Every point in the interior of a $n$-dimensional cone can be written as a positive combination of $n$ elements of this cone?

Question

This is one of those questions that make sense, but may be hard to prove.

Let $C$ be a nonempty closed convex pointed cone in a $n$-dimensional linear space $X$.

Also, assume that $int(C)\neq \emptyset$, so $span(C)=X$.

Motivation: A result by Fenchel (Lecture notes 1951, Thm. 7) tells us that any nonzero $x\in C$ can be written in the form $x=\sum_{i=1}^n\alpha_i x_i$ for some linearly independent vectors $x_1,\ldots,x_n\in C$ and some nonnegative scalars $\alpha_1,\ldots,\alpha_n$.

My question: Now, then let $x\in int(C)$. Is the following also true?

There exist linearly independent vectors $x_1,\ldots,x_n\in C$ such that $x=\sum_{i=1}^n \alpha_i x_i$, where $\alpha_i>0$, $\forall i\in \{1,\ldots,n\}$.

That is, can we always find such a representation with $n$ strictly positive coefficients, for interior points of $C$?

Thank you!

Reference: https://www.convexoptimization.com/TOOLS/Fenchel1951.pdf

Edit: Added "pointed" after a counterexample by @copper.hat.

Answer 1

Yes this is possible. Note that the set $$S^{n-1}=\left\{\sum\alpha_ix_i\mid\alpha_i\ge0,\;\sum\alpha_i=1\right\}$$ is the affine $(n-1)$-simplex with vertices $\{x_i\}_{i=1}^n$ (see for example https://en.wikipedia.org/wiki/Simplex). If we let some of the $\alpha_i$'s be $0$ then we get a set of the same form where we have basically ignored some of the $x_i$'s, in other words we get a lower dimensional simplex $S^{n-m-1}$ (where $m$ is the number of $\alpha_i$'s which are zero. Crucially note that $S^{n-1}$ is the convex hull of the set $\{x_i\}_{i=1}^n$, and $S^{n-m-1}$ lies entirely in its boundary.

Now, given that your choice of $x$ is in the interior of $C$, the set $\{x_i\}_{i=1}^n$ can be chosen such that $x$ lies in the interior of the cone $$\mathbb{R}^{\ge0}S^{n-1}=\left\{\sum\alpha_ix_i\mid\alpha_i\ge0\right\}.$$ But from what I said above, the interior of this cone is exactly the set $$\left\{\sum\alpha_ix_i\mid\alpha_i\gneq0\right\}.$$

Edit

The slightly underhand part of my answer above is that I did not justify

the set $\{x_i\}_{i=1}^n$ can be chosen such that $x$ lies in the interior of the cone $$\mathbb{R}^{\ge0}S^{n-1}=\left\{\sum\alpha_ix_i\mid\alpha_i\ge0\right\}.$$

so let me do that explicitly. The point $x$ is in the interior of of $C$, so let $B=B(x,\varepsilon)$ be a closed ball centred on $x$ contained in $C$, and let $\partial B=\mathbb{S}^{n-1}$ be its boundary. There are $n$ points $\{x_i\}_{i=1}^n$ on $\mathbb{S}^{n-1}$ which define an $(n-1)$-simplex $S^{n-1}$ containing $x$ in its interior (and clearly $\{x_i\}_{i=1}^n$ must be linearly independent since they are not co-hyperplanar).

Answer 2

Choose $x \in C^\circ$, note that $x \neq 0$. Let $L= \{x\}^\bot$ be the $n-1$ dimensional subspace orthogonal to $x$. Let $b_1,...,b_{n-1}$ be an orthogonal basis for $L$ and note that $0 = b_1+\cdots+b_{n-1} + b_n$ where $b_n=(-1) (b_1+\cdots+b_{n-1})$ (note that the $b_k$ are affinely independent). Choose $\epsilon$ such that $x+\epsilon b_k \in C^\circ$ and note that the points $x+\epsilon b_k$ are linearly independent.