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?
Best Answer
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:
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.