Prove all alternating $d$-linear function $f:V^d\rightarrow W$ is antisymmetric. Prove all  alternating $d$-linear  function $f:V^d\rightarrow W$  is antisymmetric.
Definition: Let $V_1,\ldots,V_d, W$ vector spaces, such that $f:V_1\times\cdots\times V_d\rightarrow W$ is multilinear.
Then $f$ is $d$-linear  when $V_1=\cdots=V_d$.
I don't have idea of how solve this exercise. Can someone give me a hint or how to solve this exercise? Thanks.
 A: Let $i, j \in \{1, \ldots, d\}$, $i \ne j$. We wish to prove:
$$f(x_1, \ldots, x_i, \ldots, x_j, \ldots, x_n) = -f(x_1, \ldots, x_j, \ldots, x_i, \ldots, x_n)$$
for arbitrary $(x_1, \ldots, x_n) \in V^n$.
We have:
\begin{align}
0 &= f(x_1, \ldots, x_i + x_j, \ldots, x_i + x_j, \ldots, x_n)\\ 
&= \underbrace{f(x_1, \ldots, x_i, \ldots, x_i, \ldots, x_n)}_{=0} \\
&\quad+ f(x_1, \ldots, x_i, \ldots, x_j, \ldots, x_n) \\
&\quad+ f(x_1, \ldots, x_j, \ldots, x_i, \ldots, x_n) \\
&\quad+ \underbrace{f(x_1, \ldots, x_j, \ldots, x_j, \ldots, x_n)}_{=0}\\
&= f(x_1, \ldots, x_i, \ldots, x_j, \ldots, x_n) + f(x_1, \ldots, x_j, \ldots, x_i, \ldots, x_n) \\
\end{align}
Rearranging gives:
$$f(x_1, \ldots, x_i, \ldots, x_j, \ldots, x_n) = -f(x_1, \ldots, x_j, \ldots, x_i, \ldots, x_n)$$
A: Special Case: $d=2$.
Let $f:V\times V \to W$ be alternating. Note that every $(x,y) \in V$ can be written as 
Then by assumption, we have that  $f(x,y)+f(y,x)=f(x,y)+f(x,x)+f(y,y)+f(y,x)=f(x+y,x+y)=0$, where the last equality is by assumption.
Hence, we have that $f(x,y)+f(y,x)=0$, or $f(x,y)=-f(y,x)$
Can you generalize?
