Intersection of $n+k$ subspaces of $\mathbb{R}^n$

In $\mathbb{R}^n$, $n\ge 3$ consider $n+k$ vectors $v_1,\dots, v_{n+k}$, $k\ge 1$ such that subset of cardinality $n$ of $\{v_1,\dots, v_{n+k}\}$ is composed of linearly independent vectors of $\mathbb{R}^n$. Define the subsets $W_i$ of $\mathbb{R}^n$ as $W_i:=\{w\in \mathbb{R}^n|w\cdot v_i\le 0\}$, $i=1,\dots, n+k$.

What is a sufficient condition on $v_1,\dots,v_{n+k}$ to guarantee $\cap_{i=1}^{n+k}W_i=\{0\}$?

In the meantime I had an idea:

A sufficient condition for $\cap_{i=1}^{n+k}W_i=\{0\}$ could be, up to renumbering the indexes, $v_{n+1}=\alpha_1v_1+\dots \alpha_nv_n$ with $\alpha_i<0$ for all $i=1,\dots,n$. In this way, if $w\in \cap_{i=1}^{n}W_i,w\neq 0$ it's clear that $w\notin W_{n+1}$. But this is a rather unpleasant condition: for $n$ big it will be very complicated to compute the $\alpha_i$ (as one should compute the inverse of the Gram matrix associated to the basis). Do you know a better condition?

Note that your condition is the same as saying that 0 is in the positive cone generated by the $v_i$, namely in $C=\{\sum \alpha_i v_i \mid \alpha_i>0\}$ (you can actually increase it a little bit and only require that at least $n$ of the coefficients are nonzero). Indeed, if $w\cdot v_i\leq 0$ for all $i$ we get that $0=w\cdot \bar{0}=\sum \alpha_i w\cdot v_i$ so $w \cdot v_i$ must be zero for every $i$ and since the $v_i$ span the whole space you obtain that $w=0$.

This condition is also necessary. If $C$ doesn't contain zero, then its closure is not the whole space, since otherwise $C$ would contain points which are very close to $\pm e_i$, and therefore contain their positive cone which contains zero. Assume that $z\notin \bar{C}$. Using the hyperplane separation theorem (which we can since both $\{z\}$ and $\bar{C}$ are closed and convex, and $\{z\}$ is compact), we can find a nonzero vector $w$ and a constant $c$ such that $w\cdot z> c >w\cdot v$ for all $v\in \bar{C}$. If $v\in \bar{C}$ then $rv\in \bar{C}$ for every $r>0$, so we cannot have that $w\cdot v>0$. We conclude that $w\cdot v\leq 0$ for every $v\in \bar{C}$.

If you're looking for a way to actually compute this $w$, then you can multiply your entire system by a suitable invertible matrix and assume that $v_i=e_i$ for $i=1,...,n$. Then the condition is reduced to the nonnegative cone of $v_{n+1},...,v_{n+k}$ should contain a vector such that all of its coordinates are negative. So if $k$ is much smaller than $n$ you will have a much easier task.

====EDIT====

If $g\in GL_n(\mathbb{R})$, then $$\{w \mid w\cdot g(v_i)=0\}=\{w \mid g^t(w)\cdot v_i=0\}=(g^{-1})^t(\{w \mid w\cdot v_i=0\}).$$ It then follows that $(g^{-1})^t(\bigcap W_i)=\bigcap (g^{-1})^t(W_i)$, so that $\bigcap W_i =\{0\}$ for the set $v_1,...,v_{n+k}$ if and only if the same is true for $g(v_1),...,g(v_{n+k})$. So the first reduction will be to choose such $g$ for which $v_i=e_i$.

The second step is to note that $\sum \alpha_i e_i + \sum \beta_j v_{n+j}=0$ where $\alpha_i, \beta_j>0$ is the same as being able to find in the positive cone of $v_{n+1},...,v_{n+k}$ a vector with all negative entries. Moreover, if you can find such a vector in the nonnegative cone (allow zero coefficients), then there is such a vector in the positive cone becuase you can add a very small epsilon to each coefficient without changing the signs of the entries of the resulting vector.