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I'm doing the exercise below and I'm stuck on how to show what is being asked for.

Given an $m \times n$ matrix $A$ and a convex set $ C \in \mathbb{R}^n$, show that the set $\{Ax |x \in C\}$ is convex.

In this case, I also want to show or give a counter example if I replace the word convex with open, closed and compact. However using the definition of convex set $\alpha x + (1 - \alpha)y \in A$ for the first case, I tried to do the following:

  1. Like $x \in C$ then $\alpha Ax + (1 - \alpha)y \in C$

But I don't know if this is correct or how to continue.

Can someone help me?

Robby the Belgian
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Danilo Perl
  • 15
  • To show ${Ax,|,x\in C}=A(C)$ is convex, take two elements of $a_1,a_2\in A(C)$, and show that $\alpha a_1+(1-\alpha)a_2\in A(C)$, with $0\leq \alpha \leq 1$. If $a_1,a_2\in A(C)$ then $a_1=A(c_1)$ and $a_2=A(c_2)$ for some $c_1,c_2\in C$. Try continue from there. – snulty Oct 06 '20 at 15:07

2 Answers2

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No, you want $\alpha A x + (1-\alpha) A y \in A(C)$ if $x$ and $y$ are in $C$ and $0 \le \alpha \le 1$.

Hint: $\alpha A x + (1-\alpha) A y = A(\ldots)$.

Robert Israel
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You want to show that $$\text{for }\lambda\in[0,1],\ A\tilde x,A\tilde y\in Ax:\; \lambda A\tilde x +(1-\lambda) A\tilde y\in Ax$$So: $$\lambda A\tilde x +(1-\lambda) A\tilde y=A(\lambda \tilde x +(1-\lambda)\tilde y)$$And because $\tilde x,\tilde y\in C$ and C is convex, we know that $\lambda \tilde x +(1-\lambda)\tilde y\in C$ and so $A(\lambda \tilde x +(1-\lambda)\tilde y)\in Ax$, meaning $Ax$ is convex

user619755
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