Proof Question: Determinant of Block Lower Triangular Matrix

Consider the matrix $$B=\left[\begin{array}{c|c}A&0\\ \hline C&D\end{array}\right]$$ where $$A$$ is a matrix of size $$k\times k$$ and $$D$$ is a matrix of size $$n\times n$$. I am trying to prove that $$\det(B)=\det(A)\cdot \det(D)$$ like user "Zilin J." did here.

Here is the area where I get stuck in the proof.

\begin{align*} \det B &= \sum_{\sigma\in S_{n+k}}\operatorname{sgn}\sigma\prod_{i=1}^{n+k}b[i, \sigma(i)]\\ &= \text{haven't got here yet..}\\ \end{align*}

-I understand the proof where it states if $$\sigma(i) = j$$, $$i\le k$$, and $$j > k$$, then the corresponding summand $$\prod_i b[i,\sigma(i)]$$ is $$0$$. Thus, referring to the determinant above, it is clear we could get rid of those elements in $$S_{n+k}$$ where that occurs in the lower limit of our sum above.

-But the yellow part is where I get stuck. My question is how are we able to deduce the yellow part below? Else, we consider $$\sigma(i)\neq j$$ or $$i>k$$ or $$j\leq k$$. In other words, if $$\sigma(i)=j$$, then $$i>k$$ or $$j\leq k$$. [Note I'm using $$P\implies Q \equiv \lnot P\lor Q$$ to get that other words part.]

Any permutation $$\sigma\in S_{n+k}$$ for which no such $$i$$ and $$j$$ exist can be uniquely "decomposed" into two permutations, $$\pi$$ and $$\tau$$, where $$\pi\in S_k$$ and $$\tau\in S_n$$ such that $$\sigma(i) = \pi(i)$$ for $$i \le k$$ and $$\sigma(k+i) = k+\tau(i)$$ for $$i \le n$$.

Let $$\sigma\in S_{k+n}$$ be a permutation such that for all $$i\le k$$, it is not the case that $$\sigma(i)>k$$, i.e. it takes $$\sigma(i)\le k$$.
That just defines $$\pi\in S_k$$, as $$\{1,\dots,k\}$$ is invariant under $$\sigma$$, so that we can set $$\pi:=\sigma\restriction_{\{1,\dots,k\}}$$.
And, consequently, as $$\sigma$$ is a one-to-one mapping on $$\{1,2,\dots,k+n\}$$, it must map all the remaining elements to a remaining element, i.e. $$\sigma$$ also leaves $$\{k+1,\dots,k+n\}$$ invariant, thus defining $$\tau$$.
Observe that $$\sigma=\pi\tau=\tau\pi$$.
• I appreciate the answer. Why are you only considering the case that for all $i\leq k$? Also, I haven't seen that vertical half arrow symbol before, i.e. where $\pi:=\sigma\restriction_{\{1,\dots,k\}}$. What does that arrow symbol refer to? – W. G. May 2 at 23:47
• We have $k$ elements $i\le k$. All of them are (one-to-one) mapped to elements $\le k$, that eats up all the first $k$ elements. I do care the other elements: they must also form an invariant subset. – Berci May 2 at 23:57