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May 29, 2011 at 0:49 comment added Mark Meckes Fair enough.$ $
May 25, 2011 at 7:18 comment added Roland Bacher Not really: You don't need exponentials for proving that $\det(G)$ is proportional to $1/\hbox{Volume}(G)^2$ : It is enough to stare at an orthogonal basis formed of eigenvectors for $G$. In this sense this proof is more elementary.
May 20, 2011 at 14:20 comment added Mark Meckes It's worth noting that this is secretly the same as Suvrit's answer.
May 20, 2011 at 9:22 history edited Roland Bacher CC BY-SA 3.0
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May 20, 2011 at 6:53 history answered Roland Bacher CC BY-SA 3.0