Bounding the number of nonzero coefficients in a conic combination I'm looking for a proof for the following statement in order to understand a proof about integer programming I'm reading.
Given vectors $x_1, \ldots, x_s \in \mathbb R^n$, nonnegative coefficients $\lambda_1, \ldots, \lambda_s \in \mathbb R_{\geq 0}$ and a vector $v = \sum_{i=1}^s \lambda_i x_i$, then there exist coefficients $\lambda_1', \ldots, \lambda_s' \in \mathbb R_{\geq 0}$ of which at most $n$ are nonzero with $v = \sum_{i=1}^s \lambda_i' x_i$.
In other words: given a finite set $B$ of real vectors, every vector in the polyhedral cone generated by $B$ can be written as a conical combination of at most $n$ vectors of $B$.
 A: Thanks to dineshdileep's comment I had a look at Caratheodory's theorem and its proof and adapted it to a statement about polyhedral cones instead of polytopes:
Proposition: Let $x_1, \ldots, x_s \in \mathbb R^n$, $\lambda_1, \ldots, \lambda_s \in \mathbb R_{\geq 0}$ and $v = \sum_{i=1}^s \lambda_i x_i$. Then there are $\lambda'_i \geq 0$ with $v = \sum_{i=1}^s \lambda'_i x_i$ and at most $n$ of the $\lambda_i'$ are nonzero.
Proof: Without loss of generality, assume that $\lambda_i > 0$ for all $i = 1, \ldots, s$. If $s \leq n$, there is nothing to proof, so let $s > n$. Then the set $\{\, x_i \mid i = 1, \ldots, s \,\}$ is linearly dependent. With $\mu_1, \ldots, \mu_s \in \mathbb R$ let $\sum_{i=1}^s \mu_i x_i = 0$ be a nontrivial linear combination of zero. Let
$$k := \mathop{\text{arg min}}_{i = 1, \ldots, s} \;\left\{\, \left|\frac{\lambda_i}{\mu_i}\right| : \mu_i \neq 0 \,\right\} $$
and $\alpha := \frac{\lambda_j}{\mu_j}$. For all $i = 1, \ldots, s$, we have 
$$\lambda_i - \alpha \mu_i = \lambda_i - \mathop{\text{sgn}}(\mu_i \mu_k) \left| \frac{\lambda_k}{\mu_k} \mu_i \right| $$ and two cases:
\begin{cases}
\lambda_i - \alpha \mu_i \geq \lambda_i \hphantom{{}- \frac{\lambda_i}{\mu_i}{\mu_i}} \geq 0 & \mathop{\text{sgn}}(\mu_i \mu_k) = -1 \\
\lambda_i - \alpha \mu_i \geq \lambda_i - \frac{\lambda_i}{\mu_i}{\mu_i} = 0 & \mathop{\text{sgn}}(\mu_i \mu_k) = \hphantom{-{}}1.
\end{cases}
Therefore for $i = 1, \ldots, s$ we have $\lambda_i' := \lambda_i - \alpha \mu_i \geq 0$ and in particular $\lambda'_k = 0$.
Repeating this a finite number of times completes the proof. $\square$
