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?
Best Answer
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 $tx+(1-t)y\in A(C)$.