【深度学习】动手学深度学习(PyTorch版)李沐 2.4.3 梯度【公式推导】

2.4.3. 梯度

  我们可以连接一个多元函数对其所有变量的偏导数，以得到该函数的梯度（gradient）向量。 具体而言，设函数$f:{\mathbb{R}}^n\to \mathbb{R}$的输入是一个$n$维向量 x ⃗ = [ x 1 x 2 ⋅ ⋅ ⋅ x n ] \vec x=\begin{bmatrix} x_1\\x_2\\···\\x_n\end{bmatrix} x = ​x1​x2​⋅⋅⋅xn​​ ​，输出是一个标量。 函数$f(x)$相对于$x$的梯度是一个包含$n$个偏导数的向量：
∇ x ⃗ f ( x ⃗ ) = [ ∂ f ( x ⃗ ) ∂ x 1 ∂ f ( x ⃗ ) ∂ x 2 ⋅ ⋅ ⋅ ∂ f ( x ⃗ ) ∂ x n ] \nabla_{\vec x} f(\vec x) = \begin{bmatrix}\frac{\partial f(\vec x)}{\partial x_1}\\\frac{\partial f(\vec x)}{\partial x_2}\\···\\ \frac{\partial f(\vec x)}{\partial x_n}\end{bmatrix} ∇x ​f(x )= ​∂x1​∂f(x )​∂x2​∂f(x )​⋅⋅⋅∂xn​∂f(x )​​ ​
其中${\nabla }_{x}f(x)$通常在没有歧义时被$\nabla f(x)$取代。

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假设$x$为$n$维向量，在微分多元函数时经常使用以下规则:

