TITLE: Prove that there is a vector $v\in \mathbb{R}^k$ such that $u \cdot v =0$
QUESTION [6 upvotes]: Let $u \in \mathbb{R}^k$ be a vector with one component positive, one component negative, and the remaining $k-2$ can have at most one component that is equal to zero. Then is there a vector $v \in \mathbb{R}^k$ such that all its components are strictly positive and $u \cdot v= 0$? 
Intuitively this seems to be true. But how can I go about showing this formally?

REPLY [4 votes]: Let $P>0$ be the sum of the positive components in $u$, and $N<0$ be the sum of the negative components. Define  vector $v$ to have value $1$ where $u_i\ge0$ and value $b$ where $u_i<0$, where $b$ is such that
$P-b|N|=0$. Clearly $b=P/|N|$ is positive, so the vector $v$ has all positive components. Since $\sum u_iv_i=P-b|N|=0$, we are done.