标签:center mes display 种类型 end 图片 play 数学基础 方向
目录
F
范数:设矩 \(\mathbf A=(a_{i,j})_{m\times n}\) ,则其F
范数为 \(||\mathbf A||_F=\sqrt{\sum_{i,j}a_{i,j}^{2}}\) 。F
范数等 \(\mathbf A\mathbf A^T\) 的迹的平方根 \(||\mathbf A||_F=\sqrt{tr(\mathbf A \mathbf A^{T})}\) 。三维向量的点积 \(\mathbf{\vec u}\cdot\mathbf{\vec v} =u _xv_x+u_yv_y+u_zv_z = |\mathbf{\vec u}| | \mathbf{\vec v}| \cos(\mathbf{\vec u},\mathbf{\vec v})\) 。
两个向量的并矢:给定两个向 \(\mathbf {\vec x}=(x_1,x_2,\cdots,x_n)^{T}, \mathbf {\vec y}= (y_1,y_2,\cdots,y_m)^{T}\) ,则向量的并矢记作:
\[\mathbf {\vec x}\mathbf {\vec y} =\begin{bmatrix}x_1y_1&x_1y_2&\cdots&x_1y_m\\ x_2y_1&x_2y_2&\cdots&x_2y_m\\ \vdots&\vdots&\ddots&\vdots\\ x_ny_1&x_ny_2&\cdots&x_ny_m\\ \end{bmatrix}\] 也记 \(\mathbf {\vec x}\otimes\mathbf {\vec y}\) 或 \(\mathbf {\vec x} \mathbf {\vec y}^{T}\) 。
Hadamard product
(又称作逐元素积):Kronnecker product
:\(\mathbf {\vec x},\mathbf {\vec a},\mathbf {\vec b},\mathbf {\vec c}\) \(n\) 阶向量 \(\mathbf A,\mathbf B,\mathbf C,\mathbf X\) \(n\) 阶方阵,则有:
\[\frac{\partial(\mathbf {\vec a}^{T}\mathbf {\vec x}) }{\partial \mathbf {\vec x} }=\frac{\partial(\mathbf {\vec x}^{T}\mathbf {\vec a}) }{\partial \mathbf {\vec x} } =\mathbf {\vec a}\] \[\frac{\partial(\mathbf {\vec a}^{T}\mathbf X\mathbf {\vec b}) }{\partial \mathbf X }=\mathbf {\vec a}\mathbf {\vec b}^{T}=\mathbf {\vec a}\otimes\mathbf {\vec b}\in \mathbb R^{n\times n}\] \[\frac{\partial(\mathbf {\vec a}^{T}\mathbf X^{T}\mathbf {\vec b}) }{\partial \mathbf X }=\mathbf {\vec b}\mathbf {\vec a}^{T}=\mathbf {\vec b}\otimes\mathbf {\vec a}\in \mathbb R^{n\times n}\] \[\frac{\partial(\mathbf {\vec a}^{T}\mathbf X\mathbf {\vec a}) }{\partial \mathbf X }=\frac{\partial(\mathbf {\vec a}^{T}\mathbf X^{T}\mathbf {\vec a}) }{\partial \mathbf X }=\mathbf {\vec a}\otimes\mathbf {\vec a}\] \[\frac{\partial(\mathbf {\vec a}^{T}\mathbf X^{T}\mathbf X\mathbf {\vec b}) }{\partial \mathbf X }=\mathbf X(\mathbf {\vec a}\otimes\mathbf {\vec b}+\mathbf {\vec b}\otimes\mathbf {\vec a})\] \[\frac{\partial[(\mathbf A\mathbf {\vec x}+\mathbf {\vec a})^{T}\mathbf C(\mathbf B\mathbf {\vec x}+\mathbf {\vec b})]}{\partial \mathbf {\vec x}}=\mathbf A^{T}\mathbf C(\mathbf B\mathbf {\vec x}+\mathbf {\vec b})+\mathbf B^{T}\mathbf C(\mathbf A\mathbf {\vec x}+\mathbf {\vec a})\] \[\frac{\partial (\mathbf {\vec x}^{T}\mathbf A \mathbf {\vec x})}{\partial \mathbf {\vec x}}=(\mathbf A+\mathbf A^{T})\mathbf {\vec x}\] \[\frac{\partial[(\mathbf X\mathbf {\vec b}+\mathbf {\vec c})^{T}\mathbf A(\mathbf X\mathbf {\vec b}+\mathbf {\vec c})]}{\partial \mathbf X}=(\mathbf A+\mathbf A^{T})(\mathbf X\mathbf {\vec b}+\mathbf {\vec c})\mathbf {\vec b}^{T} \] \[\frac{\partial (\mathbf {\vec b}^{T}\mathbf X^{T}\mathbf A \mathbf X\mathbf {\vec c})}{\partial \mathbf X}=\mathbf A^{T}\mathbf X\mathbf {\vec b}\mathbf {\vec c}^{T}+\mathbf A\mathbf X\mathbf {\vec c}\mathbf {\vec b}^{T}\]
假 \(\mathbf U= f(\mathbf X)\) 是关 \(\mathbf X\) 的矩阵值函数 \(f:\mathbb R^{m\times n}\rightarrow \mathbb R^{m\times n}\) ), \(g(\mathbf U)\) 是关 \(\mathbf U\) 的实值函数 $g:\mathbb R^{m\times n}\rightarrow \mathbb R $ ),则下面链式法则成立:
\[\frac{\partial g(\mathbf U)}{\partial \mathbf X}=
\left(\frac{\partial g(\mathbf U)}{\partial x_{i,j}}\right)_{m\times n}=\begin{bmatrix}
\frac{\partial g(\mathbf U)}{\partial x_{1,1}}&\frac{\partial g(\mathbf U)}{\partial x_{1,2}}&\cdots&\frac{\partial g(\mathbf U)}{\partial x_{1,n}}\\
\frac{\partial g(\mathbf U)}{\partial x_{2,1}}&\frac{\partial g(\mathbf U)}{\partial x_{2,2}}&\cdots&\frac{\partial g(\mathbf U)}{\partial x_{2,n}}\\
\vdots&\vdots&\ddots&\vdots\\
\frac{\partial g(\mathbf U)}{\partial x_{m,1}}&\frac{\partial g(\mathbf U)}{\partial x_{m,2}}&\cdots&\frac{\partial g(\mathbf U)}{\partial x_{m,n}}\\
\end{bmatrix}\\
=\left(\sum_{k}\sum_{l}\frac{\partial g(\mathbf U)}{\partial u_{k,l}}\frac{\partial u_{k,l}}{\partial x_{i,j}}\right)_{m\times n}=\left(tr\left[\left(\frac{\partial g(\mathbf U)}{\partial \mathbf U}\right)^{T}\frac{\partial \mathbf U}{\partial x_{i,j}}\right]\right)_{m\times n}\]
?
本文转载自华校专老师博客,博客地址:http://www.huaxiaozhuan.com/
标签:center mes display 种类型 end 图片 play 数学基础 方向
原文地址:https://www.cnblogs.com/guoyaohua/p/9803027.html