Need help by the notation:
For $i \in \{1,...,n\}$
$\displaystyle \det(A)=\sum_{\sigma \in S_n}\operatorname{sgn} (\sigma)\cdot \prod_{k=1}^{n}a_{k,\sigma(k)}=\sum_{j=1}^{n}\sum_{\sigma \in S_n, \atop \sigma(i)=j}\operatorname{sgn}(\sigma)\cdot \prod_{k \in \{1,...,n\}, \atop k \ne i}a_{k,\sigma(k)}\cdot a_{i,j}$
So, if the first equality gives the definition of the determinant, then the notation of the second one is a problem for me at the moment. How should I put $i$-s in it?
I take e.g. an $A \in K^{3 \times 3}$ matrix, so I have $S_3=\{ \operatorname{id} , (1,2), (1,3), (2,3), (1,2,3), (3,2,1)\}$. $\operatorname{sgn}(1,2)=\operatorname{sgn}(1,3)=\operatorname{sgn}(2,3)=-1$. Then I can rewrite the formula:
$\displaystyle \det(A)=\ldots =\sum_{\sigma \in S_3, \atop \sigma(i)=1}\operatorname{sgn}(\sigma)\cdot \prod_{k \in \{1,2,3\}, \atop k \ne i}a_{k,\sigma(k)}\cdot a_{i,1}+\sum_{\sigma \in S_3, \atop \sigma(i)=2}\operatorname{sgn}(\sigma)\cdot \prod_{k \in \{1,2,3\}, \atop k \ne i}a_{k,\sigma(k)}\cdot a_{i,2}+ \sum_{\sigma \in S_3, \atop \sigma(i)=3}\operatorname{sgn}(\sigma)\cdot \prod_{k \in \{1,2,3\}, \atop k \ne i}a_{k,\sigma(k)}\cdot a_{i,3}$
But what about the $i$-s? Should I take all the $i$-s $\in \{1,2,3\}$ or should I pick up only one $i$? Anyway, how do I compute next to get a more simple formula?
As we can write $\sum_{k=1}^{n}a_n$ as $a_1 + a_2 + \ldots + a_n$, I want to do the same thing here. But, anyway, I would be glad to understand what the notations means and how it should be computed with the formula.
Thanks.
UPD: where the second formula comes from at all?
Let $i\in\left\{1,\dots,n\right\}$ be fixed. Define for $j\in\left\{1,\dots,n\right\}$ the set $$E_j:=\left\{\sigma\in\mathcal S_n \mid \sigma\left(i\right)=j\right\}.$$ Since all the elements of $\mathcal S_n$ are bijections, the sets $E_j$ are pairwise disjoint and their union is $\mathcal S_n$. If $\sigma$ belongs to $E_j$, then $$\prod_{k=1}^{n}a_{k,\sigma(k)}= a_{i,j}\cdot\prod_{k \in \{1,...,n\}, \atop k \ne i}a_{k,\sigma(k)}.$$