Producting alternating or not? A product $[., .]:K^n \times K^n \rightarrow K$ is alternating (=the form has a value of $0$ whenever two of its arguments are equal) if 
$$\forall v, w \in K^n: [v, w]+[w, v]=0$$
(so $[v, w]$ means $vw$)
($K$ is a field, $n$ ist the dimension, so for example $\mathbb{R}^2$)
I don't know if the statement is true or false and would appreciate some help/solution. Maybe it helps considering $char(K)=2$.
Thanks!
 A: As far as I understand you define being alternating as being equal to $0$ when the arguments are equal and you want to show that $\forall v,w\in K^n, [v,w]+[w,v]=0$. This is of course false if you don't have any assumption on $[.,.]$. But if you know that $[.,.]$ is bilinear and char$(K)\neq 2$, then the answer is yes, they are equivalent : 
If $[v,w]+[w,v]=0$ for all $v,w$ then $$0=[v,v+w]+[v+w,v]=[v,v]+[v,w]+[v,v]+[w,v]=2\cdot [v,v]$$
(Hence if char$(K)\neq 2$ you can divide by $2$ to get $[v,v]=0$)
If $[v,v]=0$ for all $v$, then $$0=[v+w,v+w]=[v,v]+[v,w]+[w,v]+[w,w]=[v,w]+[w,v]$$ (Note that I did not use anything about the characteristic of the field here, i.e. every alternating form satisfies the required property)
Edit : If for all $v,w$ you have $[v,w]+[w,v]=0$, the bilinear form is called skew-symmetric. When char$(K)\neq 2$, alternating and skew-symmetric forms are the same but if char$(K)=2$ then every alternating form is skew-symmetric whereas there are skew-symmetric forms that are not alternating. Here is an example :
Set $V=\mathbb{F}_2$ and the map $F:V\times V\rightarrow \mathbb{F}_2$ by $[a,b]=a\cdot b$. Then $[0,0]=[1,0]=[0,1]=0$ hence the form is skew-symmetric however it is not alternating since $1=[1,1]\neq 0$.
