[example: Standard Symplectic Vector Space]
Let $V=\mathbb R^{2n}$ with coordinate functions $(q_1,\dots,q_n,p_1,\dots,p_n)$, and define
\begin{align*}
\omega_0=\sum_{i=1}^n dq_i\wedge dp_i.
\end{align*}
We verify directly that $\omega_0$ is alternating and nondegenerate. For tangent vectors
\begin{align*}
u=\sum_{i=1}^n a_i\frac{\partial}{\partial q_i}+b_i\frac{\partial}{\partial p_i}
\end{align*}
and
\begin{align*}
v=\sum_{i=1}^n c_i\frac{\partial}{\partial q_i}+d_i\frac{\partial}{\partial p_i},
\end{align*}
the identity $(\alpha\wedge\beta)(u,v)=\alpha(u)\beta(v)-\alpha(v)\beta(u)$ gives
\begin{align*}
(dq_i\wedge dp_i)(u,v)=dq_i(u)dp_i(v)-dq_i(v)dp_i(u)=a_i d_i-c_i b_i.
\end{align*}
Therefore
\begin{align*}
\omega_0(u,v)=\sum_{i=1}^n(a_i d_i-b_i c_i).
\end{align*}
Interchanging $u$ and $v$ gives
\begin{align*}
\omega_0(v,u)=\sum_{i=1}^n(c_i b_i-d_i a_i)=-\sum_{i=1}^n(a_i d_i-b_i c_i)=-\omega_0(u,v),
\end{align*}
so $\omega_0$ is alternating.
To prove nondegeneracy, suppose $\omega_0(u,v)=0$ for every $v\in V$. Taking $v=\partial/\partial p_j$ gives $c_i=0$ for all $i$, $d_j=1$, and $d_i=0$ for $i\ne j$, hence
\begin{align*}
0=\omega_0\left(u,\frac{\partial}{\partial p_j}\right)=a_j.
\end{align*}
Taking $v=\partial/\partial q_j$ gives $c_j=1$, $c_i=0$ for $i\ne j$, and $d_i=0$ for all $i$, hence
\begin{align*}
0=\omega_0\left(u,\frac{\partial}{\partial q_j}\right)=-b_j.
\end{align*}
Thus $a_j=b_j=0$ for every $j$, so $u=0$. Hence $(\mathbb R^{2n},\omega_0)$ is a symplectic vector space, and the formula shows explicitly that each $q_i$-direction pairs only with its corresponding $p_i$-direction.
[/example]