How to prove conv(M)+u=conv(M+u) and Aff(M)+u=Aff(M+u)

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) ?

• "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. Mar 22, 2016 at 21:34
• @user251257 I added definition and my solution, would you have a look? Mar 22, 2016 at 21:45

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*}