向量求导运算法则

前言

矩阵,向量的求导经常碰到和用到,但是老是忘记,在网上收集总结一下。

1.矩阵对元素的求导

矩阵对元素的求导比较简单,就是对矩阵的每个元素分别进行求导。

\[若:Y= \begin{pmatrix} y_{11} &\cdots & y_{1n} \\\ \vdots &\cdots & \vdots \\\ y_{m1} &\cdots & y_{mn} \end{pmatrix} \]

\[则:{\partial{Y} \over \partial{x}}= \begin{pmatrix} \partial{y_{11}} \over \partial{x} &\cdots & \partial{y_{1n}} \over \partial{x} \\\ \vdots &\cdots & \vdots \\\ \partial{y_{m1}} \over \partial{x} &\cdots & \partial{y_{mn}} \over \partial{x} \end{pmatrix} \]

2.元素对矩阵求导

\[{ \partial{x} \over \partial{Y}}= \begin{pmatrix} \partial{x} \over \partial{y_{11}} &\cdots & \partial{x} \over \partial{y_{1n}} \\\ \vdots &\cdots & \vdots \\\ \partial{x} \over \partial{y_{m1}} &\cdots & \partial{x} \over \partial{y_{mn}} \end{pmatrix} \]

3.行向量对列向量求导

\[若:Y= \begin{bmatrix} y_{1} &\cdots & y_{n} \end{bmatrix}, X= \begin{bmatrix} x_{1} \\\ \vdots\\\ x_{m} \end{bmatrix} \]

\[则: { \partial{Y} \over \partial{X}}= \begin{pmatrix} \partial{y_1} \over \partial{x_{1}} &\cdots & \partial{y_n} \over \partial{x_{1}} \\\ \vdots &\cdots & \vdots \\\ \partial{y_1} \over \partial{x_{m}} &\cdots & \partial{y_n} \over \partial{x_{m}} \end{pmatrix} \]

4.行向量对行向量求导

\[若:Y= \begin{bmatrix} y_{1} &\cdots & y_{n} \end{bmatrix}, X= \begin{bmatrix} x_{1} & \cdots & x_{p} \end{bmatrix} \]

\[则得到一个超级大的行向量: { \partial{Y} \over \partial{X}}= \begin{bmatrix} \partial{Y} \over \partial{x_{1}} &\cdots & \partial{Y} \over \partial{x_{p}} \end{bmatrix} \]

5.列向量对列向量求导

\[若:Y^T= \begin{bmatrix} y_{1} &\cdots & y_{n} \end{bmatrix}, X^T= \begin{bmatrix} x_{1} & \cdots & x_{p} \end{bmatrix} \]

\[则得到一个超级大的列向量: { \partial{Y} \over \partial{X}}= \begin{bmatrix} \partial{Y} \over \partial{x_{1}} &\cdots & \partial{Y} \over \partial{x_{p}} \end{bmatrix}^T \]

6.矩阵对行向量求导\行向量对矩阵求导

\[若:Y= \begin{pmatrix} y_{11} &\cdots & y_{1n} \\\ \vdots &\cdots & \vdots \\\ y_{m1} &\cdots & y_{mn} \end{pmatrix} X= \begin{bmatrix} x_{1} & \cdots & x_{p} \end{bmatrix} \]

\[则: { \partial{Y} \over \partial{X}}= \begin{bmatrix} \partial{Y} \over \partial{x_{1}} &\cdots & \partial{Y} \over \partial{x_{p}} \end{bmatrix}^T\\\ { \partial{X} \over \partial{Y}}= \begin{pmatrix} \partial{X} \over \partial{y_{11}} &\cdots & \partial{X} \over \partial{y_{1n}} \\\ \vdots &\cdots & \vdots \\\ \partial{X} \over \partial{y_{m1}} &\cdots & \partial{X} \over \partial{y_{mn}} \end{pmatrix} \]

7.矩阵对列向量求导\列向量对矩阵求导

\[若:Y= \begin{pmatrix} y_{11} &\cdots & y_{1n} \\\ \vdots &\cdots & \vdots \\\ y_{m1} &\cdots & y_{mn} \end{pmatrix} X= \begin{bmatrix} x_{1} \\\ \vdots \\\ x_{p} \end{bmatrix} \]

\[则: { \partial{Y} \over \partial{X}}= \begin{pmatrix} \partial{y_1} \over \partial{X} &\cdots & \partial{y_n} \over \partial{X} \\\ \vdots &\cdots & \vdots \\\ \partial{y_1} \over \partial{X} &\cdots & \partial{y_n} \over \partial{X} \end{pmatrix}\\\ { \partial{X} \over \partial{Y}}= \begin{bmatrix} \partial{X_1} \over \partial{Y} \\\ \vdots \\\ \partial{X_p} \over \partial{Y} \end{bmatrix} \]

8.矩阵对矩阵求导

\[若:Y= \begin{bmatrix} y_{11} &\cdots & y_{1n} \\\ \vdots &\cdots & \vdots \\\ y_{m1} &\cdots & y_{mn} \end{bmatrix} = \begin{bmatrix} y_{1} \\\ \vdots \\\ y_{m} \end{bmatrix}; X= \begin{bmatrix} x_{11} &\cdots & x_{1q} \\\ \vdots &\cdots & \vdots \\\ x_{p1} &\cdots & x_{pq} \end{bmatrix} = \begin{bmatrix} x_{1} & \cdots & x_{m} \end{bmatrix} \]

则:

\[{ \partial{Y} \over \partial{X}}= \begin{bmatrix} \partial{Y} \over \partial{x_1} & \cdots & \partial{Y} \over \partial{x_q} \end{bmatrix} = \begin{bmatrix} \partial{y_1} \over \partial{X} \\\ \vdots \\\ \partial{y_m} \over \partial{X} \end{bmatrix} = \begin{bmatrix} \partial{y_1} \over \partial{x_1} & \cdots & \partial{y_1} \over \partial{x_q} \\\ \vdots & \cdots & \vdots \\\ \partial{y_m} \over \partial{x_1} & \cdots & \partial{y_m} \over \partial{x_q} \end{bmatrix} \]

原文地址:https://www.cnblogs.com/liujshi/p/5658120.html