一、对于所有$A\in {\mathbb{R}}^{m\times n}$，都有${\nabla }_{x}Ax=A^{\top }$；

证明：设$A_{(m,n)}$ = [ a 1 , 1 a 1 , 2 ⋅ ⋅ ⋅ a 1 , n a 2 , 1 a 2 , 2 ⋅ ⋅ ⋅ a 2 , n ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ a m , 1 a m , 2 ⋅ ⋅ ⋅ a m , n ] \begin{bmatrix} a_{1,1}&a_{1,2}&···&a_{1,n} \\ a_{2,1}&a_{2,2}&···&a_{2,n} \\ ··· & ··· & ··· & ··· \\ a_{m,1} & a_{m,2} &···&a_{m,n} \end{bmatrix} ​a1,1​a2,1​⋅⋅⋅am,1​​a1,2​a2,2​⋅⋅⋅am,2​​⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅​a1,n​a2,n​⋅⋅⋅am,n​​ ​，
则$A{x}_{(m,1)}$ = [ a 1 , 1 x 1 + a 1 , 2 x 2 + ⋅ ⋅ ⋅ + a 1 , n x n a 2 , 1 x 1 + a 2 , 2 x 2 + ⋅ ⋅ ⋅ + a 2 , n x n ⋅ ⋅ ⋅ a m , 1 x 1 + a m , 2 x 2 + ⋅ ⋅ ⋅ + a m , n x n ] \begin{bmatrix} a_{1,1}x_1+a_{1,2}x_2+···+a_{1,n}x_n \\ a_{2,1}x_1+a_{2,2}x_2+···+a_{2,n}x_n \\ ··· \\ a_{m,1}x_1+a_{m,2}x_2+···+a_{m,n}x_n \end{bmatrix} ​a1,1​x1​+a1,2​x2​+⋅⋅⋅+a1,n​xn​a2,1​x1​+a2,2​x2​+⋅⋅⋅+a2,n​xn​⋅⋅⋅am,1​x1​+am,2​x2​+⋅⋅⋅+am,n​xn​​ ​,
${\nabla }_{x}Ax$= [ ∂ A x ⃗ ∂ x 1 ∂ A x ⃗ ∂ x 2 ⋅ ⋅ ⋅ ∂ A x ⃗ ∂ x n ] \begin{bmatrix}\frac{\partial A\vec x}{\partial x_1}\\\frac{\partial A\vec x}{\partial x_2}\\···\\ \frac{\partial A\vec x}{\partial x_n}\end{bmatrix} ​∂x1​∂Ax ​∂x2​∂Ax ​⋅⋅⋅∂xn​∂Ax ​​ ​
= [ ∂ a 1 , 1 x 1 + a 1 , 2 x 2 + ⋅ ⋅ ⋅ + a 1 , n x n ∂ x 1 ∂ a 2 , 1 x 1 + a 2 , 2 x 2 + ⋅ ⋅ ⋅ + a 2 , n x n ∂ x 1 ⋅ ⋅ ⋅ ∂ a m , 1 x 1 + a m , 2 x 2 + ⋅ ⋅ ⋅ + a m , n x n ∂ x 1 ∂ a 1 , 1 x 1 + a 1 , 2 x 2 + ⋅ ⋅ ⋅ + a 1 , n x n ∂ x 2 ∂ a 2 , 1 x 1 + a 2 , 2 x 2 + ⋅ ⋅ ⋅ + a 2 , n x n ∂ x 2 ⋅ ⋅ ⋅ ∂ a m , 1 x 1 + a m , 2 x 2 + ⋅ ⋅ ⋅ + a m , n x n ∂ x 2 ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ∂ a 1 , 1 x 1 + a 1 , 2 x 2 + ⋅ ⋅ ⋅ + a 