Subset of an n-dimensional inner product space containing 2n elements

Let $V$ be an inner product space of finite dimension $n$ over its field of scalars. Show that there exists a subset $\{v_1,...,v_{2n}\}$ of $V$ satisfying the conditions that $\langle v_i, v_j \rangle \leq 0$ for all $1 \leq i \neq j \leq 2n$.

I know that if the inner product is equal to zero than the two vectors are orthogonal. Also, correct me if I'm wrong, but for the inner product to be $<0$ that would imply that the angle between them is greater than 90 degrees.

That's about as far as I'm getting here. Is there a need for the Gram-Schmidt Process?

• In $\Bbb R^3$, think of an octahedron. – pjs36 Nov 28 '15 at 22:25
• @I.Cavey What is the inner product of $(1,1)$ and $(-1,-1)$? – David C. Ullrich Nov 28 '15 at 22:34
• @DavidC.Ullrich Ah, thank you for letting me realize my own mistake. – I. Cavey Nov 28 '15 at 22:38

Pick an orthonormal basis for your space, call it $B = \{e_1, e_2, \ldots, e_n\}$. Such a thing exists, of course, thanks to Gram-Schmidt process.
Now for each $e_i$, let $f_i = -e_i$. You can verify that $\langle e_i, f_i \rangle = -1$, while all $e_i$ are orthogonal to all $f_j$ when $j \neq i$. Of course the various $f_i$ are related to each other in the same way the various $e_i$ are.
In $\Bbb R^n$, the convex hulls of these $2n$ vectors are the $n$-dimensional cross-polytopes, dual to well-known $n$-dimensional cubes.