Determinant derivatives

Consider a matrix $A$ , we want find a simple way to express derivatives of the its determinant respect to the matrix elements $a_{ij}$ . The determinant can be expanded in the elements $a_{ij}$ :

$$\det{A}=\left\vert A \right\vert=\sum_j{a_{ij}}(-1)^{i+j}C_{ji}$$ (D.1)

where $C_{ij}$ is the minor of the matrix $A$ respect to the element $a_{ij}$ and therefore does not depend explicitly by the elements of the raw $i^{th}$ . So the derivative with respect to $a_{kj}$ will be:

$$\frac{\partial \left\vert A \right\vert}{\partial a_{kj}} = (-1)^{k+j}C_{kl}$$ (D.2)

and for the logarithmic derivative we have:

$$\frac{\partial \ln \left\vert A \right\vert}{\partial a_{kj}} = (-1)^{k+j}C_{kl}\frac{1}{\left\vert A \right\vert}=A^{-1}_{kj}$$ (D.3)

We want to find a simple relation to evaluate second derivatives of the determinant. We write the relation

$$\sum_j a_{ij}A^{-1}_{jk} =\sum_j a_{ij} \frac{1}{\left\vert A \right\vert} \frac{\partial \left\vert A \right\vert}{\partial a_{kj}} = \delta_{ik}$$ (D.4)

if we derive this equation for $a_{ln}$ we obtain:

$$\frac{\partial}{\partial a_{ln}}\left ( \sum_j a_{kj}A^{-1}_{ji} \right) = \sum_j a_{kj}\left ( \frac{-1}{\left\vert A \right\vert^2} \frac{... ...elta_{nj} \frac{1}{A} \frac{\partial \left\vert A \right\vert}{\partial a_{ji}}$$

$$= \sum_j a_{kj}\left ( \frac{-1}{\left\vert A \right\vert^2} \frac{... ...tial a_{lj}} \frac{\partial \left\vert A \right\vert}{\partial a_{ni}} \right )$$

where we substitute the $\delta_{lk}$ with the eq. D.4. Because of this equation is zero for all $a_k$ this means that the expression in parentheses is zero, and this yields to:

$$\frac{1}{\left\vert A \right\vert}\frac{\partial^2 \left\vert A \... ...\partial a_{kn}\partial a_{jm}} = A^{-1}_{nk}A^{-1}_{mj}-A^{-1}_{mk}A^{-1}_{nj}$$ (D.5)