# Linear transformation maps the first quadrant to a closed set

My question is

Let $$A: \mathbb{R}^n \to \mathbb{R}^m$$ be a linear transformation, how can one prove that $$\{Ay|y\geq 0\}$$ is a closed set?

Here $$y\geq 0$$ means each component of $$y$$ $$\geq 0$$.

I can simplify the question in two ways

1. Since the image of $$A$$ is a closed linear subspace of $$\mathbb{R}^m$$, by replacing $$\mathbb{R}^m$$ with this linear subspace, we may assume A is surjective.

2. After a change of basis of the codomain and a reordering of the basis of the domain, we may assume $$A$$ is of the form $$[I|B]$$

Any help or hints are appreciated, thank you.

Ideas:

Let $$v_i=Ae_i$$, then my original statement is equivalent to saying the set $$\{\sum a_iv_i |a_i\geq 0\}$$ is a closed set. So I should now prove the following:

For any $$v_1,...,v_k \in \mathbb{R}^n$$, the set $$S=\{\sum a_iv_i |a_i\geq 0\}$$ is a closed subset of $$\mathbb{R}^n$$.

I suspect that the set $$S$$ is an intersection of half spaces, from this the closedness will be obvious.

Equivalently, I suspect that for each $$x\notin S$$, there exists a half space $$H$$ (cut by a codimension 1 linear subspace)s.t. $$S\subset H$$.

Is this statement true? I guess this should be true after drawing some examples.

New ideas: I am proving that the set $$\{\sum a_iv_i|a_i\geq 0 \}$$ is closed, and this is easy if all the $$v_i$$'s are indeed independent, so I would want to prove the following:

Let $$S=\{$$linearly independent subsets of $$\{v_i\}\}$$, then $$\{\sum a_iv_i|a_i\geq 0\}=\cup_{T\subset S}\{$$non negative linear combinations of $$T\}$$

And this union is closed since it is a finite union of closed set.

Equivalently, let $$x=\sum a_iv_i,a_i\geq 0$$, I want to write $$x=\sum b_ie_i$$, where $$b_i\geq 0$$ and $$\{e_i\}\subset \{v_i\}$$ is linearly independent.

My proof is as follows: I can replace $$\{v_i\}$$ with a smaller subset of itself s.t. $$x$$ cannot be expressed by non negative linear combinations of any strictly smaller subset of it. Now I claim that this set is linearly independent.

Suppose not, then we can write $$c_1v_1+c_2v_2+\cdots + c_nv_n=0$$, with some $$c_i$$'s are positive(since we can multiply the whole expression by $$-1$$) and I assume that $$c_i>0$$ for $$i=1,2,\dots ,k$$ and $$c_i \leq 0$$ for $$i>k$$

Now I assume $$b=a_1/c_1=\min _{i=1,2,\dots ,k}{a_i/c_i}$$, Then I have $$x=\sum a_iv_i-b(\sum c_iv_i)= (a_2-bc_2)v_2 +(a_3-bc_3)+\cdots + (a_n-bc_n)v_n$$ Which is a non negative linear combinations of fewer vectors, contradicts to my assumption and hence the minimal set $$\{v_i\}$$ is linearly independent.

• What do you know about continuous maps? Is $x \mapsto Ax$ continuous? Mar 13, 2017 at 1:32
• Yes, linear transformations between Euclidean spaces must be continuous. And then? Mar 13, 2017 at 1:38
• What do you know about continuous maps and open sets (and about closed sets)? Mar 13, 2017 at 2:25
• I know preimage of open sets are open and preimage of closed sets are closed, but I am considering the image of a closed set here Mar 13, 2017 at 2:30
• The proof after 'New ideas' is fine imho.
– daw
Mar 16, 2017 at 12:56

Let $(x_k)$ in $S$ be given, such that $x_k\to x$.
The set $D_k:=\{ y\ge 0: \ Ay = x_k\}$ is non-empty, closed, and convex. Hence for each $k$ there is $y_k\in D_k$ satisfying $$\|y_k\|_2 = \inf_{y\in D_k}\|y\|_2.$$ If $(y_k)$ contains a bounded subsequence, then we are done: $y_{k_n}\to y$ implies $y\ge0$, $Ay=x$, and $x\in S$. It remains to consider the case $\|y_k\|_2\to \infty$.
Denote $v_k:=\frac1{\|y_k\|_2}y_k$. By compactness, it contains a converging subsequence. W.l.o.g. let $(v_k)$ be converging to $v$ with $\|v\|_2=1$, $v\ge 0$. Moreover, it holds $Av =0$.
The idea is to prove that $y_k-v$ is in $D_k$. Obviously $A(y_k-v)=x_k$. Denote $I:=\{i: v_i>0\}$. Then $y_{k,i}=\|y_k\|_2\cdot v_{k,i}\to\infty$ for $i\in I$. And there is an index $K$ such that $y_{k,i}-v_i \ge0$ for all $i\in I$ and $k>K$. For $i\not\in I$, it holds $y_{k,i}-v_i =y_{k,i}\ge0$. Hence, $y_k-v\in D_k$ for all such $k>K$. Since $0\le y_k -v\le y_k$ and $v\ne 0$, we have $\|y_k-v\|< \|y_k\|_2$ due to the strict convexity of $\|\cdot\|_2$, a contradiction to the minimality of $y_k$. Thus the unbounded case does not happen.