,}\\ & A_{3\times 2}{B}_{2\times 4}=(AB{)}_{3\times 4}.\end{array}$$

中间相等（判断两个矩阵是否可以相乘），取两头（乘机矩阵的形状）

$$\left(\begin{array}{ccc}1& 1& 1\\ 0& 1& 1\end{array}\right)\left(\begin{array}{cccc}1& 0& -1& 0\\ 2& 0& 0& 0\\ 1& 1& 4& 5\end{array}\right)=\left(\begin{array}{cccc}4& 1& 3& 5\\ 3& 1& 4& 5\end{array}\right)$$

1. 矩阵的乘法不满足交换律
   AB有意义，BA可能无意义
   $A_{2\times 3}{B}_{3\times 4}=C_{2\times 4}=B_{3\times 4}不能乘A_{2\times 3}$

当AB,BA都有意义时，AB,BA不一定相等.

$$\left(\begin{array}{cc}1& 1\\ 0& 0\end{array}\right)\left(\begin{array}{cc}1& 0\\ 1& 0\end{array}\right)=\left(\begin{array}{cc}2& 0\\ 0& 0\end{array}\right),\left(\begin{array}{cc}1& 0\\ 1& 0\end{array}\right)\left(\begin{array}{cc}1& 1\\ 0& 0\end{array}\right)=\left(\begin{array}{cc}1& 1\\ 1& 1\end{array}\right)$$ $$可得：AB\ne BA$$

2. 矩阵的乘法不满足消去律
3. 两个非零矩阵的乘积可能是零矩阵
   若AB=0，推不出A=0或B=0

矩阵的乘法不满足交换律，矩阵乘积要注意次序，分左乘还是右乘。
AB——A左乘B
BA——A右承B

单位矩阵

AE=A,EA=A;其中 E 为单位矩阵
单位矩阵E在矩阵乘法中相当于数乘1

$$\left(\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\\ 7& 8& 9\end{array}\right)\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)=\left(\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\\ 7& 8& 9\end{array}\right)$$

矩阵的转置

将矩阵A的各行依次变为列后得到的矩阵，即为A的转置矩阵$A^T$。

$$\text{若}A=\left(\begin{array}{ccc}1& 2& 3\\ 0& -1& 4\end{array}\right),\text{则}{A}^T=\left(\begin{array}{cc}1& 0\\ 2& -1\\ 3& 4\end{array}\right)$$

矩阵对向量变换

Scale Matrix

Linear Transforms

线性变换 Linear Transforms(of the same dimension) important

缩放矩阵

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S

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$$\left[\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right]=\left[\begin{array}{cc}s& 0\\ 0& s\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]$$

展开

$$\begin{array}{c}{x}^{\prime }=sx+0\cdot y=sx\\ y^{\prime }=0\cdot x+sy=sy\end{array}$$

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渲染矩阵

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$${\mathbf{R}}_{\theta }=\left[\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right]$$

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以上这些变换都是 Linear Transforms
可以写成

$$\begin{gathered} \begin{array}{rl}{x}^{\prime }& =ax+by\\ y^{\prime }& =cx+dy\\ & \\ \left[\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right]& =\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]\\ \\ & \end{array} \end{gathered}$$ $${\mathbf{x}}^{\prime }=\text{M x}$$

我们发现有一个问题
Translation¹ cannot be represented in matrix form
平移 无法用一个矩阵表示。这不是一个线性变换（一个向量乘以一个矩阵的形式）

$$\left[\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right]=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]+\left[\begin{array}{c}{t}_x\\ t_y\end{array}\right]$$

引入 Homogeneous Coordinates （齐次坐标）

Homogeneous Coordinates

add a third coordinate (w-coordinate)

$$\begin{array}{rl}2\text{D point}& =(x,y,1{)}^{\mathrm{T}}\\ 2\text{D vector}& =(x,y,0{)}^{\mathrm{T}}\end{array}$$

Matrix representation of translations 平移的矩阵表示形式

$$\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\\ w^{\prime }\end{array}\right)=\left(\begin{array}{ccc}1& 0& t_x\\ 0& 1& t_y\\ 0& 0& 1\end{array}\right)\cdot \left(\begin{array}{c}x\\ y\\ 1\end{array}\right)=\left(\begin{array}{c}x+t_x\\ y+t_y\\ 1\end{array}\right)$$

scale

$$\mathbf{S}(s_x,s_y)=\left(\begin{array}{ccc}{s}_x& 0& 0\\ 0& s_y& 0\\ 0& 0& 1\end{array}\right)$$

rotation
$\mathbf{R}(\alpha )=\left(\begin{array}{ccc}\cos \alpha & -\sin \alpha & 0\\ \sin \alpha & \cos \alpha & 0\\ 0& 0& 1\end{array}\right)$
translation
$\mathbf{T}(t_x,t_y)=\left(\begin{array}{ccc}1& 0& t_x\\ 0& 1& t_y\\ 0& 0& 1\end{array}\right)$

Transform Ordering Matter

Matrix multiplication is not commutative
矩阵的乘法不满足交换律

$$R_{45}\cdot T_{(1,0)}\ne T_{(1,0)}\cdot R_{45}$$

例如这两个矩阵$T_{(1,0)}和R_{45}$不能交换位置。先平移，再旋转。
$T_{(1,0)}\cdot R_{45}\left[\begin{array}{c}x\\ y\\ 1\end{array}\right]=\left[\begin{array}{ccc}1& 0& 1\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{ccc}\cos 45^{\circ }& -\sin 45^{\circ }& 0\\ \sin 45^{\circ }& \cos 45^{\circ }& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{c}x\\ y\\ 1\end{array}\right]$

Footnotes

1. formal or technical the process of moving something from one place to another; the translation of the relics of St Thomas of Canterbury. 坎特伯雷圣托马斯的遗骸移葬。 ↩