Find a linear functional such that vanishes on given vectors Suppose that $v_1,v_2,...,v_k\in\mathbb{R}^n\setminus \{0\}.$ Show that there exists a linear functional $f:\mathbb{R}^n\rightarrow \mathbb{R}$ such that $f(v_j)\neq 0$ for all $j.$ 
I just taking $f(x)=\langle x,y\rangle$ for fixed $y\neq 0 $. This map is linear. But I am not sure whether it will work.

The set of vectors orthogonal to $v_1$ forms a subspace $V_1$. Because $v_1\neq0$ the subspace $V_1$ is not all of $\Bbb{R}^n$. Similarly for all $i$, the set of vectors orthogonal to $v_i$ forms a proper subspace $V_i$, $i=1,2,\ldots,k$.
At this point I refer you to the latter part of this answer, where I show that the union $\bigcup_{i=1}^kV_i$ is not all of $\Bbb{R}^n$ either. Can you do the rest with the inner product idea?

The argument works even if we had countably infinitely many vectors $v_i,i\in\Bbb{N}$.
