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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!

Xsy
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    What if $f$ is symmetric ? – Rene Schipperus Jun 19 '14 at 22:41
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    This is false, as it implies the bilinear form is antisymmetric, which is not true in general. – Cheerful Parsnip Jun 19 '14 at 22:42
  • @ReneSchipperus I'm a bit confused now. If a billinear form is symmetric, does it mean that in any basis the matrix will also be symmetric? Another question in continuation to what GrumpyParsnip wrote, Is it possible that the condition will indeed work for a certain basis for every billinear form, but the billinear form itself be symmetric? – Xsy Jun 19 '14 at 23:50
  • it seems you missed the obvious condition $v_i \ne v_j$. – achille hui Jun 20 '14 at 00:03

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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 $$f(x,y)=\sum_{i,j} r_is_j f(v_i,v_j)=-\sum_{i,j} r_is_j f(v_j,v_i)=-f(y,x).$$

Cheerful Parsnip
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  • Thank's @GrumpyParsnip, I understand my initial error now. But I still do not understand how to prove the statement. Because it refers to all billinear forms, and specifically to symmetric one, how can a symmetric form carry out that equality? Is the question botched? – Xsy Jun 20 '14 at 07:39
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    @Xsy: you can't prove the statement because it's false. (Botched question?!) – Cheerful Parsnip Jun 20 '14 at 08:39
  • Heh, well at least I didn't completely waste my time working on that one, learned a few things. Though I do have to wonder how the professor could put this question on a test, pretty sloppy. Anyways, Thank's for the clarification :) – Xsy Jun 20 '14 at 08:46
  • @XSy: what's the source of this problem? Two other people asked it on MSE. It seems that if you include the condition $v_i\neq v_j$, then it is indeed possible to find such a basis. – Cheerful Parsnip Jun 21 '14 at 22:34
  • The source is from a collection of tests from a professor in my university, which is in Hebrew. The condition you mentioned is not in the question description, but isn't it true for every basis anyway? – Xsy Jun 22 '14 at 05:50
  • @Xsy thanks for the info. If you assume that $f(v_i,v_j)=-f(v_j,v_i)$ only when $i\neq j$, then the form is not necessarily antisymmetric, so my proof doesn't apply. – Cheerful Parsnip Jun 22 '14 at 06:15
  • So if I assume such a condition, then a suitable basis should be one where the matrix doesn't necessarily has a zero diagonal, but how can you find such a basis for every $f$? – Xsy Jun 22 '14 at 07:05