TITLE: Find a vector NOT perpendicular to a given set of vectors
QUESTION [9 upvotes]: Let us suppose to have a finite set of vectors $S=\{v_1,\ldots,v_m\}$ in $\mathbb{R}^n$ (with $m \gg n$ in general). I need to find a vector $x \in \mathbb{R}^n$ that is NOT perpendicular to any vector in $S$. The existence of such a vector $x$ is guaranteed, moreover almost every vector in $\mathbb{R}^n$ satisfies this property. But I need to find an algorithm to determine a vector with these properties, I cannot close my eyes and choose.

Any ideas?
EDIT1: in my case, the vectors in $S$ have some "symmetries" in the sense that they are generated by permutation and change of signs of a few vectors in $S$
EDIT2: $v_1+\ldots+v_m=0 \in \mathbb{R}^n$
EDIT3: I simplified the solution proposed by Tom Collinge. We define a sequence $\{x_i\}_{i=1}^m$ such that $x_m=x$ is what we are looking for. First define $x_1=v_1$, so $x_1 \cdot v_1 \ne 0$ and they are not perpendicular (obviously). Then, recursively, define for $k \in \{2,\ldots,m\}$
$$
x_k=\begin{cases}
x_{k-1}+2v_k & \text{if }x_{k-1}=-v_k\\
x_{k-1}+v_k & \text{otherwise}
\end{cases}.
$$
By construction we have that $x_k \cdot v_i \ne 0$ for $i \le k$, then $x_m \cdot v_i \ne 0$ for all $v_i \in S$, so $x=x_m$ is not perpendicular to every element in $S$. Can it works?

EDIT4: As pointed out, the algorithm proposed in EDIT3 does not work

REPLY [2 votes]: I don't think it matters whether $m > n$ or not, and clearly we assume that $s_i \ne 0$ for any $i$ 
Define a sequence of vectors $(x_j)_ {j = 1}^m$ and constants $(\alpha_i)_{i = 1}^m$  as follows and then $x = x_m$ satisfies your requirement.
$\alpha_1 = 1; x_1 = \alpha_1 s_1$ and so $x_1$ is not orthogonal to $s_1$.
Given $k < m$ and $x_k = \sum_{i = 1}^k \alpha_i s_i$ and $x_k$ not orthogonal to any $\{s_i\}_{i = 1, k}$
Then for $i = 1, k + 1$ let $c_i = x_k \cdot s_i$  so that by non-orthogonality  $c_i \ne 0$ for $i = 1, k$.
If $c_{k+1} \ne 0 $ then put $\alpha_{k+1} = 0 \implies x_{k+1} = x_k$ and $x_{k+1}$ is not orthogonal to $\{s_i\}_{i = 1}^{k+1}$ 
Otherwise, chose $\alpha_{k+1} \ne 0$ and such that $\alpha_{k+1} s_{k+1} \cdot s_i \ne - c_i$ for $i = 1, k$
(which can be done even if $s_{k+1}\cdot s_i = 0$, since $c_i \ne 0$ for $i = 1, k$).
Then for $i = 1, k$,
$$
\begin{split}
x_{k+1} \cdot s_i 
  &= (x_k \cdot s_i) + \alpha_{k+1}(s_{k+1} \cdot s_i) \\
  &= c_i + \alpha_{k+1}(s_{k+1}.s_i) \\
  &\ne 0
\end{split}
$$
And for $i = k+1$,
$$
\begin{split}
x_{k+1} \cdot s_i
 &= (x_k \cdot s_i) + \alpha_{k+1}(s_{k+1} \cdot s_i) \\
 &= (x_k \cdot s_{k+1}) + \alpha_{k+1}(s_{k+1} \cdot s_{k+1}) \\
 &= 0 + \alpha_{k+1}(s_{k+1} \cdot s_{k+1}) \\
 &\ne 0
\end{split}
$$
So, $x_{k+1}$ is not orthogonal to $\{s_i\}_{i = 1, k+1}$