Face of face of a cone is a face. Proof?

1. Dec 16, 2013

Silversonic

I could really do with some help. I'm trying to show that the face of a face of a convex polyhedral cone is again a face of that polyhedral cone. I have spent a couple hours thinking about this and CAN'T show it. The following apparently gives a proof of this, but it's surely invalid

http://img30.imageshack.us/img30/4752/vsqc.png [Broken]

The bit I have underlined. I can see literally no reason why $\langle u, v \rangle \geq 0$ would mean that $\langle u, v \rangle = 0$. Can anyone help?

Last edited by a moderator: May 6, 2017
2. Dec 17, 2013

Office_Shredder

Staff Emeritus
This text is really hard to read because some stuff doesn't render or is mis type-setted, and they flip the meaning of v and w in the proof of 3. I believe the claim is using that v is contained in $\check{\sigma}$.

We know that $\left<v,w \right>$ is non-negative because v is in $\check{sigma}$ and w is in $\sigma$. Furthermore, p is non-negativve and $\left<u,w \right>$ is non-negative as well (for the same reason as $\left<v,w\right>$. So we are adding two non-negative things together and getting zero. The only way this can occur is if both non-negative things were zero to begin with.

3. Dec 17, 2013

ChrisVer

I would also try to answer that, but why is p non negative? except for if $R_{+}$ notation means positive reals... I interpreted it at first as the real numbers supplied by the action of summation.

in the 2nd (3) and 2nd - it confused me more about it

4. Dec 17, 2013

Office_Shredder

Staff Emeritus
Chris, $\mathbb{R}_+$ does mean positive reals.

5. Dec 17, 2013

Silversonic

Thanks for the replies, yeah I noticed the text was quite hard to read but it was the only proof I could find after a long search on google.

I have actually figured it out (after harder searching) and did it before I saw this thread. I've attached in case anyone wants a look.

http://img202.imageshack.us/img202/4387/2x7r.png [Broken]

Last edited by a moderator: May 6, 2017