I didn't want to bloat the title, i'll also add that the basis is in ${\mathbb{R}^n}$. This question seems kind of easy, but its from a test so I assume there is a catch here somewhere.
I tried to prove it by contradiction. Assume by contradiction that there exists no basis that satisfies the condition. But as known, every bilinear form can by identified by a matrix so that $A_{ij} = f(v_i,v_j)$. If there is no basis so $f(v_i,v_j) = -f(v_j,v_i)$ it means there is no skew-symmetric matrix which can identify a bilinear form over $\mathbb{R}$ which is obviously not true( I guess an example can be given here, I think ).
This proof just seems to simple to be true, but I can't seem to figure what could be wrong with it. I'll greatly appreciate any tips, thanks!
Antisymmetry is a property that doesn't depend on a basis. Suppose you found a basis for which $f(v_i,v_j)=-f(v_j,v_i)$. Now, given arbitrary vectors $x$ and $y$, $x=\sum r_iv_i$ and $y=\sum s_i v_i$, with $r_i,s_i$ scalars. Then $$f(x,y)=\sum_{i,j} r_is_j f(v_i,v_j)=-\sum_{i,j} r_is_j f(v_j,v_i)=-f(y,x).$$