Customary layout of $\phi_i(v_j)$

Question

What is the customary layout of $\phi_i(v_j)$? Is it $$\pmatrix{\phi_1(v_1)&\phi_1(v_2) & \cdots \\ \phi_2(v_1) & \phi_2(v_2) & \cdots\\ \vdots &\vdots & \ddots},$$ or$$\pmatrix{\phi_1(v_1)&\phi_2(v_1) & \cdots \\ \phi_1(v_2) & \phi_2(v_2) &\cdots \\ \vdots &\vdots & \ddots}?$$

This question comes from the following problem:

Show that whenever $\phi_1, \ldots, \phi_k \in V^*$, and $v_1, \ldots, v_k \in V$, then $$\phi_1 \wedge \cdots \wedge \phi_p (v_1, \ldots, v_) = \frac{1}{k!}\det[\phi_i(v_j)].$$

Thanks!

Answer 1

The general consensus seems to be that $i$ denotes the row and $j$ the column, so your first variant is correct. Of course the determinant of the transpose is the same so it wouldn't really matter here for the result.