Let $f: G= \mbox{GL}(n,\mathbb{R}) \to \mathbb{R}$ be the determinant function.
Then $\mbox{Hess} (f)$ is a two linear map on $M\_{n}(\mathbb{R})\simeq T\_{e}(G)$ where $e$ is the neutral element of $G$, the identity matrix.
What is an explicit formula for this Hessian? (In terms of matrix terminologies)



Is there an explicit formula for the hessian of "Determinant"?
The formula you're looking for can be obtained by differentiating [Jacobi's formula](https://en.wikipedia.org/wiki/Jacobi%27s_formula)
$$ \frac{\mathrm{d}}{\mathrm{d}t} \det A(t) = \det A(t) \cdot \operatorname{tr}\left( A^{-1} \frac{\mathrm{d}A}{\mathrm{d}t} \right) $$
with respect to a second parameter, say $s$:
\begin{multline}
\frac{\partial^2}{\partial s \partial t} \det A(s,t) =
\det A(s,t) \cdot \bigg[
\operatorname{tr}\left( A^{-1} \frac{\partial A}{\partial s} \right) \operatorname{tr}\left( A^{-1} \frac{\partial A}{\partial t} \right) \\
+ \operatorname{tr}\left( A^{-1} \frac{\partial^2 A}{\partial s \partial t} \right)
- \operatorname{tr}\left( A^{-1} \frac{\partial A}{\partial s} A^{-1} \frac{\partial A}{\partial t} \right)
\bigg]
\end{multline}
Now take $s = A\_{ij}$ and $t = A\_{kl}$, so $\frac{\partial A}{\partial s} = E\_{ij}$ is the matrix with a one in its $(i,j)$-entry, and zeros elsewhere. Similarly $\frac{\partial A}{\partial t} = E\_{kl}$, and $\frac{\partial^2 A}{\partial s \partial t} = 0$. The desired Hessian is then
\begin{align\*}
\operatorname{Hess}(\det)\_A(U,V) &= U\_{ij} V\_{kl} (\det A)\bigg[
\operatorname{tr}\left( A^{-1} E\_{ij} \right)
\operatorname{tr}\left( A^{-1} E\_{kl} \right) - \operatorname{tr}\left( A^{-1} E\_{ij} A^{-1} E\_{kl} \right)
\bigg] \\
&= U\_{ij} V\_{kl} (\det A)\bigg[
(A^{-1})\_{mn} (E\_{ij})\_{nm} (A^{-1})\_{pq} (E\_{kl})\_{qp} \\
&\hspace{4cm} - (A^{-1})\_{mn} (E\_{ij})\_{np} (A^{-1})\_{pq} (E\_{kl})\_{qm}
\bigg] \\
&= U\_{ij} V\_{kl} (\det A)\bigg[
(A^{-1})\_{mn} \delta\_{in} \delta\_{jm} (A^{-1})\_{pq} \delta\_{kq} \delta\_{lp} \\
&\hspace{4cm} - (A^{-1})\_{mn} \delta\_{in} \delta\_{jp} (A^{-1})\_{pq} \delta\_{kq} \delta\_{lm}
\bigg] \\
&= U\_{ij} V\_{kl} (\det A)\bigg[
(A^{-1})\_{ji} (A^{-1})\_{lk} - (A^{-1})\_{li} (A^{-1})\_{jk}
\bigg] \\
&= \det A \bigg[
U\_{ij} (A^{-1})\_{ji} V\_{kl} (A^{-1})\_{lk} - U\_{ij} (A^{-1})\_{jk} V\_{kl} (A^{-1})\_{li}
\bigg] \\
&= \det A \bigg[
\operatorname{tr}(U A^{-1}) \operatorname{tr}(V A^{-1}) - \operatorname{tr}(U A^{-1} V A^{-1})
\bigg] \\
\end{align\*}
with the Einstein summation convention in full force throughout. By evaluating this formula at $A = e,$ the $n \times n $ identity matrix, we obtain the desired special case
$$ \operatorname{Hess}(\det)\_e(U,V) = \operatorname{tr}(U) \operatorname{tr}(V) - \operatorname{tr}(U V). $$


