# Bivector

#### smile

##### New member
Hello everyone

Can anyone help me to solve the following problem

Prove that for any bivector $v\in \Lambda^2 (V)$ there is a basis $\{e_1,e_2,...,e_n\}$ of $V$ such that $v=e_1\Lambda e_2 +e_3\Lambda e_4 +...+e_{k-1}\Lambda e_k$

I did it in this way, since the out product $e_i\Lambda e_j$ forms a bivector in $\Lambda^2 (V)$ then, it forms a basis for $v$. However, I am not sure about the dimension of the bivector $v$ how many basis do we need to form the bivector $v$?

Thanks