$n$-linear alternating form with $\dim{V}Prove that every $n$-linear alternating form on a vector space of dimension
less than $n$ is the zero form.
 A: Let $k$ be a field, and let $V$ be a vector space over $k$, with $\dim V <n$.  Let $\omega$ be an alternating $n$-form on $V$.  Then $\omega : V^n \to k$ is an alternating, multilinear map.  For general input $v_1,\ldots,v_n$, there exists a linear dependence $\sum c_i v_i=0$ among the $v_i$, by dimension of $V$.  If $c_n \neq 0$ (e.g.), we can scale $c_n=1$.  Then solving for $v_n$, we may write
$$\omega(v_1,\ldots,v_n)=\omega\left(v_1,v_2,\ldots,v_{n-1},-\sum_{i<n} c_iv_i\right)=-\sum_{i<n}c_i \omega\left(v_1,v_2,\ldots,v_{n-1},v_i\right),$$
in which the last step uses multilinearity.  Since $\omega$ is alternating, switching the $i$th and $n$th inputs negates the value of $\omega$.  Yet here $\omega(v_1,\ldots,v_{n-1},v_i)$ is clearly invariant under such a permutation, hence
$$\omega(v_1,\ldots,v_{n-1},v_i)=0$$
for all $i$.  Thus $\omega$ is the zero map, as desired.
In general, we can show (by constructing a basis) that the dimension of the space of alternating $n$-forms on a $d$-dimensional space is given by $\binom{d}{n}$, which vanishes for $n > d$.
A: Just use the definition of alternating, along with the fact that multilinearity says that the function is totally determined by what happens when you put basis vectors in each slot.
A: Let $\omega$ be an $n$-multilinear, alternating form on $V$, where $\dim(V)=m<n$. Let $\{e_1,\ldots,e_m\}$ be any basis of $V$. Because $\omega$ is multilinear, $\omega$ is determined entirely by its values on $n$-tuples of elements from $\{e_1,\ldots,e_m\}$ (apply $\omega$ to an $n$-tuple of arbitrary vectors; write each of them as a linear combination of the $e_i$'s, and use the linearity of $\omega$ in each entry). But $n>m$, so every such $n$-tuple has a repeat...
