# Separate the trivial partition by a linear hyperspace

Let $$e=[1,1,\ldots,1]\in\mathbb{Z}^n$$. I am looking for a way to find a vector $$a\in\mathbb{Z}^n$$ such that:

1. $$\langle a,e\rangle=0$$ and
2. for every nonnegative $$v\in\mathbb{Z}^n$$ such that $$\langle e,v\rangle=n$$ and $$v\neq e$$ we have $$\langle a,v\rangle\neq 0$$

Here the vector $$v$$ is nonnegative if it has nonnegative entries. The $$\langle a,v\rangle$$ denotes the standard scalar product. I am looking for the $$a$$'s whose lengths would polynomially depend on $$n$$ (actually my cost is the maximum of the absolute values $$|\langle a,v\rangle|$$ over all $$v$$'s as above).

The question is motivated by my attempts to find a convenient way to cancel out, in calculations, the inputs of nontrivial partitions, leaving only the one coming from $$1+1+\ldots+1=n$$.

There is no chance you can keep $$a$$ of polynomial size. Indeed, let $$a$$ be any vector with the sum of absolute values less than $${n\choose [n/2]}/2$$. Then among the sums $$\sum_{k\in I}a_k$$ where $$I\subset\{1,\dots,n\}$$, $$|I|=[n/2]$$ there are two equal, meaning that there is a non-trivial vector $$w$$ consisting of $$0,\pm1$$ such that $$\langle a,w\rangle=\langle e,w\rangle=0$$. Then $$v=e+w$$ is your killer.