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[06] 矩阵计算

2021-11-08 23:01:51  阅读:152  来源: 互联网

标签:frac bmatrix mathbf 矩阵 ym 计算 y1 06 partial


主要是关于矩阵的求导。

  1. ∂ y ∂ x \frac{\partial y}{\partial \mathbf{x}} ∂x∂y​ [ y y y是标量, x \mathbf{x} x是列向量,导数是行向量]
    x = [ x 1 x 2 ⋮ x n ] , ∂ y ∂ x = [ ∂ y ∂ x 1 , ∂ y ∂ x 2 , ⋯   , ∂ y ∂ x n ] \mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix}, \quad \frac{\partial y}{\partial \mathbf{x}} = [\frac{\partial y}{\partial x_1}, \frac{\partial y}{\partial x_2}, \cdots, \frac{\partial y}{\partial x_n}] x=⎣⎢⎢⎢⎡​x1​x2​⋮xn​​⎦⎥⎥⎥⎤​,∂x∂y​=[∂x1​∂y​,∂x2​∂y​,⋯,∂xn​∂y​]
    例如: ∂ ∂ x x 1 2 + 2 x 2 2 = [ 2 x 1 , 4 x 2 ] \frac{\partial}{\partial \mathbf{x}}x_1^2 + 2x_2^2 = [2x_1, 4x_2 ] ∂x∂​x12​+2x22​=[2x1​,4x2​]

    一些样例

    y y y a a a a u au au sum ( x ) \text{sum}(x) sum(x) ∥ x ∥ 2 \|\mathbf{x}\|^2 ∥x∥2
    ∂ y ∂ x \frac{\partial y}{\partial \mathbf{x}} ∂x∂y​ 0 T \mathbf{0}^T 0T a ∂ u ∂ x a\frac{\partial u}{\partial \mathbf{x}} a∂x∂u​ 1 T \mathbf{1}^T 1T 2 x T 2\mathbf{x}^T 2xT
    y y y u + v u+v u+v u v uv uv < u , v > <\mathbf{u}, \mathbf{v}> <u,v>
    ∂ y ∂ x \frac{\partial y}{\partial \mathbf{x}} ∂x∂y​ ∂ u ∂ x + ∂ v ∂ x \frac{\partial u}{\partial \mathbf{x}} + \frac{\partial v}{\partial \mathbf{x}} ∂x∂u​+∂x∂v​ ∂ u ∂ x v + ∂ v ∂ x u \frac{\partial u}{\partial \mathbf{x}} v+ \frac{\partial v}{\partial \mathbf{x}} u ∂x∂u​v+∂x∂v​u u T ∂ v ∂ x + v T ∂ u ∂ x \mathbf{u}^T \frac{\partial \mathbf{v}}{\partial \mathbf{x}} + \mathbf{v}^T \frac{\partial \mathbf{u}}{\partial \mathbf{x}} uT∂x∂v​+vT∂x∂u​
  2. ∂ y ∂ x \frac{\partial \mathbf{y}}{\partial x} ∂x∂y​ [ y \mathbf{y} y是列向量, x x x是标量,导数是列向量]
    y = [ y 1 y 2 ⋮ y m ] , ∂ y ∂ x = [ ∂ y 1 x ∂ y 2 x ⋮ ∂ y m x ] \mathbf{y} = \begin{bmatrix} y_1 \\ y_2 \\ \vdots \\ y_m \end{bmatrix}, \quad \frac{\partial \mathbf{y}}{\partial x} = \begin{bmatrix} \frac{\partial y_1}{x} \\ \frac{\partial y_2}{x} \\ \vdots \\ \frac{\partial y_m}{x} \end{bmatrix} y=⎣⎢⎢⎢⎡​y1​y2​⋮ym​​⎦⎥⎥⎥⎤​,∂x∂y​=⎣⎢⎢⎢⎡​x∂y1​​x∂y2​​⋮x∂ym​​​⎦⎥⎥⎥⎤​

