I am trying to prove that if $A \in M_{n\times n}(F)$, then
$$\det(A)=\sum_{\sigma\in S_n}\operatorname{sgn}(\sigma) \prod_{i=1}^n a_{i \sigma(i)}\tag 1$$
The question asks me to use the following theorem:
If $\delta: M_{n×n}(F) → F$ is an alternating $n-$linear function such that $\delta(I) = 1$, then $\delta(A) = \det(A)$ for every $A ∈ M_{n×n}(F)$.
The way I am interpreting the question is that I need to show that it satisfies the multilinear (n-linear), alternating, and normalized properties (the formula maps the identity matrix to 1) of the determinant.
Then the second part of the question asks to show that if a single column of A is completely comprised of $0$'s, then $\det(A)=0$. I feel that the solution to second part of the question should be clearer after solving for the first part.
How do I go about doing this?
For the normalised property and the second part, here's a hint: $\prod_{i=1}^n a_{i,\sigma(i)}$ can't contain more than one entry from the same row or the same column.
– Randy Marsh Jul 29 '20 at 21:18