Let $f$ a alternating function. prove $f(v_1,..,v_d)=0$ Let $f:V^d\rightarrow W$ a d-linear alternating function. Prove if $\{w_1,...,w_d\}$ is linearly dependent, then $f(w_1,...,w_d)=0$.
I try this:
Let $(w_1,...w_d)\in V^d$ and $\beta_i\in\mathbb{K}$ with $1\leq i \leq d$ as $\{w_1,...,w_d\}$ is linearly dependent, then
$w_i=\beta_1 w_1+.+\beta_{i-1}w_{i-1}+\beta_i+1w_{i+1}+..+\beta_dw_d$
Moreover,
$f(w_1,\cdots,w_i,\cdots,w_d) = f(w_1,\cdots,\sum_{j=1, j\neq i} \beta_j w_j, \cdots,w_d)$
As $f$ is lineal then
$f(w_1,\cdots,\sum_{j=1, j\neq i} \beta_j w_j, \cdots,w_n)=\sum_{j=1, j\neq i} \beta_jf(w_1,\cdots,w_j,\cdots,w_d)$
In this step i'm stuck, can someone help me with the next step? Thanks.
 A: Suppose $w_1 = \sum_{k >1} \alpha_k w_k$ then
$f(w_1,...,w_n) = \sum_{k >1} \alpha_k f(w_k,...,w_n)$.
Show that $f(w_k,...,w_n) = 0$ for $k >1$.
A: Up to a permutation of indices, there exists $\lambda_1,\ldots,\lambda_{n-1}\in\mathbb{K}$ such that: 
$$w_n:=\sum_{i=1}^{n-1}\lambda_iw_i.$$ Therefore, one gets:
$$f(w_1,\ldots,w_n)=\sum_{i=1}^{n-1}\lambda_if(w_1,\ldots,w_{n-1},w_i).$$
Let $i\in\{1,\ldots,n-1\}$, permuting indices $i$ and $n$, one has:
$$f(w_1,\ldots,w_{n-1},w_i)=-f(w_1,\ldots,w_{n-1},w_i).$$
Assuming you work over a field of characteristic different from $2$, one gets:
$$f(w_1,\ldots,w_{n-1},w_i)=0.$$
Whence the result.
A: You've done well up to the point where you stopped. You are asking why your last sum is zero. This is because all those $f (w_1,\ldots,w_j,\ldots,w_n) $ are zero.
Note that your notation is a bit misleading: $f (w_1,\ldots,w_j,\ldots,w_n) $ is actually $f (w_1,\ldots,w_j,\ldots,w_j,\ldots,w_n) $, i.e. $w_j $ appears twice, once at index $i $ and again at index $j $. It is well-known that alternating forms become zero in that case.