1 , n x n ∂ x n ∂ a 2 , 1 x 1 + a 2 , 2 x 2 + ⋅ ⋅ ⋅ + a 2 , n x n ∂ x n ⋅ ⋅ ⋅ ∂ a m , 1 x 1 + a m , 2 x 2 + ⋅ ⋅ ⋅ + a m , n x n ∂ x n ] \begin{bmatrix}\frac{\partial a_{1,1}x_1+a_{1,2}x_2+···+a_{1,n}x_n}{\partial x_1}& \frac{\partial a_{2,1}x_1+a_{2,2}x_2+···+a_{2,n}x_n}{\partial x_1}&···&\frac{\partial a_{m,1}x_1+a_{m,2}x_2+···+a_{m,n}x_n}{\partial x_1}\\ \frac{\partial a_{1,1}x_1+a_{1,2}x_2+···+a_{1,n}x_n}{\partial x_2}& \frac{\partial a_{2,1}x_1+a_{2,2}x_2+···+a_{2,n}x_n}{\partial x_2}&···&\frac{\partial a_{m,1}x_1+a_{m,2}x_2+···+a_{m,n}x_n}{\partial x_2}\\ ···&···&···&···\\ \frac{\partial a_{1,1}x_1+a_{1,2}x_2+···+a_{1,n}x_n}{\partial x_n}& \frac{\partial a_{2,1}x_1+a_{2,2}x_2+···+a_{2,n}x_n}{\partial x_n}&···&\frac{\partial a_{m,1}x_1+a_{m,2}x_2+···+a_{m,n}x_n}{\partial x_n}\end{bmatrix} ​∂x1​∂a1,1​x1​+a1,2​x2​+⋅⋅⋅+a1,n​xn​​∂x2​∂a1,1​x1​+a1,2​x2​+⋅⋅⋅+a1,n​xn​​⋅⋅⋅∂xn​∂a1,1​x1​+a1,2​x2​+⋅⋅⋅+a1,n​xn​​​∂x1​∂a2,1​x1​+a2,2​x2​+⋅⋅⋅+a2,n​xn​​∂x2​∂a2,1​x1​+a2,2​x2​+⋅⋅⋅+a2,n​xn​​⋅⋅⋅∂xn​∂a2,1​x1​+a2,2​x2​+⋅⋅⋅+a2,n​xn​​​⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅​∂x1​∂am,1​x1​+am,2​x2​+⋅⋅⋅+am,n​xn​​∂x2​∂am,1​x1​+am,2​x2​+⋅⋅⋅+am,n​xn​​⋅⋅⋅∂xn​∂am,1​x1​+am,2​x2​+⋅⋅⋅+am,n​xn​​​ ​
= [ a 1 , 1 a 2 , 1 ⋅ ⋅ ⋅ a m , 1 a 1 , 2 a 2 , 2 ⋅ ⋅ ⋅ a m , 2 ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ a 1 , n a 2 , n ⋅ ⋅ ⋅ a m , n ] \begin{bmatrix} a_{1,1} & a_{2,1} & ··· & a_{m,1}\\ a_{1,2} & a_{2,2} & ··· & a_{m,2} \\ ···&···&···&··· \\ a_{1,n}&a_{2,n}&···&a_{m,n} \end{bmatrix} ​a1,1​a1,2​⋅⋅⋅a1,n​​a2,1​a2,2​⋅⋅⋅a2,n​​⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅​am,1​am,2​⋅⋅⋅am,n​​ ​=$A^{\top }$

