逆矩阵的导数

hbghlyj

矩阵$M=(m_{ij})$对θ的导数定义为(对各个分量求导)$$\frac{\partial \mathbf{M}}{\partial \theta}=\left(\begin{array}{cccc}\frac{\partial m_{11}}{\partial \theta} & \frac{\partial m_{12}}{\partial \theta} & \ldots & \frac{\partial m_{1 d}}{\partial \theta} \\ \frac{\partial m_{21}}{\partial \theta} & \frac{\partial m_{22}}{\partial \theta} & \ldots & \frac{\partial m_{2 d}}{\partial \theta} \\ \vdots & \vdots & \ddots & \vdots \\ \frac{\partial m_{n 1}}{\partial \theta} & \frac{\partial m_{n 2}}{\partial \theta} & \ldots & \frac{\partial m_{n d}}{\partial \theta}\end{array}\right)$$求证$$\frac{\partial}{\partial \theta} \mathbf{M}^{-1}=-\mathbf{M}^{-1} \frac{\partial \mathbf{M}}{\partial \theta} \mathbf{M}^{-1}$$证明(除了矩阵乘法不满足交换律之外,和$(f(x)^{-1})'=-f'(x)·f(x)^{-2}$的证明没有什么差别)
$$
0=I'=({\bf M}{\bf M}^{-1})'={\bf M}'{\bf M}^{-1}+{\bf M}({\bf M}^{-1})'.
$$
将$({\bf M}^{-1})'$解出,
$$
{\bf M}({\bf M}^{-1})'=-{\bf M}'{\bf M}^{-1}\implies({\bf M}^{-1})'=-{\bf M}^{-1}{\bf M}'{\bf M}^{-1}.
$$

hbghlyj

如果我们有一个 $d$ 维向量 $\mathbf{x}$ 的 $n$ 维向量值函数 $\mathbf{f}$，导数可表示为雅可比矩阵
$$
\mathbf{J}(\mathbf{x})=\frac{\partial \mathbf{f}(\mathbf{x})}{\partial \mathbf{x}}=\left(\begin{array}{ccc}
\frac{\partial f_{1}(\mathbf{x})}{\partial x_{1}} & \ldots & \frac{\partial f_{1}(\mathbf{x})}{\partial x_{d}} \\
\vdots & \ddots & \vdots \\
\frac{\partial f_{n}(\mathbf{x})}{\partial x_{1}} & \ldots & \frac{\partial f_{n}(\mathbf{x})}{\partial x_{d}}
\end{array}\right)
$$考虑一个矩阵 $\bf M$ 和一个与向量 $\bf x$ 无关的向量 $\bf y$。可以通过写出分量来验证此类矩阵的以下向量导数恒等式：
$$
\begin{aligned}
\frac{\partial}{\partial \mathbf{x}}[\mathbf{M x}] &=\mathbf{M} \\
\frac{\partial}{\partial \mathbf{x}}\left[\mathbf{y}^{t} \mathbf{x}\right] &=\frac{\partial}{\partial \mathbf{x}}\left[\mathbf{x}^{t} \mathbf{y}\right]=\mathbf{y} \\
\frac{\partial}{\partial \mathbf{x}}\left[\mathbf{x}^{t} \mathbf{M x}\right] &=\left[\mathbf{M}+\mathbf{M}^{t}\right] \mathbf{x}
\end{aligned}
$$