Existence of a nonzero vector to form Let $ f: \mathbb {R}^m\times \mathbb {R}^m \rightarrow \mathbb {R}^m $ an alternate form of grade two.
If $ m $ is odd, prove that there exists $ v\neq 0 $ such that $ f (u, v) = 0 $, for all $ u \in \mathbb {R}^m$.
Thanks for any suggestions.
 A: Let $M = (f(e_i, e_j))_{1 \leqslant i,j \leqslant m}$ be the matrix associated to $f$ (where $(e_1, \dots, e_m)$ is the canonical basis of $\mathbb{R}^m$). It is a skew-symmetric matrix (why?). What can you derive on $\det M$?
Edit: responding to your comment. Let $v \in \mathbb{R}^m$ be a nonzero vector such $M \,v = 0$ (why does this exist?). Show that $v$ answers your question.
A few comments: Here are a couple general facts that are probably useful for you to know:
Let $V$ be a vector space with a given basis $(e_1, \dots, e_n)$.
If $f$ is a bilinear form on $V$, it has an associated matrix defined as above.
For any two vectors $x, y \in V$, we have the relation
$$f(x,y) = X^T \, M \,Y$$
Where $X$ and $Y$ are the column vectors associated to $x$ and $y$ using the given basis (show this relation).
Vectors $x$ (resp. $y$) satisfying $f(x, y) = 0$ for any $y$ (resp. $x$) in $V$ are the elements of what is called the left (resp. right) nullspace of $f$. If $f$ is symmetric or alternate, the left and right nullspaces are the same (show this).
In general, using the given basis, the left nullspace corresponds to the left nullspace (or cokernel) of $M$ while the right nullspace corresponds to the kernel of $M$ (show this).
