Consider the problem from my assignment on tensor forms:
Problem: We say that a 2-form n on a vector space V is non-degenerate of for $u\in V$ , n(u,v)=0 for all v$\in V$ implies u=0. Let w be a non-degenerate 2-form on a vector space V.
Then show that:(a) Show that V is even dimensional.
(b) If dim(V)= 2n, show that there exists a basis {$e^{*}_1,...,e^{*}_{2n}$} of $V^{*}$ such $w=e^{*}_{1} \bigwedge e^{*}_{n+1} + e^{*}_2 \bigwedge e^{*}_{n+2} + e^{*}_n \bigwedge e^{*}_{2n}$.
Thoughts:(a)Let on the contrary V is odd dimensional. But I am not sure what result should I use to proceed towards contradiction.
(b) I have to construct such a basis. $V^{*}$ will also have a dimension 2n. I have studied wedge product and it's properties from my class notes but I am still not sure how can I proceed towards such a construction.
Kindly give some hints.
Thanks!
To show the space is even dimensional, let $v_1,\dots, v_n$ be a basis, define $ \omega_{ij} = \omega(v_i,v_j)$ so you can think this a matrix coefficient. Show this matrix is skew-symmetric (i.e $A^T = -A $) . Show that this matrix invertible using non-degeneracy. Then proof is reduced to showing every skew-symmetric matrix of odd dimension is singular. Use $\det(A) = \det(A^T)$ and being skew-symmetric.
Constructing such basis of dual space is same as finding a basis such that $$\omega(e_i,e_{i+n}) = 1 $$ and $$\omega(e_i,e_j) = 0 \text{ when }|j-i|\neq n$$ Pick a basis than apply gram-schmidt but instead of inner product use the 2 form