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)

  • 3
    $\begingroup$ I think you can find what you're looking for by differentiating Jacobi's formula. Not sure if the answer will be nicely expressible in terms of standard matrix operations, though. $\endgroup$ – David Zhang Sep 6 '16 at 22:59
  • $\begingroup$ @DavidZhang Thanks for your comment. $\endgroup$ – Ali Taghavi Sep 6 '16 at 23:15

The formula you're looking for can be obtained by differentiating Jacobi's 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 $$ \operatorname{Hess}(\det)_e(U,V) = \operatorname{tr}(U) \operatorname{tr}(V) - \operatorname{tr}(U V). $$

| cite | improve this answer | |
  • $\begingroup$ and for your very elegant formula. $\endgroup$ – Ali Taghavi Sep 7 '16 at 11:06
  • $\begingroup$ Thank you again for your computations and very interesting formula. The formula is invariant under the natural action of $G$ on $\mathfrak{g}\times \mathfrak{g}$.Do you think that $\endgroup$ – Ali Taghavi Sep 12 '16 at 8:00
  • $\begingroup$ ....it is natural in the following sense: If a function is constant on every conjugacy class of a Lie group, then its Hessian at e is invariant under the natural action of $G$ on $\mathfrak{g} \times \mathfrak {g}$? $\endgroup$ – Ali Taghavi Sep 12 '16 at 8:04
  • $\begingroup$ @AliTaghavi I am not sure. Perhaps you can ask this as a separate question. $\endgroup$ – David Zhang Sep 12 '16 at 16:19
  • $\begingroup$ Yes I try to ask it. BTW, your formula define a pseudo Riemannian metric on $Gl(n, \mathbb{R}$, since the two form is non degenerate.Even more: $SO(n)$ is seudo Riemannian submanifold. Is this special structure studied already?What can be said about the signature of this metric?Is $SO(n)$ a complet submanifold?Are you interested in such type of questions? $\endgroup$ – Ali Taghavi Sep 12 '16 at 18:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.