Symplectic vector space

A vector space $V$ together with a non-degenerate alternating form

The alternating form plays the role of pairing the coordinates of the space. The canonical example is ${\mathbb{R}}^{2n}$ with coordinates $(q^1,\dots ,q^n,p_1,\dots ,p_n)$ and

so that each $q^i$ is paired with its $p_i$, and the space splits into the distinguished planes ${\mathrm{\Pi }}_i=⟨q^i,p_i⟩$.

From a vector to a covector. Fix $v\in V$. The map

is linear, so it is a covector. Its kernel is a hyperplane of $V$ (equivalently, its level sets are a family of parallel hyperplanes, the picture of a covector in visualization of k-forms). This way, to each vector the form associates a hyperplane, and therefore a covector. Since $\omega$ is non-degenerate, a nonzero vector gives a nonzero covector, and the correspondence is an isomorphism $V\cong V^{\ast }$, similar to the musical isomorphism in dual vector spaces.

Intuition.
The form is a device that measures parallelograms: given $u,v\in V$, the number $\omega (u,v)$ is computed by projecting the parallelogram spanned by $u$ and $v$ onto each of the distinguished planes ${\mathrm{\Pi }}_i$ and adding up the (signed) areas of these projections. This is the measurement picture of visualization of k-forms.
When we fix a vector $u$, vectors $v$s such that $\omega (u,v)=0$ are those who after projection of $u$ and $v$, are proportional to $u$. In the canonical example above, if $u={\partial }_{q_1}$, the $v$s are precisely all the ones whose component in ${\partial }_{p_1}$ is zero.

Related: symplectic form, dual vector space.