A为$m \times n$的矩阵，x为$n \times 1$的矩阵，则Ax为$m \times 1$的列向量，记$\vec{y}=A.\vec{x}$
令，
$
A=
\left[
\begin{matrix}
a_{11} & a_{12} & \cdots & a_{1n} \\
a_{21} & a_{22} & \cdots & a_{2n} \\
\vdots & \vdots & \ddots & \vdots \\
a_{m1} & a_{m2} & \cdots & a_{mn}
\end{matrix}
\right]
$
$
\vec{x}=
\left[
\begin{matrix}
x_{1} \\
x_{2} \\
\vdots \\
x_{n} \\
\end{matrix}
\right]
A.\vec{x}=
\left[
\begin{matrix}
a_{11}x_{1}+a_{12}x_{2}+…+a_{1n}x_{n} \\
a_{21}x_{1}+a_{22}x_{2}+…+a_{2n}x_{n} \\
\vdots \\
a_{m1}x_{1}+a_{m2}x_{2}+…+a_{mn}x_{n}
\end{matrix}
\right]
$
因此有，
$
\frac{\partial{\vec{y}}}{\partial{x}}=\frac{\partial{A\vec{x}}}{\partial{x}}
=\left[
\begin{matrix}
a_{11} & a_{21} & \cdots & a_{m1} \\
a_{12} & a_{22} & \cdots & a_{m2} \\
\vdots & \vdots & \ddots & \vdots \\
a_{1n} & a_{2n} & \cdots & a_{mn}
\
\end{matrix}
\right]
=A^{T}
$
进一步推广可得：
$
\frac{\partial{A\vec{x}}}{\partial{\vec{x}}}=A^{T} \\
\frac{\partial{A^{T}\vec{x}}}{\partial{\vec{x}}}=A \\
\frac{\partial{A\vec{x}}}{\partial{\vec{x^{T}}}}=A
$
