Determinant of Lower Triangular Block Matrix (Proof Question: Decomposition) I am wanting to prove the determinant of a block lower triangular matrix is the product of its diagonals. [Note, I am looking at Zilin J's answer here and asked the following question yesterday about it. However, I am still stuck on the part as to why there are two "decompositions" of functions here where $\sigma=\pi \tau$ (see the third bullet down).] 
\begin{eqnarray}\det B &=& \sum_{\sigma\in S_{n+k}}\operatorname{sgn}\sigma\prod_{i=1}^k b[i,\sigma(i)]\prod_{ i=k+1}^{k+n} b[i,\sigma(i)] \tag{1}\\
\end{eqnarray}
Looking at equation $(1)$, if $i\leq k$ and $\sigma(i)>k$, then we have a zero summand as $b[i,\sigma(i)]=0$. 
-That means we $\underline{\text{only}}$ consider values of $\sigma$ where $k<i$ or $\sigma(i)\leq k$ holds true. 
$\bullet$ [Show $\pi \in S_k$.]
Now, let $\pi(i):=\sigma(i)$ for $i\leq k$. Since $i\leq k$, we know $\sigma(i)\leq k$ must hold true which means $\pi(i)\leq k$. So, $\pi\in S_k$.
$\bullet$ [Show $\tau \in S_n$.]
Now, let $\tau(i):=\sigma(k+i)-k$ for $i\leq n$. Since $\sigma(k+i)\leq k+n$, we know $\tau(i)=\sigma(k+i)-k\leq k+n-k=n$. Thus, $\tau \in S_n$.
$\bullet$ [Show $\operatorname{sgn}\sigma=\operatorname{sgn}\tau \operatorname{sgn}\pi$. ] 

Where do I go from here?

 A: First, we know a bit more about $\sigma$. For every $i$ with $i \le k$, we know $\sigma(i) \le k$, so $\sigma$ maps the values $\{1,2,\dots,k\}$ to $\{1,2,\dots,k\}$ (in some order). But this "uses up" all the values in that range as possible values of $\sigma(i)$. So $\sigma$ must map the values $\{k+1,k+2,\dots,n\}$ to $\{k+1,k+2,\dots,n\}$ (in some order). In other words, if $i>k$, we know $\sigma(i)>k$, too.
The parity $\operatorname{sgn}(\sigma)$ can be defined in two ways:


*

*As $(-1)^x$ where $x$ is the number of inversions in $\sigma$: pairs $(i,j)$ with $i<j$ but $\sigma(i) > \sigma(j)$.

*As $(-1)^y$ where $y$ is length of a representation of $\sigma$ as a product of transpositions (length-$2$ cycles).


Both of these can be used to show that $\operatorname{sgn}(\sigma) = \operatorname{sgn}(\tau)\operatorname{sgn}(\pi)$, so you get two proofs in one answer.


*

*For every pair $(i,j)$ with $i \le k$ and $j > k$, we have $\sigma(i) \le k$ and $\sigma(j) > k$, so no such pairs are inversions. Therefore the inversions in $\sigma$ are pairs $(i,j)$ with $i<j\le k$ and $\sigma(i) > \sigma(j)$ - the inversions in $\pi$ - and pairs $(i,j)$ with $k < i < j$ and $\sigma(i) > \sigma(j)$ - the inversions in $\tau$. If there are $x_1$ inversions in $\pi$ and $x_2$ inversions in $\tau$, then $$\operatorname{sgn}(\sigma) = (-1)^{x_1 + x_2} = (-1)^{x_1} (-1)^{x_2} = \operatorname{sgn}(\pi) \operatorname{sgn}(\tau).$$

*If we represent $\pi$ as a product of $y_1$ transpositions and $\tau$ as a product of $y_2$ transpositions, then we can find a representation of $\sigma$ as a product of $y_1 + y_2$ transpositions: the transpositions representing $\pi$, together with a translation to the range $k+1, \dots, n$ of the transpositions representing $\tau$. Therefore $$\operatorname{sgn}(\sigma) = (-1)^{y_1 + y_2} = (-1)^{y_1} (-1)^{y_2} = \operatorname{sgn}(\pi) \operatorname{sgn}(\tau).$$
A: I'm just adding the following as notes.
\begin{eqnarray}\det B &=& \sum_{\sigma\in S_{n+k}}\operatorname{sgn}\sigma\prod_{i=1}^k b[i,\sigma(i)]\prod_{ i=k+1}^{k+n} b[i,\sigma(i)] \tag{1}\\
\end{eqnarray}
Looking at equation $(1)$, if $i\leq k$ and $\sigma(i)>k$, then we have a zero summand as $b[i,\sigma(i)]=0$. 
-That means we $\underline{\text{only}}$ consider values of $\sigma$ where $k<i$ or $\sigma(i)\leq k$ holds true. 
$\bullet$ [Show $\pi \in S_k$.]
Assume $k<i$ or $\sigma(i)\leq k$ holds true.