二、对于所有$A\in {\mathbb{R}}^{n\times m}$，都有${\nabla }_x{x}^{\top }A=A$；

证明：设$A_{(n,m)}$= [ a 1 , 1 a 1 , 2 ⋅ ⋅ ⋅ a 1 , m a 2 , 1 a 2 , 2 ⋅ ⋅ ⋅ a 2 , m ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ a n , 1 a n , 2 ⋅ ⋅ ⋅ a n , m ] \begin{bmatrix} a_{1,1}&a_{1,2}&···&a_{1,m} \\ a_{2,1}&a_{2,2}&···&a_{2,m} \\ ··· & ··· & ··· & ··· \\ a_{n,1} & a_{n,2} &···&a_{n,m} \end{bmatrix} ​a1,1​a2,1​⋅⋅⋅an,1​​a1,2​a2,2​⋅⋅⋅an,2​​⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅​a1,m​a2,m​⋅⋅⋅an,m​​ ​，
则$x^{\top }A$=
$\left[\begin{array}{cccc}{a}_{1,1}{x}_1+a_{2,1}{x}_2+\cdot \cdot \cdot +a_{n,1}{x}_n& a_{1,2}{x}_1+a_{2,2}{x}_2+\cdot \cdot \cdot +a_{n,2}{x}_n& \cdot \cdot \cdot & a_{1,m}{x}_1+a_{2,m}{x}_2+\cdot \cdot \cdot +a_{n,m}{x}_n\end{array}\right]$,
${\nabla }_x{x}^{\top }A$= [ ∂ x ⃗ ⊤ A ∂ x 1 ∂ x ⃗ ⊤ A ∂ x 2 ⋅ ⋅ ⋅ ∂ x ⃗ ⊤ A ∂ x n ] \begin{bmatrix}\frac{\partial \vec x^\top A}{\partial x_1}\\\frac{\partial \vec x^\top A}{\partial x_2}\\···\\ \frac{\partial \vec x^\top A}{\partial x_n}\end{bmatrix} ​∂x1​∂x ⊤A​∂x2​∂x ⊤A​⋅⋅⋅∂xn​∂x ⊤A​​ ​
= [ ∂ a 1 , 1 x 1 + a 2 , 1 x 2 + ⋅ ⋅ ⋅ + a n , 1 x n ∂ x 1 ∂ a 1 , 2 x 1 + a 2 , 2 x 2 + ⋅ ⋅ ⋅ + a n , 2 x n ∂ x 1 ⋅ ⋅ ⋅ ∂ a 1 , m x 1 + a 2 , m x 2 + ⋅ ⋅ ⋅ + a n , m x n ∂ x 1 ∂ a 1 , 1 x 1 + a 2 , 1 x 2 + ⋅ ⋅ ⋅ + a n , 1 x n ∂ x 2 ∂ a 1 , 2 x 1 + a 2 , 2 x 2 + ⋅ ⋅ ⋅ + a n , 2 x n ∂ x 2 ⋅ ⋅ ⋅ ∂ a 1 , m x 1 + a 2 , m x 2 + ⋅ ⋅ ⋅ + a n , m x n ∂ x 2 ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ∂ a 1 , 1 x 1 + a 2 , 1 x 2 + ⋅ ⋅ ⋅ + a n , 1 x n ∂ x n ∂ a 1 , 2 x 1 + a 2 , 2 x 2 + ⋅ ⋅ ⋅ + a n , 2 x n ∂ x n ⋅ ⋅ ⋅ ∂ a 1 , m x 1 + a 2 , m x 2 + ⋅ ⋅ ⋅ + a n , m x n ∂ x n ] \begin{bmatrix}\frac{\partial a_{1,1}x_1+a_{2,1}x_2+···+a_{n,1}x_n}{\partial x_1}& \frac{\partial a_{1,2}x_1+a_{2,2}x_2+···+a_{n,2}x_n}{\partial x_1}&···&\frac{\partial a_{1,m}x_1+a_{2,m}x_2+···+a_{n,m}x_n}{\partial x_1}\\ \frac{\partial a_{1,1}x_1+a_{2,1}x_2+···+a_{n,1}x_n}{\partial x_2}& \frac{\partial a_{1,2}x_1+a_{2,2}x_2+···+a_{n,2}x_n}{\partial x_2}&···&\frac{\partial a_{1,m}x_1+a_{2,m}x_2+···+a_{n,m}x_n}{\partial x_2}\\ ···&···&···&···\\ \frac{\partial a_{1,1}x_1+a_{2,1}x_2+···+a_{n,1}x_n}{\partial x_n}& \frac{\partial a_{1,2}x_1+a_{2,2}x_2+···+a_{n,2}x_n}{\partial x_n}&···&\frac{\partial a_{1,m}x_1+a_{2,m}x_2+···+a_{n,m}x_n}{\partial x_n}\end{bmatrix} ​∂x1​∂a1,1​x1​+a2,1​x2​+⋅⋅⋅+an,1​xn​​∂x2​∂a1,1​x1​+a2,1​x2​+⋅⋅⋅+an,1​xn​​⋅⋅⋅∂xn​∂a1,1​x1​+a2,1​x2​+⋅⋅⋅+an,1​xn​​​∂x1​∂a1,2​x1​+a2,2​x2​+⋅⋅⋅+an,2​xn​​∂x2​∂a1,2​x1​+a2,2​x2​+⋅⋅⋅+an,2​xn​​⋅⋅⋅∂xn​∂a1,2​x1​+a2,2​x2​+⋅⋅⋅+an,2​xn​​​⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅​∂x1​∂a1,m​x1​+a2,m​x2​+⋅⋅⋅+an,m​xn​​∂x2​∂a1,m​x1​+a2,m​x2​+⋅⋅⋅+an,m​xn​​⋅⋅⋅∂xn​∂a1,m​x1​+a2,m​x2​+⋅⋅⋅+an,m​xn​​​ ​
= [ a 1 , 1 a 1 , 2 ⋅ ⋅ ⋅ a 1 , m a 2 , 1 a 2 , 2 ⋅ ⋅ ⋅ a 2 , m ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ a n , 1 a n , 2 ⋅ ⋅ ⋅ a n , m ] \begin{bmatrix} a_{1,1} & a_{1,2}&···&a_{1,m}\\ a_{2,1}&a_{2,2}&···&a_{2,m} \\ ···&···&···&···\\ a_{n,1}&a_{n,2}&···&a_{n,m} \end{bmatrix} ​a1,1​a2,1​⋅⋅⋅an,1​​a1,2​a2,2​⋅⋅⋅an,2​​⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅​a1,m​a2,m​⋅⋅⋅an,m​​ ​=$A$

三、对于所有$A\in {\mathbb{R}}^{n\times n}$，都有${\nabla }_x{x}^{\top }Ax=(A+A^{\top })x$；

