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I don't know how to prove that if $ M \subseteq R^n, \forall u \in R^n $ then conv(M)+u=conv(M+u) and Aff(M)+u=Aff(M+u). conv is convex hull and Aff is affine hull.

Yes it is a homework question, but I'm totally stuck. Mainly because of ambiguity of defined terms. My teachers seems to interchange vectors and points in definitions. I'm totally confused with this. How can be convex hull of points from space be equal to convex hull of vectors? I know this can be proved from definitions but can someone explain it?

EDIT-my solution

Def: Convex hull $X \subseteq R^n $ is intersection of all convex sets containing X

Theorem: Convex hull of X is equal to set of convex combinations $ \{ \alpha_{0} a_0 + ... + \alpha_{k} a_k | k \in N, a_i \in X , \alpha_i \ge 0 , \sum \alpha_i = 1 \} $

My solution is therefore just adding u into convex combinations $ \{ \alpha_{0} a_0 +... + \alpha_{k} + \alpha_{k+1} u \ | \ k \in N, a_i \in X , \alpha_i \ge 0 , \sum \alpha_i = 1 \} $ and....that is conv(M+u) ?

Martin
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  • "Points" and vectors are the same in linear spaces... Now, what are your definitions of convex / affine hull? Depending on the definition, it is straight forward. – user251257 Mar 22 '16 at 21:34
  • @user251257 I added definition and my solution, would you have a look? – Martin Mar 22 '16 at 21:45

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I will prove it for the affine hull. For the convex hull it is basically the same.

Let $x\in \operatorname{aff}(M) + u$. Then there exists $\alpha_0,\dotsc,\alpha_k\in\mathbb R$ and $a_0,\dotsc,a_k M$ with $\alpha_0+\dotsb+\alpha_k = 1$ and $x = \alpha_0 a_0 +\dotsb + \alpha_k a_k + u$. This follows, \begin{align*} x &= \alpha_0 a_0 +\dotsb + \alpha_k a_k + u \\ &=\alpha_0 a_0 +\dotsb + \alpha_k a_k + (\alpha_0+\dotsb+\alpha_k)u \\ &=\alpha_0 a_0 +\dotsb + \alpha_k a_k + \alpha_0u +\dotsb+\alpha_k u \\ &=\alpha_0 (\underbrace{a_0 + u}_{\in M + u}) +\dotsb + \alpha_k (\underbrace{a_k + u}_{\in M + u})\in \operatorname{aff}(M+u) . \end{align*}

user251257
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