Let $A_{1},...,A_{k}$ be $n\times n$ matrices. When does there exist a vector $x\geq,\neq 0$ such that $(\sum_{i}A_{i}\alpha_{i})x\neq 0$ for all $\alpha_{1},...,\alpha_{k}\geq 0$ such that $\sum_{i}\alpha_{i}=1$?
Any help would be appreciated. This problem sort of looks like a Theorem of the Alternative kind of problem...
First of all observe that if such $x$ exist, then $x \in \left(A_1^\bot \oplus A_2^\bot \oplus \cdots \oplus A_k^\bot \right)^\bot$ ($A_i^\bot$ being the column null space of $A_i$). Since if $\exists i: x \in A_i^\bot$ then set $a_i = 1$ and $\forall j,j \neq i: a_j=0$ and you get $\left(\sum_{i}A_{i}\alpha_{i}\right)x=0$.
In general we need $\left(\sum_{i}A_{i}\alpha_{i}\right)x\neq0$ but we can restrain it more (since there is the assumption $\forall i: a_i \ge 0$ it has some convex cone-like structure) that $\left(\sum_{i}A_{i}\alpha_{i}\right)x\ge 0$ or $\left(\sum_{i}A_{i}\alpha_{i}\right)x\le 0$. Hopefully we can find a solution in this case but its not a general solution. We consider the positive case. Assume $P_i$ be the subspace spanned by the eigenvectors corresponding to the positive eigenvalues of $A_i$. Then if there exist $x\in \left(P_1 \cap P_2 \cap \cdots \cap P_k \right)$ we can assure that $\left(\sum_{i}A_{i}\alpha_{i}\right)x\ge 0$.
On the other hand if $N_i$ be the subspace subspace spanned by the eigenvectors corresponding to the negative eigenvalues of $A_i$, then $x \in \left(N_1 \cap N_2 \cap \cdots \cap N_k \right)$ guarantees that $\left(\sum_{i}A_{i}\alpha_{i}\right)x\le 0$ which is another solution.