行列式の微分

$ \mathrm d\det\pmb A=(\det\pmb A){{\pmb A}^\top}^{-1}:\mathrm d\pmb A

2次元のとき

$ \mathrm d\det\pmb A=\mathrm d(A_{00}A_{11}-A_{01}A_{10})

$ =(\mathrm d A_{00})A_{11}+A_{00}\mathrm dA_{11}-(\mathrm d A_{01})A_{10}-A_{01}\mathrm dA_{10}

$ = \begin{pmatrix}A_{11}&-A_{10}\\-A_{01}&A_{00}\end{pmatrix}:\mathrm d\begin{pmatrix}A_{00}&A_{01}\\A_{10}&A_{11}\end{pmatrix}

$ = (\det\pmb A){{\pmb A}^\top}^{-1}:\mathrm d\pmb A

一般の導出

固有値を使った方法

$ \mathrm d\det\pmb A=\det(\pmb A+\mathrm d\pmb A)-\det\pmb A

$ =\det(\pmb A\cdot(\pmb I+{\pmb A}^{-1}\cdot\mathrm d\pmb A))-\det\pmb A

$ = (\det\pmb A)(\det(\pmb I+\pmb A^{-1}\cdot\mathrm d\pmb A)-1)

$ = (\det\pmb A)(\det(\pmb I+\pmb A^{-1}\cdot(\pmb 0+\mathrm d\pmb A))-\det(\pmb I+\pmb A^{-1}\cdot\pmb0))

$ = (\det\pmb A)\left.\frac{\mathrm d}{\mathrm d\epsilon}\det(\pmb I+\pmb A^{-1}\cdot\epsilon d\pmb A)\right|_{\epsilon=0}

$ = (\det\pmb A)\left.\frac{\mathrm d}{\mathrm d\epsilon}\left(\prod_i\left(1+\epsilon\lambda_i^{{\pmb A}^{-1}\cdot\mathrm d\pmb A}\right)-1\right)\right|_{\epsilon=0}

$ = (\det\pmb A)\left.\sum_j\lambda_j^{{\pmb A}^{-1}\cdot\mathrm d\pmb A}\prod_{i\neq j}\left(1+\epsilon\lambda_i^{{\pmb A}^{-1}\cdot\mathrm d\pmb A}\right)\right|_{\epsilon=0}

$ = (\det\pmb A)\sum_j\lambda_j^{{\pmb A}^{-1}\cdot\mathrm d\pmb A}

$ = (\det\pmb A)\mathrm{tr}\left({\pmb A}^{-1}\cdot\mathrm d\pmb A\right)

$ \underline{=(\det\pmb A){{\pmb A}^\top}^{-1}:\mathrm d\pmb A\quad}_\blacksquare