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Jun 11, 2016 at 6:22 vote accept polfosol
Jun 8, 2016 at 8:10 comment added Suvrit @Dirk yes, indeed. This example was written only for a single fixed $m$...
Jun 8, 2016 at 3:05 answer added Christian Remling timeline score: 3
Jun 7, 2016 at 16:56 answer added Ilya Bogdanov timeline score: 2
Jun 7, 2016 at 15:31 comment added Dirk @Suvrit Ok, but how do you continue with the $u_m$? Setting $u_m=0$ from that point on will give $K_m=0$, and continuing with $u_m = e_n$ will drive the inverse down to zero… Seems like and interesting question to me.
Jun 7, 2016 at 14:57 comment added Suvrit Say $m > n$. Let $u_1,\ldots,u_{n-1}$ be canonical basis vectors $(e_i)_{i=1}^{n-1}$. Let $u_{n},\ldots,u_{m-1}=0$ and let $u_m=e_n$. Then $\sum_{i=1}^m u_iu_i^T$=$I_n$ and $\|K_m\|=1$.
Jun 7, 2016 at 14:51 history edited GH from MO
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Jun 7, 2016 at 14:08 history edited polfosol CC BY-SA 3.0
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Jun 7, 2016 at 9:52 review First posts
Jun 7, 2016 at 11:31
Jun 7, 2016 at 9:48 history asked polfosol CC BY-SA 3.0