Let $\mu$ be a $2$-form in a vector space $V$ with dimension $2n$. Prove that $\mu$ is non-degenerate $\Leftrightarrow \mu^{\wedge n}=\mu\wedge...\wedge \mu\neq 0$.
I can prove $\Rightarrow$, by using the fact that there exists a basis $\{x_1,y_1,...,x_n, y_n \}$ for $V$ such that $\mu(x_i, y_j)=\delta_{ij}$ and $\mu(x_i,x_j)=0$, so that $\mu^{\wedge n}(x_1,y_1,...,x_n, y_n)\neq 0$. But I could not prove $\Leftarrow$.
I find it strange, because I can only assume $\mu^{\wedge n}(x_1,...,x_{2n})\neq 0$ for some $2n$-uple $(x_1,...,x_{2n})$, which is a specific case, but then I have to prove that for every $x\in V-\{0\}$ there is a $y\in V$ such that $\mu(x, y)\neq 0$, which is a general case.
How can I do that?
Suppose by contradiction that exists a vector $u \in V\setminus\{0\}$ such that $\mu(u,v) = 0$ for every vector $v$. As you said, there exist a basis $\{x_1, \ldots, x_{2n}\}$ such that $\mu^{\wedge n}(x_1, \ldots, x_{2n}) = 1$ (if it is necessary you can rescale). Since those vectors form a basis, we have that $u = \sum\limits_{i=1}^{2n}\beta_i x_i$. Wlog, suppose $\beta_1 \neq 0$, it is straightforward that $\mu^{\wedge n}(u,x_2 \ldots, x_{2n}) = \beta_1$. On the other hand, it can be shown by induction that $\mu^{\wedge k}(u, v_1, \ldots, v_{2k-1}) = 0$, since $\mu(u,v) = 0$. Finally, these two facts leads you to a contradiction.