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

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!

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

• If $x \neq 0$ then since $x \in C$ you have $x=1 \cdot x$. If $n=1$ and you take the $C=\mathbb{R}$ then $x=0$ cannot be written in the above fashion. Aug 26 '21 at 16:22
• @copper.hat You're right in your second statement, my question was not well formulated. I just added an assumption that $C$ is pointed as well, so $0\notin int(C)$. About your first statement, note that we're looking for a combination of exactly $n$ l.i. points of $C$.
– Myth
Aug 26 '21 at 17:02

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).

• Hey, thank you! I was trying to follow the same path as you, but the hard part for me was to prove that "the set $\{x_i\}^n_{i=1}$ can be chosen such that x lies in the interior of the cone $\mathbb{R}^{\geq 0} S^n$". If there's no such simplex, then $x$ must be at the boundary of $C$, right? But how can we prove that? I also added the "pointed" assumption so $x=0$ is not an interior point of $C$.
– Myth
Aug 26 '21 at 17:06
• I realised I got the dimension of the simplex wrong, so I fixed that, and added justification of the point you queried. Is that now convincing? Aug 26 '21 at 17:26
• Yes, thank you! It's really a pity that I cannot accept more than one answer, since the other one is also really good.
– Myth
Aug 26 '21 at 17:32

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.

• Thank you, this is really clever.
– Myth
Aug 26 '21 at 17:29