Prove that there is a vector $v\in \mathbb{R}^k$ such that $u \cdot v =0$

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.