Dimension of the space of alternating n-linear maps Given a vector space $V$ with dimension $k$ over a field $F$, I have to show that the dimension of the space of alternating n-linear functions from V to F is $k \choose n$. How would I go about doing that? What important property of alternating multilinear maps do I have to consider?  I'm struggling to approach the problem. 
 A: Let $V^{k}$ stand for the k-fold cartesian product of $V$. Suppose $V$ has dimension $n$ and a basis $e_1,\dots,e_n$ of $V$ and let $\epsilon_1,\ldots,\epsilon_n$ be the corresponding dual basis of $V^{*}$.
The space of multilinear maps $V^{k} \to F$ has a basis elements of the form $\epsilon_{i_1,\ldots,i_k}(v_1,\ldots,v_k) = \epsilon_{i_{1}}(v_1)\ldots\epsilon_{i_{k}}(v_k)$ and thus has dimension $n^k$. The coordinates of a multilinear map with respect to this basis the values the map takes on a basis element. 
The space of alternating multilinear maps is a subspace of this. All you need to do is work out how the additional information $A$ is alternating allows you to deduce what it's coordinates are without evaluating it on all of the basis vectors. You will find that you can deduce the coordinates by evaluating $A$ on $k \choose n$ basis vectors. Proving that the dimension of the space is $n \choose k$ is not too hard from this point.
As a hint, alternating multilinear maps are antisymmetric. 
A: Let $L^{n}(V_{k}, \mathbb{K})$ be the space of all $n$-multilinear forms. In Dimension of vector space of multilinear form it is shown that $ \dim L^{n}(V_{k}, \mathbb{K}) = k^{n}$. Let $L_{a}^{n}(V_{k}, \mathbb{K})$ vector space of all alternating $n$-multilinear forms and let $f \in L^{n}_{a}(V_{k}, \mathbb{K})$. We claim that
$$\dim L^{n}_{a}(V_{k}, \mathbb{K}) = \binom{k}{n} = \frac{k!}{(k - n)! n!}.$$
To show that consider a basis $\{a_{1}, \dots, a_{k}\}$ of $V$, then we can write
$$
x_{j} = x_{1j}a_{1} + x_{2j} a_{2} + \dots + x_{kj}a_{k} = \sum_{i_{j} = 1}^{k} x_{i_{j} j} a_{i_{j}}.
$$
Then, by the linearity of $f$ at each component we have that
$$
f(x_{1}, \dots, x_{n}) = \sum_{i_{1}, \dots, i_{n} = 1}^{k} x_{i_{1}1} x_{i_{2}2} \dots x_{i_{n} n}f(a_{i_{1}}, \dots, a_{i_{n}}).
$$
Let $\mathcal{F}$ denote the set of all functions from $\{1, \dots, n\}$ to $\{1, \dots, k\}$. If $t \in \mathcal{F}$ is the function $t(j) = i_{j}$, let's write
$$
x_{t} := x_{t(1)1}x_{t(2)2} \dots x_{t(n)n}, \quad 
a_{t}:= (a_{t(1)}, \dots, a_{t(n)}).
$$
Hence, we can express $f$ as
$$
f = 
\sum_{t \in \mathcal{F}} x_{t}f(a_{t}).
$$
Since $f$ is an alternating multilinear form, then $f(a_{t}) = 0$ if $t$ is not inyective (in this case $f$ would have two equal components). Hence,
$$
f = \sum_{t \in \text{injec}(\mathcal{F})} x_{t} f(a_{t})
$$
where $\text{injec}(\mathcal{F})$ denotes the subset of all injective functions on $\mathcal{F}$ (there are $k!/(k-n)!$ functions on this set). Let $\mathcal{G}_{n}$ the set of all permutations of $\{1, \dots, n\}$, and let $\sigma \in \mathcal{G}_{n}$, then we define the function $\sigma f$ by
$$
\sigma f(x_{1}, \dots, x_{n}) = f(x_{\sigma(1)}, \dots, x_{\sigma(n)}).
$$
By definition, since $f$ is an alternating function, then $\sigma f = \text{sig}(\sigma)f$, where $\text{sig}(\sigma) = 1$ if $\sigma$ is an even permutation and $-1$ otherwise. Note that $t \circ \sigma \in \text{injec}(\mathcal{F})$ and
$$
f(a_{t \circ \sigma}) = \sigma f(a_{t}) = \text{sig}(\sigma) f(a_{t}).
$$
Now, consider the set $A_{n,k}$ of all increasing functions from $\{1, \dots, n\}$ to $\{1, \dots, k\}$ (there are $\binom{k}{n} = k!/[(k-n)!n!]$ elements in this set.). In addition, note that
$$
\text{injec}(\mathcal{F}) = 
\{t \circ \sigma : t \in A_{n, k}, \sigma \in \mathcal{G}_{n}\}.
$$
Then we can write
$$
f =
\sum_{t \in \text{injec}(\mathcal{F})}
f(a_{t})x_{t} =
\sum_{t \in A_{n,k}} 
\sum_{\sigma \in \mathcal{G}_{n}}
f(a_{t \circ \sigma}) x_{t \circ \sigma} = 
\sum_{t \in A_{n,k}} 
f(a_{t})
\left(
\sum_{\sigma \in \mathcal{G}_{n}}
\text{sig}(\sigma) x_{t \circ \sigma} 
\right)
$$
The final step is to show that
$$
v_{t}(x_{1}, \dots, x_{n}) = 
\sum_{\sigma \in \mathcal{G}_{n}}
\text{sig}(\sigma) x_{t \circ \sigma} 
$$
is an alternating $n$-multilinear form for each $t \in A_{n, k}$ and that the set $\{v_{t}: t \in A_{n, k}\}$ is linearly independent and then it is basis of $L_{a}^{n}(V_{k}, \mathbb{K})$ with $\binom{k}{n}$ elements.
