笔记：矩阵行列式求导以及矩阵的逆的求导
一、结论
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A=(a_{ij}(t))_{n\times n}
A=(aij(t))n×n，
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\begin{aligned} &\frac{d |A|}{dt} =|A|tr\left(A^{-1}\frac{dA}{dt}\right)\\ &\frac{d A^{-1}}{dt} = -A^{-1} \frac{d A}{dt} A^{-1} \end{aligned}
dtd∣A∣=∣A∣tr(A−1dtdA)dtdA−1=−A−1dtdAA−1
以下两节给一个简要的推导，可能并不严格。
二、矩阵的行列式求导
根据链式法则，
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\frac{d |A|}{dt} = \sum_{i}\sum_{j}\frac{\partial |A|}{\partial a_{ij}}\frac{d a_{ij}}{dt}
dtd∣A∣=i∑j∑∂aij∂∣A∣dtdaij我们注意到，
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\frac{\partial |A|}{\partial a_{ij}} = \lim_{\varepsilon \rightarrow 0} \frac{|A + \varepsilon I_{ij}| - |A|}{\varepsilon} = \lim_{\varepsilon \rightarrow 0} \frac{\varepsilon A_{ij}}{\varepsilon} = A_{ij}
∂aij∂∣A∣=ε→0limε∣A+εIij∣−∣A∣=ε→0limεεAij=Aij其中，
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Iij为只有
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Aij表示矩阵
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(i,j)处的代数余子式。于是链式法则求得的式子变为如下形式，
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\frac{d |A|}{dt} = \sum_{i}\sum_{j}A_{ij}\frac{d a_{ij}}{dt}
dtd∣A∣=i∑j∑Aijdtdaij又由，
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\begin{aligned} &\frac{da_{ij}}{dt} = \left(\frac{dA}{dt}\right)_{ij} \\ &A^*A = |A|I_n \Rightarrow A^* = |A|A^{-1}\Rightarrow A_{ij} = |A|\left(A^{-1}\right)_{ji} \end{aligned}
dtdaij=(dtdA)ijA∗A=∣A∣In⇒A∗=∣A∣A−1⇒Aij=∣A∣(A−1)ji其中
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A∗为
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A的伴随矩阵，则有，
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\begin{aligned} \frac{d |A|}{dt} &= \sum_{i}\sum_{j}A_{ij}\frac{d a_{ij}}{dt}\\ &=\sum_{j}\sum_{i}|A|\left(A^{-1}\right)_{ji}\left(\frac{dA}{dt}\right)_{ij}\\ &= |A|\sum_{j}\left(A^{-1}\frac{dA}{dt}\right)_{jj}\\ &=|A|tr\left(A^{-1}\frac{dA}{dt}\right) \end{aligned}
dtd∣A∣=i∑j∑Aijdtdaij=j∑i∑∣A∣(A−1)ji(dtdA)ij=∣A∣j∑(A−1dtdA)jj=∣A∣tr(A−1dtdA)
三、矩阵的逆的导数
首先，
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\mathbf{0}_{n\times n} = \frac{\partial I_n}{\partial t} = \frac{\partial AA^{-1}}{\partial t} = \frac{\partial A}{\partial t}A^{-1} + A\frac{\partial A^{-1}}{\partial t}
0n×n=∂t∂In=∂t∂AA−1=∂t∂AA−1+A∂t∂A−1则有，
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\frac{d A^{-1}}{dt} = -A^{-1} \frac{d A}{dt} A^{-1}
dtdA−1=−A−1dtdAA−1
参考资料：