Let $A= (a_{ij}) \in M_n (F)$ be of the form \begin{align*} A = \begin{pmatrix} B & C \\ O & D \end{pmatrix}, \end{align*} where $B = (b_{ij}) \in M_r (F), D = (d_{ij}) \in M_s (F)$ and $C = (c_{ij}) \in M_{r \times s}(F)$ and $r+s =n $.
I'm reading a textbook where the author proves that $\det(A) = \det(B) \det(D)$. He uses the Leibniz formula to show this. I'll write his proof and then explain what I don't understand.
Proof: Note that for $1 \leq i, j \leq r$ we have $a_{ij} = b_{ij}$, for $1 \leq i, j \leq s $ we have $a_{i+r,j+r} = d_{ij}$, and if $i > r$ and $j \leq r$, then $a_{ij} = 0$. From the definition of the determinant we have \begin{align*} \det(A) = \sum_{\sigma \in \mathcal{S}_n} (sgn(\sigma)) a_{1 \sigma(1)} \ldots a_{r \sigma(r)} a_{r+1 \sigma(r+1)} \ldots a_{n \sigma (n)}. \end{align*} By definition, each $\sigma \in \mathcal{S}_n$ is just a rearrangement (i.e. a permutation) of the $n$ elements in $\mathcal{S}_n$. This means that for each $\sigma \in \mathcal{S}_n$ with the property that $\sigma(i) > r$ for some $i \leq r$, there must be some $i' > r$ such that $\sigma(i') \leq r$. Then for this $i'$ we have $a_{i' \sigma(i')} = 0$, and hence each term in $\det(A)$ that contains one of these factors is zero. Therefore each nonzero term in the above sum must be over only those permutations $\sigma$ such that $\sigma(i) > r$ if $i > r$ (block $D$), and $\sigma(i) \leq r$ if $i \leq r$ (block $B$).
To separate the action of the allowed $\sigma$ on the blocks $B$ and $D$, we define the permutations $\alpha \in \mathcal{S}_r$ and $\beta \in \mathcal{S}_s$ by $\alpha(i) = \sigma(i)$ for $1 \leq i \leq r$, and $\beta(i) = \sigma(i+r)-r$ for $1 \leq i \leq s$ (emphasis mine). In other words, each of the allowed $\sigma$ is just some rearrangement $\alpha$ of the values of $i$ for $1 \leq i \leq r$ along with some rearrangement $\beta$ of the values of $i$ for $r < i \leq r + s = n$, and there is no mixing between these blocks. The permutations $\alpha$ and $\beta$ are thus independent, and since $sgn(\sigma)$ is defined by the number of transpositions involved, this number simply seperates into the number of transpositions in $\alpha$ plus the number of transpositions in $\beta$. Therefore $sgn(\sigma) = sgn(\alpha)sgn(\beta)$. The result is that all nonzero terms in $\det(A)$ are of the form \begin{align*} (sgn(\alpha)) b_{1 \alpha(1)} \ldots b_{r \alpha(r)} (sgn(\beta)) d_{1 \beta(1)} \ldots d_{s \beta(s)}. \end{align*} Furthermore, every term in the expression for $\det(A)$ is included, and hence $\det(A) = \det(B) \det(D).$
Q.E.D
I understand everything until he divides $\sigma$ in two permutations $\alpha$ and $\beta$. Why do we have to do this, and why does he define them like that? It seems to me that $\beta(i) = \sigma(i+r)-r$ for $1 \leq i \leq s$ is pretty random. Why does he define these permutations like this? Also, he doesn't state how the result follows; he just gives it. How did he deduce the result?
Please help me understanding this proof!
Edit: Deduction of the last step:
Hence it follows that we have \begin{align*} \det(A) &= \sum_{\alpha \in \mathcal{S}_r} \sum_{\beta \in \mathcal{S}_s} sgn(\alpha) sgn(\beta) \bigg( \prod_{i=1}^r b_{i \alpha(i)} \bigg) \bigg( \prod_{i = 1}^{n-r} d_{i \beta(i)} \bigg) \\ &= \bigg( \sum_{\alpha \in \mathcal{S}_r} sgn(\alpha) \prod_{i=1}^r b_{i \alpha(i)} \bigg) \bigg(\sum_{\beta \in \mathcal{S}_s} sgn(\beta) \prod_{i =1}^{n-r} d_{i \beta(i)} \bigg) \\ &= \det(B) \det(D). \end{align*}