Example 22.2 Lee $\omega=\sum_{i=1}^n \alpha^i \wedge \beta^i$ is a symplectic form In Example 22.2 of "Introduction to smooth manifolds" by Lee, I'm trying to understand why $\omega$ is a symplectic form by performing explicitly the necessary passages, but I think I don't manage wedge product and differential forms.
Let $V$ be a $2n$-vector space with basis $ (A_1,B_1, \dots, A_n,B_n )$ and let $(\alpha_1,\beta_1, \dots, \alpha_n,\beta_n )$ be the corresponding dual basis for $V^*$. Let $\omega \in (\Lambda^2(V^*))$
$$\omega=\sum_{i=1}^n \alpha^i \wedge \beta^i$$
If $ (A_1,B_1, \dots, A_n,B_n )=(\frac{\partial}{\partial x_1},\frac{\partial}{\partial y_1},\dots,\frac{\partial}{\partial x_n}, \frac{\partial}{\partial y_n})$ and if $v=a^i\frac{\partial}{\partial x_i}+b^i\frac{\partial}{\partial y_i} \in V$, why $\omega(v,\frac{\partial}{\partial x_i})=-b^i$?
 A: From a well known property I also think is mentioned in Lee's book (Edit: Prop. 14.11 up to a transpose), the wedge product of covectors can be computed explicitly as
$$\alpha^i \wedge \beta^i\left ( \frac{\partial}{\partial x_k},\frac{\partial}{\partial x_j} \right )=\text{det}\begin{bmatrix}
\alpha^i\left ( \frac{\partial}{\partial x_k} \right ) & \alpha^i\left ( \frac{\partial}{\partial x_j} \right )\\ 
\beta^i\left (  \frac{\partial}{\partial x_k}\right ) & \beta^i\left ( \frac{\partial}{\partial x_j} \right )
\end{bmatrix} = 
\text{det}\begin{bmatrix}
\delta_{ik}& \delta_{ij}\\ 
0 & 0
\end{bmatrix}=0$$
And
$$\alpha^i \wedge \beta^i\left ( \frac{\partial}{\partial y_k},\frac{\partial}{\partial x_j} \right )=\text{det}\begin{bmatrix}
\alpha^i\left ( \frac{\partial}{\partial y_k} \right ) & \alpha^i\left ( \frac{\partial}{\partial x_j} \right )\\ 
\beta^i\left (  \frac{\partial}{\partial y_k}\right ) & \beta^i\left ( \frac{\partial}{\partial x_j} \right )
\end{bmatrix} = 
\text{det}\begin{bmatrix}
0& \delta_{ij}\\ 
\delta_{ik} & 0
\end{bmatrix}=-\delta_{ij}\delta_{ik}$$
Use multilinearity to reduce to these cases. As you can see, only the term $\alpha^i \wedge \beta^i\left ( \frac{\partial}{\partial y_i},\frac{\partial}{\partial x_i} \right )=-1$ won't vanish.
