# Second exterior power canonical form

There is a canonical form: there is a basis $e_1,\cdots,e_n$ of $V$ and a $k$ such that $w=e_1 \wedge e_2 + \cdots + e_{2k−1} \wedge e_{2k}$.
Why is this true? I wasn't able to come up with a proof. My knowledge of exterior algebra is the most basic. If I understand correctly, vectors $e_{i_1} \wedge e_{i_2}$ form the basis of $\bigwedge^2(V)$, but wouldn't that mean that every $w \in \bigwedge^2(V)$ could be written as $a_1(e_1 \wedge e_2) + a_2(e_1 \wedge e_3) + \cdots$, meaning that not only $e_k \wedge e_{k + 1}$ terms are present, but also $e_k \wedge e_{k + p}$? I'm very confused about this.
• Writing this up is a pain in the ass. Basically, take a matrix that translates the standard basis into the symplectic basis. Either this matrix or its inverse will send $w$ to $e_1 \wedge e_2 + \cdots + e_{2k-1} \wedge e_{2k}$. – darij grinberg Dec 11 '17 at 5:56
• @darijgrinberg thank you, but unfortunately I'm still unable to understand. Why does this matrix (or its inverse, which is it?) send $w$ to the representation in question? Would you maybe know the book where I could look this up if expanding this idea is too time-consuming for you at the moment? – wouldnotliketo Dec 11 '17 at 6:19