矩陣的微商

關於矩陣微商的定義都是建立在一個矩陣對一個標量微商的定義之上.

設$\mathbf{A}$和$\mathbf{B}$是$n\times m$矩陣，$\mathbf{C}$是$m\times k$矩陣， $\beta$是標量.

$\mathbf{A}$和$\beta$的偏微商定義為
\begin{equation}\label{eq1}
\frac{\partial \mathbf{A}}{\partial \beta}=\left[\frac{\partial a_{ij}}{\partial \beta}\right],
\end{equation}
由定義易知
\begin{equation}\label{eq2}
\frac{\partial }{\partial \beta}(\mathbf A+\mathbf B)=\frac{\partial \mathbf A}{\partial \beta}+\frac{\partial\mathbf B}{\partial \beta},
\end{equation}

\begin{equation}\label{eq3}
\frac{\partial}{\partial\beta}({\mathbf {AC}})=\frac{\partial\mathbf A}{\partial \beta}\mathbf C+\mathbf A\frac{\partial \mathbf C}{\partial \beta}.
\end{equation}

設$\mathbf A$為$n$階可逆方陣， $\mathbf I$ 為$n$階單位陣.
由$\mathbf {AA}^{-1}=\mathbf I$知
\begin{equation}\label{eq4}
\frac{\partial \mathbf A^{-1}}{\partial\beta}=-\mathbf A^{-1}\frac{\partial \mathbf A}{\partial\beta}\mathbf A^{-1}.
\end{equation}

現在討論對向量的微商.

設$\mathbf x=[x_1,x_2,\dots,x_n]^T$，其中$T$表示轉置，$Q$為一標量，$Q$對$\mathbf{x}$的微商定義為
\begin{equation}\label{eq5}
\frac{\partial Q}{\partial\mathbf x}=
\begin{bmatrix}
\frac{\partial Q}{\partial x_1}\\ \vdots \\ \frac{\partial Q}{\partial x_n}
\end{bmatrix}
\end{equation}

設$\mathbf A$是$n\times n$數量矩陣，$Q=\mathbf x^T \mathbf A \mathbf x$，考慮$\dfrac{\partial Q}{\partial\mathbf x}$.

先計算對標量$x_k$的微商.
\begin{align*}
\frac{\partial Q}{\partial x_k}
= \frac{\partial(\mathbf x^T \mathbf A \mathbf x)}{\partial x_k}
=\frac{\partial\mathbf x^T}{\partial x_k}(\mathbf A \mathbf x)+\mathbf x^T \mathbf A\frac{\partial\mathbf x}{\partial x_k}
= \mathbf e_k^T \mathbf A\mathbf x+\mathbf x^T \mathbf A \mathbf e_k ,
\end{align*}
其中
\[
\frac{\partial\mathbf x^T}{\partial x_k}=[0,\cdots,0,1,0,\cdots,0]=\mathbf e_k^T,
\]
注意到$\mathbf x^T \mathbf A \mathbf e_k$是一個標量，所以
\[
\mathbf x^T \mathbf A\mathbf e_k=(\mathbf x^T \mathbf A \mathbf e_k)^T=\mathbf e_k^T \mathbf A^T\mathbf x,
\]
因此
\begin{equation}\label{eq6}
\frac{\partial Q}{\partial x_k}=\mathbf e_k^T(\mathbf A+\mathbf A^T)\mathbf x.
\end{equation}
從而有
\begin{equation}\label{eq7}
\frac{\partial Q}{\partial\mathbf x}=
\begin{bmatrix}
\frac{\partial Q}{\partial x_1}\\ \vdots \\ \frac{\partial Q}{\partial x_n}
\end{bmatrix}=
\begin{bmatrix}
\mathbf e_1^T(\mathbf A+\mathbf A^T)\mathbf x\\ \vdots \\ \mathbf e_n^T(\mathbf A+\mathbf A^T)\mathbf x
\end{bmatrix}=
\mathbf I(\mathbf A+\mathbf A^T)\mathbf x=(\mathbf A+\mathbf A^T)\mathbf x.
\end{equation}

若$\mathbf A$為$n$階對稱方陣，則進一步有
\begin{equation}\label{eq8}
\frac{\partial Q}{\partial \mathbf x}=2\mathbf A\mathbf x.
\end{equation}