Question about Alternating forms 
So I understand the definition of an alternating form on $\mathbb{R}^m$, but I don't really understand the proof of the lemma. Could someone explain the first observation?  Why is it so?
 A: Hint for (1): Consider $e_{j_1} = (0, \dots, 0, 1, 0, \dots, 0)$ where the $1$ appears in the $j_1^{\ \text{th}}$ entry. As per the definition of $dx^I$, we need to know what $e_{j_1}^{i_1}, \dots, e_{j_1}^{i_k}$ are as they form the entries in the first row of the matrix. Almost all the components of $e_{j_1}$ are zero, the only one which isn't is $e_{j_1}^{j_1}$ which is $1$. So, if $j_1$ is not one of the entries in $I$, then the first row of the matrix consists of zeroes only so $dx^I(e_{j_1}, \dots, e_{j_k}) = 0$. 
The same analysis above applies to $e_{j_2}$ and the second row, $e_{j_3}$ and the third row and so on. The final piece of information you need to use is that $I, J \in \mathcal{I}_k$ which is the set of all ordered $k$-tuples with strictly increasing components.
Hint for (2): The implication left to right is trivial, for the other direction note that $\omega = 0 \Leftrightarrow \omega(\xi_1, \dots, \xi_k) = 0$ for all $\xi_1, \dots, \xi_k \in \mathbb{R}^m$. Use the linearity of $\omega$ to write $\omega(\xi_1, \dots, \xi_k)$ in terms of $\omega(e_{i_1}, \dots, e_{i_k})$.
