Boundary of Convex set under Linear transformation

Question

In $\mathbb{R}^n$, any line goes to line and any convex set goes to convex set under linear trasformation. But will boundary of convex set also will go to boundary of range of convex set?

Answer 1

Not if you allow domain and range of $A$ to be of different dimensions. Choose e.g. $$A:\Bbb R^2\to\Bbb R,\quad (x,y)\mapsto x.$$ Take the convex set $[0,1]^2\subset\Bbb R^2$. We have that

$$A(\partial_{\Bbb R^2}[0,1]^2)=[0,1]\not=\{0,1\}=\partial_{\Bbb R}[0,1]=\partial_{\Bbb R}A[0,1]^2.$$

Here $\partial_{\Bbb R^n}C$ means the boundary of $C$ with respect to the standard topology on $\Bbb R^n$.

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However, if you meant $A:\Bbb R^n\to\Bbb R^n$, then your statement is true. Let $C\subseteq\Bbb R^n$ be convex. If $A$ is not invertible, then $A(C)$ has no inner points, hence all points of $\partial C$ get mapped to boundary points $\partial A(C)=A(C)$.

But if $A$ is invertible, then $A^{-1}$ is linear too. Linear maps are continuous, hence $A$ is a homeomorphism of $\Bbb R^n$ to itself. And homeomorphisms preseves boundary points.

Answer 2

Since in convex analysis, we often talk about relative interior points, if the OP also considers relative boundary of a convex set $C$, i.e. the boundary of $C$ within its affine hull, then even when the linear mapping $A$ is from $\mathbb{R}^n\rightarrow\mathbb{R}^n$, the relative boundary of $C$ may not be mapped to the relative boundary of the image $A(C)$. This happens when $A$ is not injective.

For example, consider a linear mapping $A: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ defined as $(x,y)\mapsto (x,0)$ (the ''same'' map as in the previous answer, except its image is now in $\mathbb{R}^2$), then again we have $A(\partial [0,1]^2)=[0,1]\ne \{0,1\}=\partial A([0,1]^2)$, where $\partial$ means the relative boundary.