most recent 30 from http://mathoverflow.net 2013-06-20T06:04:29Z http://mathoverflow.net/feeds/question/102321 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/102321/have-derivatives-of-determinants-along-1-psgs-ever-been-coherently-computed-vi J. Martel 2012-07-15T23:53:31Z 2012-10-30T12:22:00Z

<p>Suppose $\mathfrak{p}$ denotes all the symmetric matrices in $\mathfrak{sl}_{2n} \mathbb{R}$. </p> <p>Then for each parameterized 1-dimensional linear subspace $\xi=\xi(t)$ of $\mathfrak{p}$ we get a 1-parameter subgroup $e^{\xi(t)}$ in $SL_{2n}\mathbb{R}$.</p> <p>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.</p> <p>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?</p> <p>There is at least one obvious strategy: we shall determine those $\xi$ for which the differential $d \tilde {\Xi}$ is positive. </p> <p>As one tool, there is the so-called Jacobi formula. This very general formula tells us $$d \tilde{\Xi}=tr(adj~( \Xi(t))~~ d \Xi).$$</p> <p>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). </p> <p>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? </p>

http://mathoverflow.net/questions/102321/have-derivatives-of-determinants-along-1-psgs-ever-been-coherently-computed-vi/108622#108622 William Sit 2012-10-02T11:01:21Z 2012-10-02T11:01:21Z

<p>I refer you to W. Kahan's article: <a href="http://www.cs.berkeley.edu/~wkahan/MathH110/jacobi.pdf" rel="nofollow">http://www.cs.berkeley.edu/~wkahan/MathH110/jacobi.pdf</a></p>

http://mathoverflow.net/questions/102321/have-derivatives-of-determinants-along-1-psgs-ever-been-coherently-computed-vi/109809#109809 Deane Yang 2012-10-16T12:21:26Z 2012-10-16T12:21:26Z

<p>The formula $d(\det A) = (\det A)\operatorname{tr} A^{-1}dA$ or equivalently $d(\log\det A) = \operatorname{tr}A^{-1}dA$ is <em>extremely</em> 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.</p>