TITLE: Second Derivative of a Determinant
QUESTION [12 upvotes]: How does one evaluate the second derivative of the determinant of a square matrix?
Jacobi's formula tells us how to evaluate the first derivative but I can't find anything for the second. This is my attempt:

We can start using the partial derivative formulation of Jacobi's formula, assuming A is invertible:
\begin{equation}\frac{\partial}{\partial \alpha}\det A = (\det A) \text{tr}\left( A^{-1} \frac{\partial}{\partial \alpha} A \right)\end{equation}
        Taking a second derivative:
        \begin{equation}\frac{\partial^2}{\partial \alpha^2}\det A = \frac{\partial}{\partial \alpha}\left[(\det A) \text{tr}\left( A^{-1} \frac{\partial}{\partial \alpha} A \right)\right]\end{equation}
        Applying product rule:
        \begin{equation}= \frac{\partial}{\partial \alpha}(\det A) \cdot \text{tr}\left( A^{-1} \frac{\partial}{\partial \alpha} A \right) +
    (\det A) \frac{\partial}{\partial \alpha} \text{tr}\left( A^{-1} \frac{\partial}{\partial \alpha} A \right)\end{equation}
Replacing $\frac{\partial}{\partial\alpha}\det A$ with Jacobi's formula:
        \begin{equation}= (\det A) \text{tr}\left( A^{-1} \frac{\partial}{\partial \alpha} A \right) \cdot \text{tr}\left( A^{-1} \frac{\partial}{\partial \alpha} A \right) +(\det A) \text{tr}\left( \frac{\partial}{\partial \alpha}\left(A^{-1} \frac{\partial}{\partial \alpha} A \right)\right)\end{equation}
        Factoring out $\det(A)$:
        \begin{equation}= (\det A) \left[\text{tr}^2\left( A^{-1} \frac{\partial}{\partial \alpha} A \right) +
     \text{tr}\left( \frac{\partial}{\partial \alpha}\left(A^{-1} \frac{\partial}{\partial \alpha} A \right)\right)\right]\end{equation}
         Another product rule:
        \begin{equation}= (\det A) \left[\text{tr}^2\left( A^{-1} \frac{\partial}{\partial \alpha} A \right) +
     \text{tr}\left(\frac{\partial}{\partial \alpha}A^{-1} \frac{\partial}{\partial \alpha} A +
     A^{-1} \frac{\partial^2}{\partial \alpha^2} A \right)\right]\end{equation}
         Finally, using $A_{\alpha}$ to denote the partial of A wrt to $\alpha$ we have
        \begin{equation}\frac{\partial^2}{\partial \alpha^2}\det A= \det(A) \left[\text{tr}^2\left( A^{-1} A_{\alpha} \right) +
     \text{tr}\left(A^{-1}_{\alpha} A_{\alpha}\right) + \text{tr} \left( A^{-1} A_{\alpha^2} \right)\right]\end{equation}
         The second trace actually reduces to N, for an NxN matrix:
         \begin{equation}\text{tr}\left(A^{-1}_{\alpha} A_{\alpha}\right)=\text{tr}(I)\end{equation}
         \begin{equation}=\sum_{diagonals}1\end{equation}
         \begin{equation}=N\end{equation}
\begin{equation}\therefore\frac{\partial^2}{\partial \alpha^2}\det A= \det(A) \left[\text{tr}^2\left( A^{-1} A_{\alpha} \right) +
     \text{tr} \left( A^{-1} A_{\alpha^2} \right)+N\right]\end{equation}

I'm not overly confident with the matrix calculus, but the result is nice enough to seem plausible. Is there another method, or is this proof valid? Thanks!

REPLY [2 votes]: As pointed out by Carl, your mistake is to permute the inverse and derivative operator:
$$\partial_\alpha(A^{-1}) \neq (\partial_\alpha A)^{-1} \, .$$
You can easily generalize it for whatever two differentiating arguments ($\alpha$ and $\beta$). Starting from Jacobi's formula:
$$ \partial_{\alpha}(\mathrm{det}(A)) = \mathrm{det}(A) \, \mathrm{tr}\left(A^{-1} \, \partial_\alpha A\right) \, $$
applying on it the chain rule:
$$ \partial_{\alpha\beta}(\mathrm{det}(A)) = \partial_\beta(\mathrm{det}(A)) \, \mathrm{tr}\left(A^{-1} \, \partial_\alpha A\right) + \mathrm{det}(A) \, \mathrm{tr}\left(\partial_\beta \left(A^{-1} \, \partial_\alpha A\right)\right) \, ,$$
using Jacobi's formula on the first $\beta$ derivative and applying the chain rule on the second:
$$ \partial_{\alpha\beta}(\mathrm{det}(A)) = \mathrm{det}(A) \, \mathrm{tr}\left(A^{-1} \, \partial_\beta A\right)\,\mathrm{tr}\left(A^{-1} \, \partial_\alpha A\right) + \mathrm{det}(A) \, \mathrm{tr}\left(\partial_\beta \left(A^{-1}\right) \, \partial_\alpha A\right)  + \mathrm{det}(A) \, \mathrm{tr}\left(A^{-1} \, \partial_{\alpha\beta} A\right)\, ,$$
and finally, factoring out the determinant and applying the relation for the inverse $\partial_\beta(A^{-1}) = -A^{-1} \, \partial_\beta A \, A^{-1}$:
$$ \partial_{\alpha\beta}(\mathrm{det}(A)) = \mathrm{det}(A) \, \left[\mathrm{tr}\left(A^{-1} \, \partial_\beta A\right)\,\mathrm{tr}\left(A^{-1} \, \partial_\alpha A\right) - \mathrm{tr}\left(A^{-1} \, \partial_\beta A \, A^{-1} \, \partial_\alpha A\right) + \mathrm{tr}\left(A^{-1} \, \partial_{\alpha\beta} A\right)\right] \, , $$
which is the same as what Carl gave for $\beta=\alpha$:
$$ \partial_{\alpha\alpha}(\mathrm{det}(A)) = \mathrm{det}(A) \, \left[\left(\mathrm{tr}\left(A^{-1} \, \partial_\alpha A\right)\right)^2 - \mathrm{tr}\left(A^{-1} \, \partial_\alpha A \, A^{-1} \, \partial_\alpha A\right) + \mathrm{tr}\left(A^{-1} \, \partial_{\alpha\alpha} A\right)\right] \, . $$