证明：设$A_{(n,n)}$= [ a 1 , 1 a 1 , 2 ⋅ ⋅ ⋅ a 1 , n a 2 , 1 a 2 , 2 ⋅ ⋅ ⋅ a 2 , n ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ a n , 1 a n , 2 ⋅ ⋅ ⋅ a n , n ] \begin{bmatrix} a_{1,1}&a_{1,2}&···&a_{1,n} \\ a_{2,1}&a_{2,2}&···&a_{2,n} \\ ··· & ··· & ··· & ··· \\ a_{n,1} & a_{n,2} &···&a_{n,n} \end{bmatrix} ​a1,1​a2,1​⋅⋅⋅an,1​​a1,2​a2,2​⋅⋅⋅an,2​​⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅​a1,n​a2,n​⋅⋅⋅an,n​​ ​，
则$x^{\top }A$=$\left[\begin{array}{cccc}{a}_{1,1}{x}_1+a_{2,1}{x}_2+\cdot \cdot \cdot +a_{n,1}{x}_n& a_{1,2}{x}_1+a_{2,2}{x}_2+\cdot \cdot \cdot +a_{n,2}{x}_n& \cdot \cdot \cdot & a_{1,n}{x}_1+a_{2,n}{x}_2+\cdot \cdot \cdot +a_{n,n}{x}_n\end{array}\right]$,
$x^{\top }Ax$=$\left[\begin{array}{c}\sum _{i=1}^n\sum _{j=1}^n(a_{i,j}{x}_i{x}_j)\end{array}\right]$,
${\nabla }_x{x}^{\top }Ax$= [ ∂ ∑ i = 1 n ∑ j = 1 n ( a i , j x i x j ) ∂ x 1 ∂ ∑ i = 1 n ∑ j = 1 n ( a i , j x i x j ) ∂ x 2 ⋅ ⋅ ⋅ ∂ ∑ i = 1 n ∑ j = 1 n ( a i , j x i x j ) ∂ x n ] \begin{bmatrix} \frac{\partial \sum\limits_{i=1}^{n}\sum\limits_{j=1}^{n} (a_{i,j}x_ix_j)}{\partial x_1} \\ \frac{\partial \sum\limits_{i=1}^{n}\sum\limits_{j=1}^{n} (a_{i,j}x_ix_j)}{\partial x_2} \\ ···\\ \frac{\partial \sum\limits_{i=1}^{n}\sum\limits_{j=1}^{n} (a_{i,j}x_ix_j)}{\partial x_n} \end{bmatrix} ​∂x1​∂i=1∑n​j=1∑n​(ai,j​xi​xj​)​∂x2​∂i=1∑n​j=1∑n​(ai,j​xi​xj​)​⋅⋅⋅∂xn​∂i=1∑n​j=1∑n​(ai,j​xi​xj​)​​ ​= [ ∑ i = 1 n ( a i , 1 + a 1 , i ) x i ∑ i = 1 n ( a i , 2 + a 2 , i ) x i ⋅ ⋅ ⋅ ∑ i = 1 n ( a i , n + a n , i ) x i ] \begin{bmatrix} \sum\limits_{i=1}^{n}(a_{i,1}+a_{1,i})x_i \\ \sum\limits_{i=1}^{n}(a_{i,2}+a_{2,i})x_i \\ ···\\ \sum\limits_{i=1}^{n}(a_{i,n}+a_{n,i})x_i \\ \end{bmatrix} ​i=1∑n​(ai,1​+a1,i​)xi​i=1∑n​(ai,2​+a2,i​)xi​⋅⋅⋅i=1∑n​(ai,n​+an,i​)xi​​ ​
= [ 2 a 1 , 1 a 1 , 2 + a 2 , 1 ⋅ ⋅ ⋅ a 1 , n + a n , 1 a 2 , 1 + a 1 , 2 2 a 2 , 2 ⋅ ⋅ ⋅ a 2 , n + a n , 2 ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ a n , 1 + a 1 , n a n , 2 + a 2 , n ⋅ ⋅ ⋅ 2 a n , n ] [ x 1 x 2 ⋅ ⋅ ⋅ x n ] \begin{bmatrix} 2a_{1,1} & a_{1,2}+a_{2,1} & ···&a_{1,n}+a_{n,1} \\ a_{2,1}+a_{1,2} & 2a_{2,2} & ···&a_{2,n}+a_{n,2} \\ ···&···&···&···\\ a_{n,1}+a_{1,n} & a_{n,2}+a_{2,n} & ···&2a_{n,n} \\ \end{bmatrix} \begin{bmatrix} x_1\\ x_2\\ ···\\ x_n \end{bmatrix} ​2a1,1​a2,1​+a1,2​⋅⋅⋅an,1​+a1,n​​a1,2​+a2,1​2a2,2​⋅⋅⋅an,2​+a2,n​​⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅​a1,n​+an,1​a2,n​+an,2​⋅⋅⋅2an,n​​ ​ ​x1​x2​⋅⋅⋅xn​​ ​=$(A+A^{\top })x$

