etnad179 said:
Just to check the \tau is the set of permutation of the elements {i_1,...i_n}?
\tau and \sigma are both permutations of {1,2,...,n}, i.e. they are bijections from that set onto itself. In the revised calculation below, \rho represents a permutation of {1,2,...,n} too.
etnad179 said:
And how can we get from this permutation \tau of the row is equal to the inverse \tau^{-1} of the column for matrix element B^{\tau(1)}_{\sigma(1)}
I think I did that step wrong. This is how I would like to handle the product B^{\tau(1)}_{\sigma(1)}\cdots B^{\tau(n)}_{\sigma(n)} today: First use the fact that real numbers commute, to rearrange the factors so that the row indices (the upper indices) appear in the order (1,2,...,n). I'll write asterisks instead of the column indices until we have figured out what we should write in those slots.
B^{\tau(1)}_{\sigma(1)}\cdots B^{\tau(n)}_{\sigma(n)}=B^1_*\cdots B^n_*
Now use the fact that for each k in {1,2,...,n} we have k=\tau(\tau^{-1}(k)), to rewrite this as
=B^{\tau(\tau^{-1}(1))}_*\cdots B^{\tau(\tau^{-1}(n))}_*.
Now just look at the product we started with and note that when the row index is \tau(k), the column index is \sigma(k). This tells us what the column indices are.
=B^{\tau(\tau^{-1}(1))}_{\sigma(\tau^{-1}(1))}\cdots B^{\tau(\tau^{-1}(n))}_{\sigma(\tau^{-1}(n))} =B^{1}_{\sigma(\tau^{-1}(1))}\cdots B^{n}_{\sigma(\tau^{-1}(n))}.
So the calculation should look like this:
<br />
\begin{align*}<br />
\det(AB) &=\sum_\sigma (\operatorname{sgn}\sigma)(AB)^1_{\sigma(1)}\cdots (AB)^n_{\sigma(n)}=\sum_\sigma (\operatorname{sgn}\sigma)\Big(\sum_{i_1}A^1_{i_1} B^{i_1}_{\sigma(1)}\Big)\cdots \Big(\sum_{i_n}A^n_{i_n}B^{i_n}_{\sigma(n)}\Big)\\<br />
&=\sum_{i_1,\dots,i_n}A^1_{i_1}\cdots A^n_{i_n}<br />
\underbrace{\sum_\sigma (\operatorname{sgn}\sigma) B^{i_1}_{\sigma(1)}\cdots B^{i_n}_{\sigma(n)}}_{=0\text{ unless there's a permutation }\tau\text{ such that }\tau(1,\dots,n)=(i_1,\dots,i_n).}\\<br />
&=\sum_\tau A^1_{\tau(1)}\cdots A^n_{\tau(n)}\sum_\sigma (\operatorname{sgn}\sigma) B^{\tau(1)}_{\sigma(1)}\cdots B^{\tau(n)}_{\sigma(n)}\\<br />
&=\sum_\tau A^1_{\tau(1)}\cdots A^n_{\tau(n)}\sum_\sigma \underbrace{(\operatorname{sgn}\sigma)}_{=(\operatorname{sgn}\tau)(\operatorname{sgn}(\sigma\circ\tau^{-1}))} B^{1}_{\sigma(\tau^{-1}(1))}\cdots B^{n}_{\sigma(\tau^{-1}(n))}\\<br />
&=\sum_\tau(\operatorname{sgn}\tau)A^1_{\tau(1)}\cdots A^n_{\tau(n)}\sum_{\rho}(\operatorname{sgn}\rho)B^{1}_{\rho(1)}\cdots B^{n}_{\rho(n)}\\<br />
&=(\det A)(\det B)<br />
\end{align*}<br />
To understand the step where I introduced the symbol \rho, you just have to stare at those two lines until you see that the sums contain the same terms.
