On the group of signed permutations? Let $B_n$ be the group of signed permutations, which is a Coxeter group acting on $\mathbb{R}^n$ with Coxeter generators $\sigma_i=(i\; i+1)\in S_n$ and the change of sign $\tau(x_1,x_2,\dots,x_n)=(-x_1,x_2,\dots,x_n)$. 
So the elements can be represented by the action on the vector $(1,2,\dots,n)$ as words $w$ in the signed alphabet $\{\pm 1,\dots,\pm n\}$ where the $|w_i|$ form a permutation. Moreover, 
$$
\operatorname{inv}(w)=|\{i<j:w(i)>w(j)\}|+|\{i<j: w(i)+w(j)<0\}|+|\{i: w(i)<0\}|.
$$
Supposedly $\operatorname{inv}(w)$ is just the minimum length of an expression for $w$ written as a product of the Coxeter generators.
How does that characterization follow from this definition? Thank you.
 A: I cannot find the proof you are looking for in the book by D. E. Taylor
that ego suggested in the comments either. But you can use the techniques
of the section "Reflections and the Strong Exchange Condition" on pages
94-96 for a proof (my answer is independent of the book, for connections
see the comment at the end).

In your case you have the set $R = \{(1\;-1)\} \stackrel{.}{\cup}
\{(i\;i+1)(-i\;-i-1) \mid i \in [n-1]\}$ of involutions generating the
Coxeter group $W = \langle R\rangle$, which is a subgroup of the symmetric
group on $[n] \cup -[n]$ (using the notation $[n] := \{1, \dots, n\}$).
For the set $T = \{wrw^{-1} \mid w \in W, r \in R\}$ of all conjugates of
the generating involutions in $R$ (defined in equation (9.20) of the book) you
should be able to check
$T = T_1 \stackrel{.}{\cup} T_2 \stackrel{.}{\cup} T_3$ for
$$T_1 := \{(i\;-i)\mid i\in[n]\}$$
$$T_2 := \{(i\;j)(-i\;-j) \mid i, j \in [n], i<j\}$$
$$T_3 := \{(i\;-j)(-i\;j) \mid i, j \in [n], i<j\}$$
$T$ consists of two different conjugacy classes:
$T_1$ are the conjugates of $(1\;-1)$), and $T_2\stackrel{.}{\cup}T_3$ the
conjugates of all the other elements of $R$.
$W$ acts on $T$ by conjugation, which induces an action on the power set
$\mathcal{P}(T)$ of $T$. With the symmetric difference as addition, the
power set $\mathcal{P}(T)$ becomes an elementary abelian $2$-group, on
which $W$ acts. Define
$D'(w) = D'_1(w) \stackrel{.}{\cup} D'_2(w) \stackrel{.}{\cup} D'_3(w)$
for $w \in W$  with
$$D'_1(w) = \{(i\;-i) \in T_1 \mid i\in [n] \mbox{ and } w(i)<0\}$$
$$D'_2(w) = \{(i\;j)(-i\;-j) \in T_2 \mid i, j\in [n], i<j\mbox{ and }
w(i)>w(j)\}$$
$$D'_3(w) = \{(i\;-j)(-i\;j) \in T_3 \mid i, j\in [n], i<j\mbox{ and }
w(i)+w(j) < 0\}$$
The cardinality of $D'(w)$ is just inv$(w)$ from your question.
Claim:
(a) $D'(r) = \{r\}$ for $r \in R$
(b) $D'(w_1w_2) = w_2^{-1}D'(w_1)w_2+D'(w_2)$ for $w_1, w_2 \in W$
The first statement is easily verified, we show the second one:
For this, observe that an element $(i\;-i)$ of the conjugacy class $T_1$
is contained in $D'(w)$ if and only if $\frac{w(i)}{i}<0$ for
$i \in [n]\cup-[n]$.
Now $(i\;-i) \in D'(w_1w_2)$ is the same as $0 > \frac{w_1(w_2(i))}{i} =
\frac{w_1(w_2(i))}{w_2(i)}\cdot \frac{w_2(i)}{i}$, which in turn is
equivalent to either $ D'(w_1) \ni (w_2(i)\;-w_2(i)) = w_2(i\;-i)w_2^{-1}$
or $(i\;-i) \in D'(w_2)$, i.e., $(i\;-i) \in w_2^{-1}D'(w_1)w_2+D'(w_2)$.
An element $(i\;j)(-i\;-j)$ of the conjugacy class $T_2\cup T_3$, where
$i, j\in [n]\cup -[n]$, $i\ne j$ and $i\ne -j$, is contained in $D'(w)$
if and only if $\frac{w(j)-w(i)}{j-i}<0$ (check this by considering the
two cases $i\cdot j>0$ and $i\cdot j<0$ and by considering what happens
if you replace $i$ and $j$ by $-i$ and $-j$).
So $(i\;j)(-i\;-j) \in D'(w_1w_2)$ is equivalent to
$0 > \frac{w_1(w_2(j))-w_1(w_2(i))}{j-i} =
\frac{w_1(w_2(j))-w_1(w_2(i))}{w_2(j)-w_2(i)}\cdot
\frac{w_2(j)-w_2(i)}{j-i}$, which is the same as
$(i\;j)(-i\;-j) \in w_2^{-1}D'(w_1)w_2+D'(w_2)$.
Having proven the claim, one can easily deduce the formula for the length by
induction:
Because of (a) we may assume that $l(w) > 1$, and write $w = s\cdot v\cdot t$
with $s, t \in R$ and $l(v) = l(w)-2$. Per induction
$|D'(v)|+1 = l(v)+1 = l(v\cdot t) = |D'(v\cdot t)| \stackrel{(b)}{=}
|tD'(v)t + D'(t)| \stackrel{(a)}{=} |tD'(v)t + \{t\}|$, i.e., $t \not\in D'(v)$.
As $v\ne w = svt$ we get $t \ne v^{-1}sv$, and hence
$t\not\in \{v^{-1}sv\} + D'(v) \stackrel{(a)}{=} v^{-1}D'(s)v + D'(v)
\stackrel{(b)}{=} D'(s\cdot v)$.
It follows $D'(w) \stackrel{(b)}{=} tD'(s\cdot v)t \stackrel{.}{\cup} \{t\}$ and
therefore the induction step.

The claim is a variant of (9.22) in Taylor's book:
For $D(w) := wD'(w)w^{-1}$ you get $D(r) = r\{r\}r = \{r\}$ as $r$ has
order 2.
Also $D(w_1w_2) = w_1D'(w_1)w_1^{-1}+w_1w_2D'(w_2)w_2^{-1}w_1^{-1} =
D(w_1)+w_1D(w_2)w_1^{-1}$ showing that D fulfills (9.22).
This condition on $D$ is quite powerful. Taylor uses it to derive the strong
exchange property (with Corollary 9.26 implying the formula for the length), and
that it is equivalent to $(W, R)$ being a Coxeter system.
