Prove that there is a vector $v\in \mathbb{R}^k$ such that $u \cdot v =0$ 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?
 A: 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.
A: Without lost of generality assume $u=(u_1,u_2,\cdots, u_k)$ satisfies $u_1>0$ and $u_2<0.$ Now:


*

*If $u_3+\cdots+u_k=0$ then $v=(-u_2,u_1,1,\cdots,1)$ works;

*If $u_3+\cdots+u_k>0$ then $(1,0,1,\cdots,1)\cdot u>0.$ Since $u_2<0$ we have that $\lim_{x\to \infty} (1,x,1,\cdots,1)\cdot u=-\infty.$ So, it must exists $x_0\in \mathbb{R}_+$ such that $(1,x_0,1,\cdots,1)\cdot u=0.$

*If $u_3+\cdots+u_k<0$ then $(0,1,1,\cdots,1)\cdot u<0.$ Since $u_1>0$ we have that $\lim_{x\to \infty} (x,1,1,\cdots,1)\cdot u=+\infty.$ So, it must exists $x_0\in \mathbb{R}_+$ such that $(1,x_0,1,\cdots,1)\cdot u=0.$

A: Let $u_i$ be the $i$th component of $u$.  Let $u_k$ be the specified negative component.  Let $v_k= -(\sum_{i\neq k} u_i)/u_k$ and all other $v_i =1$.  
