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Jul 2, 2018 at 13:18 history edited Suvrit CC BY-SA 4.0
added a stronger proof.
Jul 1, 2018 at 22:07 comment added Suvrit @AlexArvanitakis indeed the same reasoning as above with a tiny bit of care can be used to prove the inequality $\det((1-\lambda)A+\lambda B) \ge \det(A)^{1-\lambda}\det(B)^\lambda$, which is actually equivalent to Minkowski's determinant inequality (easy to verify using suitably scaled versions of the matrices $A$ and $B$)
Jun 30, 2018 at 22:35 comment added AlexArvanitakis This is very clever. I wonder if such reasoning can also prove the Minkowski determinant theorem mentioned above...
May 20, 2011 at 5:53 history edited Suvrit CC BY-SA 3.0
added link, fixed typo
May 20, 2011 at 5:47 history answered Suvrit CC BY-SA 3.0