Now, note for every $j$ with $j \leq k$, we know $\sigma(j) \leq k$, so $\sigma$ maps the values $\{1,2,\dots,k\}$ to $\{1,2,\dots,k\}$ (in some order). But this "uses up" all the values in that range as possible values of $\sigma(j)$. So $\sigma$ must map the values $\{k+1,k+2,\dots,n\}$ to $\{k+1,k+2,\dots,n\}$ (in some order). In other words, if $j>k$, we know $\sigma(j)>k$, too.

Let $(i, j)$ be an inversion in $\sigma$. 


*

*Suppose $i\leq k$ and $j>k$; show a contradiction. Thus, $\sigma(i)\leq k$ must hold true by our assumtion at the top. Also, note that $\sigma(j)>k$ holds true too (see above yellow portion). Thus, $\sigma(i)\leq k<\sigma(j)\implies\sigma(i)<\sigma(j)$. But this is a contradiction as $\sigma(j)<\sigma(i)$ by definition of an inversion. Thus, no inversions occur given this supposition.

*Else, we know $k<i$ or $j\leq k$ must hold true.
Now, let $\pi(i):=\sigma(i)$ for $i\leq k$. Since $i\leq k$, we know $\sigma(i)\leq k$ must hold true which means $\pi(i)\leq k$. So, $\pi\in S_k$.
$\bullet$ [Show $\tau \in S_n$.]
Now, let $\tau(i):=\sigma(k+i)-k$ for $i\leq n$. Since $\sigma(k+i)\leq k+n$, we know $\tau(i)=\sigma(k+i)-k\leq k+n-k=n$. Thus, $\tau \in S_n$.
$\bullet$ [Show $\operatorname{sgn}\sigma=\operatorname{sgn}\tau \operatorname{sgn}\pi$. ] 

Assumption: We showed earlier we are safe to assume $k<i$ or $\sigma(i)\leq k$ holds true; we will assume this now for the remainder of the proof.

Now, note for every $j$ with $j \leq k$, we know $\sigma(j) \leq k$, so $\sigma$ maps the values $\{1,2,\dots,k\}$ to $\{1,2,\dots,k\}$ (in some order). But this "uses up" all the values in that range as possible values of $\sigma(j)$. So $\sigma$ must map the values $\{k+1,k+2,\dots,n\}$ to $\{k+1,k+2,\dots,n\}$ (in some order). In other words, if $j>k$, we know $\sigma(j)>k$, too.

Let $(i, j)$ be an inversion in $\sigma$. 


*

*Suppose $i\leq k$ and $j>k$; show a contradiction. Thus, $\sigma(i)\leq k$ must hold true by our assumtion at the top. Also, note that $\sigma(j)>k$ holds true too (see above yellow portion). Thus, $\sigma(i)\leq k<\sigma(j)\implies\sigma(i)<\sigma(j)$. But this is a contradiction as $\sigma(j)<\sigma(i)$ by definition of an inversion. Thus, no inversions occur given this supposition.

*Else, we know $k<i$ or $j\leq k$ must hold true.


$\bullet$ Case 2.1: If $i>k$ holds true, then $\sigma(i)>k$ (yellow portion above). Also, as $i<j$ (i.e. because $(i,j)$ is an inversion here), we know $k<i<j\implies \sigma(j)>k$, too.
$\bullet$ Case 2.2: Else, if $j\leq k$ holds true, we know that $i<j\leq k$. And by the first sentence of the yellow part above,  we know $\sigma(i)\leq k$ and $\sigma(j)\leq k$.


$\textbf{How do you show for case $2.1$ that this is an inversion in $\tau$?}$
Now, define $f(i):=i-k$. [Show $f(i)>0$ and $f(i)\leq n$.] As $i>k$, we know $f(i)=i-k>0$. By way of contradiction, suppose $f(i)>n$; show a cotradiction. Well, then $i-k>n\implies i>n+k$ (a contradiction as $i\leq n+k$). Thus, $\tau(f(i))$ is well-defined for any $i>k$ in the domain of $\sigma$. This means $\tau(f(i))=\tau(i-k)=\sigma(k+(i-k))-k=\sigma(i)-k$ represents the composition of functions we want for any $i>k$ in the domain of $\sigma$. 
Now, let $(i,j)$ be an arbitrary inversion in $\sigma$ where $i>k$. [Show it is an inversion in $\tau$.] Well, $\tau(i)=\sigma(i)-k$ and $\tau(j)=\sigma(j)-k$. Since $(i,j)$ is an inversion in $\sigma$, we know $\sigma(i)<\sigma(j)$. Thus, $\sigma(i)-k<\sigma(j)-k$. Thus, $\tau(i)<\tau(j)$ which means $(i,j)$ is also an inversion in $\tau$.
$\textbf{And how do you show for case $2.2$ that this is an inversion in $\pi$?}$
Now, let $(i,j)$ be an arbitrary inversion in $\sigma$ where $j\leq k$. [Show it is an inversion in $\pi$. Note we just map these functions by $g(i):=i$ here.] Clearly, $j<k$ here. Clearly, $(i,j)$ is an inversion in $\pi$ as it was an inversion in $\sigma$ looking at how we defined $\pi$ earlier. 
