# Have derivatives of determinants along 1-psg's ever been 'coherently' computed via Jacobi's formula?

Suppose $\mathfrak{p}$ denotes all the symmetric matrices in $\mathfrak{sl}_{2n} \mathbb{R}$.

Then for each parameterized 1-dimensional linear subspace $\xi=\xi(t)$ of $\mathfrak{p}$ we get a 1-parameter subgroup $e^{\xi(t)}$ in $SL_{2n}\mathbb{R}$.

Now let us take some collection of $n$ linearly independant vectors $x_1, \ldots, x_n$ in $\mathbb{R}^{2n}$, and let $X$ be the $(2n \times n)$ matrix with $x_i$'s as columns.

Setting $\Xi(t):={}^tX e^{\xi(t)}X$ consider now the following problem: along which lines $\xi$ in $\mathfrak{p}$ is the function $\tilde{\Xi}(t):=det(\Xi(t))$ increasing?

There is at least one obvious strategy: we shall determine those $\xi$ for which the differential $d \tilde {\Xi}$ is positive.

As one tool, there is the so-called Jacobi formula. This very general formula tells us $$d \tilde{\Xi}=tr(adj~( \Xi(t))~~ d \Xi).$$

So in one sense, the Jacobi formula 'computes' our derivative. However it does nothing for us (it is a 'shallow' formula). It does nothing because I imagine there is no individual in the history of the world who could describe what the 'adjugate' of a matrix 'is' (a definition, by itself, gives no images).

But I would like to be wrong on this final point. And so my own question: can anybody refer me to an instance in the world/literature/experience where either the differential of a determinant has 'fallen out' or where Jacobi's formula has yielded something tangible?

• The adjugate of an $n \times n$ matrix is its action on the second-to-top exterior power. It tells you how the matrix acts on oriented $(n-1)$-dimensional volume elements. See, for example, mathoverflow.net/questions/89069/… . Jul 16, 2012 at 0:38
• Hm, i'd never seen such an explanation of the adjugate before. It means the Laplace expansion is an expression of n-volume as linear combination of (n-1)-volumes. But this interpretation of the adjugate convinces me even further of the futility of Jacobi's formula (which is really nothing more than an arbitrary Laplace expansion): 'Knowing' the adjugate demands more than the determinant. My question is more about computing the derivative based on knowledge of $e^\xi$'s eigenspaces and the configuration of $x_1, \ldots, x_n$.
– JHM
Jul 16, 2012 at 1:57
• $adj(A) = det(A) . A^{-1}$ on the open dense set of invertible matrices. This can help to make sense of computations. Oct 2, 2012 at 13:44

The formula $d(\det A) = (\det A)\operatorname{tr} A^{-1}dA$ or equivalently $d(\log\det A) = \operatorname{tr}A^{-1}dA$ is extremely useful. I'm surprised there aren't more answers to this question. A simple but very important example is the Bishop-Gromov inequality in Riemannian geometry.