Assume that $P:={\rm conv}\ \{u_i\}_{i=1}^m$ is a $n$-dimensional polytope in $\mathbb{R}^n$ s.t. origin is an interior point in $P$. Then there is $c_i>0$ s.t. $$ \sum_{i=1}^m c_i u_i =0 $$
Proof : Clearly $m>n$. If $A =[u_1\cdots u_m]$, then $A$ has a nontrivial kernel $c:=(c_1,\cdots,c_m)$. However, how can we prove $c_i>0$ ?
Thank you in advance.
[Add] Here $u_i$ is vertex That is $u_i$ is not in ${\rm conv}\ \{ u_j\}_{j\neq i }$. And we do not consider the case where $c_i=0$ for some $i$.
Let $p$ be an interior point contained in some polytope $P = \mathsf{conv}\,\{u_i\}_{i=1}^m$ embedded in $\mathbb{R}^n$, where we assume that $u_i$ form the vertices of $P$ (e.g. there are no redundant $u_i$). By the definition of the convex hull there exist constants $\theta_1,\theta_2\ldots,\theta_m\in [0,1]$ satisfying $\sum_{i=1}^m\theta_i = 1$ and $\sum_{i=1}^m\theta_i u_i = p.$ Now note that if we fix $m-n$ $\theta_i=0$ and take the affine hull of the rest of the vertices of $P$, this set forms a nowhere dense subspace of $\mathbb{R}^n$ (as a translation of a strict subspace of $\mathbb{R}^n$). Since $p$ was interior, it follows that for each $\theta_j$ we can represent $p = \sum_{i=1}^m\theta_{ij}u_i$ with $\theta_{jj}>0$. Take the average of all these representations $p = \frac{1}{m}\sum_{ij} \theta_{ij}u_i$ to have a convex combination with strictly positive coefficients. $\blacksquare$