  3. ∂ y ∂ x \frac{\partial \mathbf{y}}{\partial \mathbf{x}} ∂x∂y​ [ y \mathbf{y} y是列向量, x \mathbf{x} x是列向量,导数是矩阵]
    x = [ x 1 x 2 ⋮ x n ] , y = [ y 1 y 2 ⋮ y m ] , ∂ y ∂ x = [ ∂ y 1 x ∂ y 2 x ⋮ ∂ y m x ] = [ ∂ y 1 ∂ x 1 , ∂ y 1 ∂ x 2 , ⋯   , ∂ y 1 ∂ x n ∂ y 2 ∂ x 1 , ∂ y 2 ∂ x 2 , ⋯   , ∂ y 2 ∂ x n ⋮ ∂ y m ∂ x 1 , ∂ y m ∂ x 2 , ⋯   , ∂ y m ∂ x n ] \mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix}, \mathbf{y} = \begin{bmatrix} y_1 \\ y_2 \\ \vdots \\ y_m \end{bmatrix}, \frac{\partial \mathbf{y}}{\partial \mathbf{x}} = \begin{bmatrix} \frac{\partial y_1}{\mathbf{x}} \\ \frac{\partial y_2}{\mathbf{x}} \\ \vdots \\ \frac{\partial y_m}{\mathbf{x}} \end{bmatrix} = \begin{bmatrix} \frac{\partial y_1}{\partial x_1}, & \frac{\partial y_1}{\partial x_2}, & \cdots, & \frac{\partial y_1}{\partial x_n} \\ \frac{\partial y_2}{\partial x_1}, & \frac{\partial y_2}{\partial x_2}, & \cdots, & \frac{\partial y_2}{\partial x_n} \\ \vdots \\ \frac{\partial y_m}{\partial x_1}, & \frac{\partial y_m}{\partial x_2}, & \cdots, &\frac{\partial y_m}{\partial x_n} \\ \end{bmatrix} x=⎣⎢⎢⎢⎡​x1​x2​⋮xn​​⎦⎥⎥⎥⎤​,y=⎣⎢⎢⎢⎡​y1​y2​⋮ym​​⎦⎥⎥⎥⎤​,∂x∂y​=⎣⎢⎢⎢⎡​x∂y1​​x∂y2​​⋮x∂ym​​​⎦⎥⎥⎥⎤​=⎣⎢⎢⎢⎢⎡​∂x1​∂y1​​,∂x1​∂y2​​,⋮∂x1​∂ym​​,​∂x2​∂y1​​,∂x2​∂y2​​,∂x2​∂ym​​,​⋯,⋯,⋯,​∂xn​∂y1​​∂xn​∂y2​​∂xn​∂ym​​​⎦⎥⎥⎥⎥⎤​
    一些样例

    y \mathbf{y} y a \mathbf{a} a x \mathbf{x} x A x \mathbf{Ax} Ax x T A \mathbf{x^TA} xTA
    ∂ y ∂ x \frac{\partial \mathbf{y}}{\partial \mathbf{x}} ∂x∂y​ 0 \mathbf{0} 0 I \mathbf{I} I A \mathbf{A} A A T \mathbf{A}^T AT
    y \mathbf{y} y a u a\mathbf{u} au A u \mathbf{Au} Au u + v \mathbf{u}+ \mathbf{v} u+v
    ∂ y ∂ x \frac{\partial \mathbf{y}}{\partial \mathbf{x}} ∂x∂y​ a ∂ u ∂ x a\frac{\partial \mathbf{u}}{\partial \mathbf{x}} a∂x∂u​ A ∂ u ∂ x \mathbf{A}\frac{\partial \mathbf{u}}{\partial \mathbf{x}} A∂x∂u​ ∂ u ∂ x + ∂ v ∂ x \frac{\partial \mathbf{u}}{\partial \mathbf{x}} + \frac{\partial \mathbf{v}}{\partial \mathbf{x}} ∂x∂u​+∂x∂v​

标签:frac,bmatrix,mathbf,矩阵,ym,计算,y1,06,partial
来源: https://blog.csdn.net/gpx33333/article/details/121218372

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