Unpicking proof that det($AB$) = det$(A)$det$(B)$ I'm trying to understand this proof that $\text{det}(AB) =\text{det}(A)\text{det}(B)$, but I'm a little stuck on the third line. Please could someone explain to me what the mapping $\kappa$ does?
Let $\mathfrak{T}_n = {\rm Maps} (\{1, \ldots , n\}, \{1, \ldots , n\})$
\begin{eqnarray*}
{\sf det}(AB) & = & \sum_{\sigma \in \mathfrak{S}_n} {\sf sgn}(\sigma) \prod_{i=1}^n (AB)_{i\sigma(i)} \\ & = & \sum_{\sigma \in \mathfrak{S}_n} {\sf sgn}(\sigma) \prod_{i=1}^n \sum_{j=1}^n a_{ij} b_{j\sigma(i)} \\
& = & \sum_{\sigma\in \mathfrak{S}_n, \kappa \in \mathfrak{T}_n} {\sf sgn}(\sigma) a_{1\kappa(1)}b_{\kappa(1) \sigma(1)} \ldots a_{n \kappa(n)} b_{\kappa (n) \sigma (n)} \\
& = &\sum_{\kappa\in \mathfrak{T}_n} a_{1\kappa(1)} \ldots a_{n \kappa(n)} \sum_{\sigma \in \mathfrak{S}_n} {\sf sgn}(\sigma) b_{\kappa(1) \sigma(1)}\ldots b_{\kappa (n) \sigma (n)} \\
& = & \sum_{\kappa\in \mathfrak{T}_n}  a_{1\kappa(1)} \ldots a_{n \kappa(n)} {\sf det} (B_{\kappa})
\end{eqnarray*} where $B_{\kappa}$ denotes the matrix which is obtained by using the rows of $B$ labelled $\kappa(1), \ldots , \kappa (n)$ for its rows $1$ to $n$.
. Therefore: $$ \sum_{\kappa\in \mathfrak{T}_n}  a_{1\kappa(1)} \ldots a_{n \kappa(n)} {\sf det} (B_{\kappa})
 = \sum_{\kappa\in \mathfrak{S}_n}  {\sf sgn}(\kappa) a_{1\kappa(1)} \ldots a_{n \kappa(n)} {\sf det} (B) = \det (A) \det (B).$$
Thanks
 A: Sure, notice that
$$\sum_{\sigma \in \mathfrak{S}_n} {\sf sgn}(\sigma) \prod_{i=1}^n \sum_{j=1}^n a_{ij} b_{j\sigma(i)}=\sum_{\sigma \in \mathfrak{S}_n} {\sf sgn}(\sigma) \prod_{i=1}^n \left (a_{i1} b_{1\sigma(i)}+a_{i2} b_{2\sigma(i)}+\cdots +a_{in} b_{n\sigma(i)}\right ),$$
so when you unfold the product, for each one of the factors, you are choosing one of the summands to put together. So for the $i-$th factor, let's say that you pick t the $j-$th summand (i.e., $a_{ij} b_{j\sigma(i)}$). This creates a pairing $\{(i,j):1\leq i,j \leq n\},$ where $i$ in the first coordinate just appears once! So the pairing is functional that is the function $\kappa :\{1,\cdots ,n\}\longrightarrow \{1,\cdots ,n\}$ that you are looking at.
For general purposes:
In general if you have a sequence $(x_{i,j})_{i\in X,j\in Y}$ (apriori with $X,Y$ finite) then $$\prod _{i\in X}\sum _{j\in Y}x_{i,j}=\sum _{f\in Y^X}\prod _{i\in X}x_{i,f(i)},$$ where $X^Y$ denotes the functions from $Y$ to $X.$
A: A lot of times, these kinds of formal manipulations become clearer if we look at an example. Let's look at $n = 2$, so that $\mathfrak T_n$ only has four elements to be considered. We have
$$
\begin{align}
\sum_{\sigma \in \mathfrak S_2} \mathsf{sgn}(\sigma) &\prod_{i=1}^2\sum_{j=1}^2 a_{ij} b_{j\sigma(i)}  = 
\sum_{\sigma \in \mathfrak S_2} \mathsf{sgn}(\sigma) \prod_{i=1}^2
(a_{i1}b_{1\sigma(i)} + a_{i2}b_{2\sigma(i)})
\\ & = \sum_{\sigma \in \mathfrak S_2}\mathsf{sgn}(\sigma)(a_{11} b_{1\sigma(1)} + a_{12}b_{\sigma(2)})(a_{21}b_{2\sigma(1)} + a_{22}b_{2\sigma(2)})
\\ & = \sum_{\sigma \in \mathfrak S_2}\mathsf{sgn}(\sigma) 
[a_{\color{red}{1}\color{green}1} b_{\color{green}1 \sigma(\color{red}1)}
a_{\color{red}{2}\color{green}1} b_{\color{green}1 \sigma(\color{red}2)}
+ a_{\color{red}{1}\color{green}1} b_{\color{green}1 \sigma(\color{red}1)}
a_{\color{red}{2}\color{green}2} b_{\color{green}2 \sigma(\color{red}2)}
%
\\ & \qquad + a_{\color{red}{1}\color{green}2} b_{\color{green}2 \sigma(\color{red}1)}
a_{\color{red}{2}\color{green}1} b_{\color{green}1 \sigma(\color{red}2)}
+ a_{\color{red}{1}\color{green}2} b_{\color{green}2 \sigma(\color{red}1)}
a_{\color{red}{2}\color{green}2} b_{\color{green}2 \sigma(\color{red}2)}].
\end{align}
$$
In the last line, we see that for each $\kappa : \{1,2\} \to \{1,2\}$, there is a term of the form $a_{1\kappa(1)}b_{\kappa(1)\sigma(1)}a_{2\kappa(2)}b_{\kappa(2)\sigma(2)}$, with $1,2$ in red and $\kappa(1),\kappa(2)$ in green.
