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Let $C$ be a non-empty convex set with non-trivial lineality space $L$ (Lineality space of a convex set $C$ being defined as $L = \{y\,|\,y+C=C\}$). How can I prove the following conclusion? $$ dim(C\cap L^\perp) + dim(L) = dim(C) $$

So far I am able to show that $C = (C\cap L^\perp) +L$, but I don't know how to proceed from here. Can anyone give me some hints?

Vokram
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  • Does the result not simply follow from the (obvious) fact, that $C \cap L^\perp$ and $L$ are orthogonal? – gerw Oct 30 '15 at 08:43
  • By the way: The lineality space $L$ always contains ${0}$, hence it is never empty. – gerw Oct 30 '15 at 08:43
  • @gerw Right.. I can see that intuitively, but how can one show that formally? I'm stuck since $dim(C)$ and $dim(C\cup L^\perp)$ involves affine hulls, and I don't know how the intuition from subspaces translate into this case. Sorry if this is really obvious. – Vokram Oct 30 '15 at 08:52
  • @gerw You are right. I have changed it to "non-trivial". – Vokram Oct 30 '15 at 08:54

1 Answers1

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Here are some hints:

  • The affine hull of $C \cap L^\perp$ is orthogonal to $L$.
  • Take a basis in $C \cap L^\perp$ and a basis in $L$.
  • Show that the union of both bases is linearly independent and form a basis of the affine hull of $C$.

The same proof works in case $L = \{0\}$ and, in this case, the dimension formula holds trivially. So you do not need to exclude this case.

gerw
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