Does there exist a basis of a Convex Cone

Question

We are given a set $C \subset V$, where $V$ is a vector space. $C$ is a convex cone, so it has the following properties

$$
0 \in C
$$

$$
C+C \in C
$$

$$
x\cdot C\in C\mid x\in\mathbb{R} \land x\geq0
$$

My question is, does there exist a set of basis vectors that spans this convex cone and only this convex cone (no vector in the span is not in the cone)?

I'm also trying to practice my math notation, so here's the above statement rewritten

Does

$$
\exists B=\{\vec b_1, \vec b_2, \ldots, \vec b_n\} \mid \operatorname{span}(B) = C \implies c_1\vec b_1 + \cdots +c_n\vec b_n \in C \forall c_k\in \mathbb{R_+}
$$

Answer 1

It seems like we need to distinguish linear span from conical hull.

Given a set $S$ of elements of a vector space $V$ (could be the polynomials, as in your example), the linear span of $S$ is defined to be

$$ \text{span}(S)=\left\{\left.\sum_{i=1}^k\lambda_iv_i\ \right|k\in\mathbb{N},v_i\in{S},\lambda_i\in\mathbb{R}\right\}. $$

In contrast, the conical hull of $S$ is defined to be:

$$ \text{cone}(S)=\left\{\left.\sum_{i=1}^k\alpha_iv_i\ \right|k\in\mathbb{N},v_i\in{S},\alpha_i\in\mathbb{R}_+\right\}. $$

The difference between these definitions is that the scalars in the conical hull must be non-negative.

So when you say:

Does there exist a set of basis vectors that spans the convex cone?

you're asking

Does there exist a set $S$ such that $\text{span}(S)=C$?

It's easy to see that $\text{span}(S)$ is a linear subspace of the vector space $V$. So the answer to the question above is yes if and only if $C$ is a linear subspace of $V$. A linear subspace is a convex cone, but there are lots of convex cones that aren't linear subspaces. So this probably isn't what you meant.

If instead, you ask

Does there exist a set $S$ such that $\text{cone}(S)=C$?

then the answer is yes, but the set $S$ may be infinite.

Based on what you said in the comments, it sounds like you are looking for a set $S$ such that $\text{cone}(S)=C$, and a vector $v$ which is orthogonal to every element of $S$. This will only be possible if $\text{span}(C)$ is a proper subspace of $V$. In this case, you can take $v$ to be a vector in the orthogonal complement of $\text{span}(C)$. Otherwise, $\text{span}(C)=V$, and there is no vector which is orthogonal to every vector in $S$.