Looking at forms in local charts and the dimension of symplectic manifolds I am reading a very simple proof that symplectic manifolds are even dimensional. The definition I was given is: 
A smooth manifold $M$ is a symplectic manifold if it has a closed, non-degenerate 2-form $\omega$.
So now, the non-degeneracy assumption implies that $M$ is even dimensional. And this is because at every $m\in M$, the components of $\omega$ at any local chart represent an $n\times n$ skewsymmetric matrix $W$. Now the non-degeneracy assumption implies $\det(W)\neq 0$ and since $W^T = -W $, we have that $\det(W)=\det(W^T)=(-1)^n \det(W)$, therefore $n$ must be even.
The part that I don't get is how can we see the components of $\omega$ as an $n\times n$ matrix. Since $\omega$ is a section of $\Lambda^2(M)\rightarrow M$, I understand that at a point $m$, $\omega(m)$ is in $\Lambda^2(T_m M)\subset T^*_m M\otimes_{\mathbb{R}}T^*_m M$. Wouldn't this mean that I plug two $n$-dimensional vectors and I get a scalar?
I am still struggling trying to understand the objects so any help would be greatly appreciated.
 A: Like you say, a symplectic form is in particular a bilinear form (on each tangent space). Given a bilinear form $g \colon V \times V \rightarrow \mathbb{R}$ on a finite dimensional real vector space and a choice $\beta =(v_1,\dots,v_n)$ of basis for $V$, we say that the matrix $G = (g_{ij})$ whose entries are $g_{ij} := g(v_i,v_j)$ is the matrix representing $g$ with respect to the basis $\beta$. If $g$ is symmetric/skew-symmetric, the matrix $G$ will also be symmetric/skew-symmetric.
In your case, the meaning of the sentence "the components of $\omega$ at any local chart" is:


*

*You pick some local chart $(x^1,\dots,x^n)$ defined on $U \subseteq M$.

*The chart induces a basis $\beta(q) = \left( \frac{\partial}{\partial x^1}|_{q}, \dots, \frac{\partial}{\partial x^n}|_{q} \right)$ for each $T_{q} M$ (with $q \in M$). Such a basis is called a local frame of $TM$.

*You represent the bilinear form $\omega|_{q} \colon T_q M \times T_q M \rightarrow \mathbb{R}$ with respect to the basis $\beta(q)$ and get a matrix $(\omega_{ij}(q))_{i,j=1}^n$ (that depends on $q$). Since $\omega|_{q}$ is skew-symmetric, so is the matrix $(\omega_{ij}(q))_{i,j=1}^n$. The entries $\omega_{ij}(q)$ are called the local components of the bilinear form $\omega$ with respect to the local chart $(x^1,\dots,x^n)$ (because they were obtained by plugging into $\omega$ the corresponding tangent vectors induced by the local chart). 

