Timeline for Is this a metric on the Grassmannian Manifold?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Sep 8, 2013 at 1:11 | history | edited | Suvrit | CC BY-SA 3.0 |
reworded first sentence
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| Sep 7, 2013 at 20:53 | history | edited | Suvrit | CC BY-SA 3.0 |
added 74 characters in body
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| Sep 7, 2013 at 20:51 | comment | added | Suvrit | That is exactly the way to go! so Cauchy-Binet suffices, very nice, I should have thought about the problem a bit before shooting off my answer! | |
| Sep 7, 2013 at 20:40 | comment | added | user35593 | After seeing the Cauchy-Binet formula I found a proof without Schoenberg' theorem: Let $a,b,c \in \mathbb{R}^{\binom{m}{n}}$ be definded by terms from the Cauchy-Binet formula, i.e. $\det(A_{S,[n]})$. Then $\|a\|_2^2=\|b\|_2^2=\|c\|_2^2=1$. The triangle inequality for $\mathbb{R}^{\binom{m}{n}}$ gives $$\sqrt{\langle a-c,a-c \rangle}\leq \sqrt{\langle a-b,a-b \rangle}+\sqrt{\langle b-c,b-c \rangle}$$ which boils down to the desired inequality. | |
| Sep 7, 2013 at 19:17 | vote | accept | user35593 | ||
| Sep 7, 2013 at 17:40 | history | edited | Suvrit | CC BY-SA 3.0 |
fixed ref
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| Sep 7, 2013 at 17:28 | history | answered | Suvrit | CC BY-SA 3.0 |