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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?

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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,… . – Qiaochu Yuan Jul 16 '12 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 $x_1, \ldots, x_n$. – J. Martel Jul 16 '12 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. – Peter Michor Oct 2 '12 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.

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I wish I knew what you had in mind-- I am unfamiliar with this B-G inequality. The determinant, in the above phrasing of the question, represented for me some parallelepiped and I was interested in finding deformations which increased its volume. Presently I'm more interested in the question of having two transverse equal volume parallelepipeds, and finding deformations (or directions) through which the two parallelepipeds still have equal volumes (but possibly different from their initial common volumes!). – J. Martel Oct 17 '12 at 2:31
That's not what your question asks. If you really want know something about parallelepiped, then please edit your question and ask the real question you want answered. – Deane Yang Oct 17 '12 at 3:16
I find your comment hostile and unpleasant. The 'real question' I want answered would most likely be immediately closed as ''too local'' or something. And besides, I'd rather answer it myself. So I think of a related question which might actually interest or attract answers from other mathematicians. I only mentioned parallelepipeds in hopes of possibly stimulating some further discussion with you on the question, eg. maybe you could have directed me to a further reference. I'm aware what the question does and doesn't ask. Despite your unfriendly remark, I still appreciate your BG exampl – J. Martel Oct 17 '12 at 4:15
Sorry. I guess I overreacted to this: "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)." It just seemed a bit grandiose for a student to say. – Deane Yang Oct 17 '12 at 11:38

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