四、${\nabla }_x\Vert x{\Vert }^2={\nabla }_x{x}^{\top }x=2x$。

证明：${\nabla }_x\Vert x{\Vert }^2$=${\nabla }_x{\sqrt{x_1^2+x_2^2+\cdot \cdot \cdot +x_n^n}}^2$=${\nabla }_x{x}_1^2+x_2^2+\cdot \cdot \cdot +x_n^n$=${\nabla }_x{x}^{\top }x$；
${\nabla }_x\Vert x{\Vert }^2$=${\nabla }_x{\sqrt{x_1^2+x_2^2+\cdot \cdot \cdot +x_n^n}}^2$=${\nabla }_x{x}_1^2+x_2^2+\cdot \cdot \cdot +x_n^n$= [ 2 x 1 2 x 2 ⋅ ⋅ ⋅ 2 x n ] \begin{bmatrix} 2x_1\\ 2x_2\\ ···\\ 2x_n \end{bmatrix} ​2x1​2x2​⋅⋅⋅2xn​​ ​=$2x$

  同样，对于任何矩阵$X$，都有${\nabla }_X\Vert X{\Vert }_F^2=2X$。正如我们之后将看到的，梯度对于设计深度学习中的优化算法有很大用处。

五、对于任何矩阵$X$，都有${\nabla }_X\Vert X{\Vert }_F^2=2X$

证明：设$X$为$m\times n$的矩阵， X = [ x 1 , 1 x 1 , 2 ⋅ ⋅ ⋅ x 1 , n x 2 , 1 x 2 , 2 ⋅ ⋅ ⋅ x 2 , n ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ x m , 1 x m , 2 ⋅ ⋅ ⋅ x m , n ] X = \begin{bmatrix} x_{1,1}& x_{1,2}&···&x_{1,n}\\ x_{2,1}& x_{2,2}&···&x_{2,n}\\ ···&···&···&···\\ x_{m,1}& x_{m,2}&···&x_{m,n}\\ \end{bmatrix} X= ​x1,1​x2,1​⋅⋅⋅xm,1​​x1,2​x2,2​⋅⋅⋅xm,2​​⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅​x1,n​x2,n​⋅⋅⋅xm,n​​ ​，
则$\Vert X{\Vert }_F^2$=${\sqrt{\sum _{i=1}^m\sum _{j=1}^n{x}_{i,j}^2}}^2$=$\sum _{i=1}^m\sum _{j=1}^n{x}_{i,j}^2$，
${\nabla }_X\Vert X{\Vert }_F^2$= [ 2 x 1 , 1 2 x 1 , 2 ⋅ ⋅ ⋅ 2 x 1 , n 2 x 2 , 1 2 x 2 , 2 ⋅ ⋅ ⋅ 2 x 2 , n ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ 2 x m , 1 2 x m , 2 ⋅ ⋅ ⋅ 2 x m , n ] \begin{bmatrix} 2x_{1,1}& 2x_{1,2}&···&2x_{1,n}\\ 2x_{2,1}& 2x_{2,2}&···&2x_{2,n}\\ ···&···&···&···\\ 2x_{m,1}& 2x_{m,2}&···&2x_{m,n}\\ \end{bmatrix} ​2x1,1​2x2,1​⋅⋅⋅2xm,1​​2x1,2​2x2,2​⋅⋅⋅2xm,2​​⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅​2x1,n​2x2,n​⋅⋅⋅2xm,n​​ ​=$2X$

初看公式时没看懂，所以自己推了一遍加深印象，以上内容为推导过程，有问题欢迎讨论