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I want to show rigorously that the convexity remains after a linear mapping remains

Let $C \subset V$ be convex: $tx + (1 - t)y \in C$ for all $x,y \in C$ and $t \in ]0,1[$. Show that $A(C) \subset W$ is convex, if $A :V \to W$ is linear.

My proposal

Set the vector spaces $V, W$ are over a field $F$. Claim: $A(C) \subset W$ is convex where $A$ is well-defined, $F$-linear, and a bijection. For each $x,y \in C$ there exists $t \in ]0,1[$, $u \in C$ and $v \in C$, so $(u+v) \in C$. Note by the definition of $\text{Hom}(V,W)$: \begin{equation} A(u) = t_{1} x + (1-t_{1}) y, \end{equation} \begin{equation} A(v) = t_{2} x + (1-t_{2}) y, \end{equation} and considering the combination \begin{align} A(u+v) &= A( t_{1} x + (1-t_{1}) y + t_{2} x + (1-t_{2}) y ) \\ &= A( t_{1} x + (1-t_{1}) y ) + A( t_{2} x + (1-t_{2}) y ) \\ &= A(u) + A(v). \end{align}

where $A(u)$ and $A(v)$ are convex so $A(u + v)$ is convex. $\square$

Comments

My proposal is not enough rigorous, I think. The thread Linear and affine transformations preserve convexity is one about a similar situation but with affinity. I may miss some necessary properties.

How can do more rigorously the proof of convexity after a linear mapping?

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We have to prove that for any $x, y\in A(C)$ and any $t\in 0,1]$, $tx+(1-t)y\in A(C)$.

Now $x=Au$ for some $u\in C$ and $y=Av$ for some $v\in C$. By linearity, we have: $$tx+(1-t)y=tAu+(1-t)Av=A(tu+(1-t)v),$$ and $w=tu+(1-t)v\in C$ since $C$ is convex, so $tx+(1-t)y\in A(C)$.

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  • $\begingroup$ Thank you for your answer! I think A has to be here too well-defined. Right? $\endgroup$ – Léo Léopold Hertz 준영 Sep 10 '15 at 5:10
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    $\begingroup$ I used the notations of your post. $A$ is supposed to be given. Btw, to defne convexity, the base field can't be any field: it makes sense only on an ordered field, and probably (not sure) is useful only on $\mathbf R$. $\endgroup$ – Bernard Sep 10 '15 at 7:36
  • $\begingroup$ Yes, I think it can be even over an ordered field $F$ on $\mathbb{C}$ too. $\endgroup$ – Léo Léopold Hertz 준영 Sep 10 '15 at 10:10
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First some general remarks:

  • $A$ does not have to be a bijection.
  • The other post you linked to also works for your case, since any linear mapping is also an affine mapping.
  • Convexity (at least as you define it) does not make sense for vector spaces over arbitrary fields. Notice that the definition of convexity requires you to multiply the elements of the vector space by an element $t \in (0,1) \subset \mathbb R$. This is well-defined for real or complex vector spaces, but not e.g. for $\mathbb F_2$-vector spaces.

Also I feel like in general you introduce too complicated terms for an otherwise simple exercise. Finally your proof is incorrect (e.g. what does it even mean when you write $A(u)$ and $A(v)$ are convex.)

Here's how you would prove it:

Pick $x, y \in A(C)$ and arbitrary $\lambda \in (0,1)$. You need to show that $\lambda x + (1-\lambda)y \in A(C)$.

By definition of $A(C)$, there exist $u,v \in C$ such that: $x=A(u)$ and $y=A(v)$. It then holds that:

$$ \begin{aligned} \lambda x + (1-\lambda)y &= \lambda A(u) + (1-\lambda) A(v) \\ &=A(\lambda u + (1-\lambda)v) \end{aligned} $$

Notice that the second equality follows from the linearity of $A$. But since $C$ is convex, it holds that $\lambda u + (1-\lambda)v \in C$. Thus $\lambda x + (1-\lambda)y \in A(C)$, exactly as we wanted to show.

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Try this:

Give any two points $y_1, y_2\in A(C)$, there exists $x_1,x_2\in C$, such that $A(x_1)=y_1$ and $A(x_2)=y_2$. Then:

$\lambda y_1+(1-\lambda)y_2=\lambda A(x_1)+(1-\lambda)A(x_2)=A(\lambda x_1+(1-\lambda)x_2).$

Since $C$ is convex, (i.e. $\forall \lambda \in [0,1]$, $x_1+(1-\lambda)x_2\in C$), then $A(\lambda x_1+(1-\lambda)x_2)\in A(C).$

Which completes the proof that $A(C)$ is convex.

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