In $\mathbb{R}^{n}$, consider k covectors $\mathbf{E}_{i_{1}},...,\mathbf{E}_{i_{k}}$ chosen from the standard dual basis of $\mathbb{R}^{n}$ , where $k\in\mathbb{N}$ is arbitrary. Then, $\mathbf{E}_{i_{1}}\wedge\mathbf{E}_{i_{2}}\wedge...\wedge\mathbf{E}_{i_{k}}$ is in $\textrm{Alt}^{k}\left(\mathbb{R}^{n}\right)$ (it is an alternating k form); it accepts k vectors in $\mathbb{R}^{n}$ as input, and is given by the formula:$\left(\mathbf{E}_{i_{1}}\wedge\mathbf{E}_{i_{2}}\wedge...\wedge\mathbf{E}_{i_{k}}\right)\left(\mathbf{v}_{1},...,\mathbf{v}_{k}\right)=\textrm{det}\left(\begin{array}{cccc} \mathbf{E}_{i_{1}}\left(\mathbf{v}_{1}\right) & \mathbf{E}_{i_{1}}\left(\mathbf{v}_{2}\right) & \cdots & \mathbf{E}_{i_{1}}\left(\mathbf{v}_{k}\right)\\ \mathbf{E}_{i_{2}}\left(\mathbf{v}_{1}\right) & \mathbf{E}_{i_{2}}\left(\mathbf{v}_{2}\right) & \cdots & \mathbf{E}_{i_{2}}\left(\mathbf{v}_{k}\right)\\ \vdots & & \ddots\\ \mathbf{E}_{i_{k}}\left(\mathbf{v}_{1}\right) & \mathbf{E}_{i_{k}}\left(\mathbf{v}_{2}\right) & \cdots & \mathbf{E}_{i_{k}}\left(\mathbf{v}_{k}\right) \end{array}\right)$
I take this expression as the definition of the wedge product of the above covectors.
Now, let $\ell\in\mathbb{N}$ be arbitrary, let $\alpha=\mathbf{E}_{i_{1}}\wedge\mathbf{E}_{i_{2}}\wedge...\wedge\mathbf{E}_{i_{k}}$ be the form from above, and let $\beta=\mathbf{E}_{j_{1}}\wedge\mathbf{E}_{j_{2}}\wedge...\wedge\mathbf{E}_{j_{\ell}}$ be a form given by the wedge product of $\ell$ standard dual basis covectors. I know that $\alpha\wedge\beta\in\textrm{Alt}^{k+\ell}\left(\mathbb{R}^{n}\right)$. What is the formula for $\left(\alpha\wedge\beta\right)\left(\mathbf{v}_{1},...,\mathbf{v}_{k},\mathbf{w}_{1},...,\mathbf{w}_{\ell}\right)$ written as a determinant in the manner of the formula given above for $\alpha$ ?
(Please write the answer as a the determinant of a matrix in the manner used above, no abbreviations. And no patronizing for me wanting a simple formula with respect to a given basis. Thank you.)
Well, it would be the same formula (only with the indexes appropriately modified)! Namely,
$$(\alpha \wedge \beta)\left(\mathbf{v}_{1},...,\mathbf{v}_{k}, \mathbf{w}_{1},...,\mathbf{w}_{l}\right) = \\ \textrm{det}\left(\begin{array}{cccccccc} \mathbf{E}_{i_{1}}\left(\mathbf{v}_{1}\right) & \mathbf{E}_{i_{1}}\left(\mathbf{v}_{2}\right) & \cdots & \mathbf{E}_{i_{1}}\left(\mathbf{v}_{k}\right) & \mathbf{E}_{i_1}(\mathbf{w}_1) & \cdots & \mathbf{E}_{i_1}(\mathbf{w}_l) \\ \mathbf{E}_{i_{2}}\left(\mathbf{v}_{1}\right) & \mathbf{E}_{i_{2}}\left(\mathbf{v}_{2}\right) & \cdots & \mathbf{E}_{i_{2}}\left(\mathbf{v}_{k}\right) & \mathbf{E}_{i_2}(\mathbf{w}_1) & \cdots & \mathbf{E}_{i_2}(\mathbf{w}_l) \\ \vdots & \ddots & \ddots & \vdots & \ddots & \ddots & \vdots \\ \mathbf{E}_{i_{k}}\left(\mathbf{v}_{1}\right) & \mathbf{E}_{i_{k}}\left(\mathbf{v}_{2}\right) & \cdots & \mathbf{E}_{i_{k}}\left(\mathbf{v}_{k}\right) & \mathbf{E}_{i_k}(\mathbf{w}_1) & \cdots & \mathbf{E}_{i_k}(\mathbf{w}_l) \\ \mathbf{E}_{j_{1}}\left(\mathbf{v}_{1}\right) & \mathbf{E}_{j_{1}}\left(\mathbf{v}_{2}\right) & \cdots & \mathbf{E}_{j_{1}}\left(\mathbf{v}_{k}\right) & \mathbf{E}_{j_1}(\mathbf{w}_1) & \cdots & \mathbf{E}_{j_1}(\mathbf{w}_l) \\ \mathbf{E}_{j_{2}}\left(\mathbf{v}_{1}\right) & \mathbf{E}_{j_{2}}\left(\mathbf{v}_{2}\right) & \cdots & \mathbf{E}_{j_{2}}\left(\mathbf{v}_{k}\right) & \mathbf{E}_{j_2}(\mathbf{w}_1) & \cdots & \mathbf{E}_{j_2}(\mathbf{w}_l) \\ \vdots & \ddots & \ddots & \vdots & \ddots & \ddots & \vdots \\ \mathbf{E}_{j_{l}}\left(\mathbf{v}_{1}\right) & \mathbf{E}_{j_{l}}\left(\mathbf{v}_{2}\right) & \cdots & \mathbf{E}_{j_{l}}\left(\mathbf{v}_{k}\right) & \mathbf{E}_{j_l}(\mathbf{w}_1) & \cdots & \mathbf{E}_{j_l}(\mathbf{w}_l) \\ \end{array}\right).$$