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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?

satokun
  • 677

3 Answers3

4

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.

grand_chat
  • 38,951
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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 $(1,x_0,1,\cdots,1)\cdot u=0.$
mfl
  • 29,399
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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 $v_i =1$.

B. Goddard
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