Reference Request: Length of a reflection in a Coxeter group can be achieved by symmetric word In a given coxeter group $(W,S)$, a reflection is an element of $W$ that can be written with a symmetric word in the generators $S$.
In multiple sources, I found the following formula:
$$
    \mathrm{dp}(\alpha) = \frac{1}{2}(l(t_\alpha) + 1)
$$
where $\alpha$ is a positive root, $t_\alpha$ the corresponding reflection and the depth $\mathrm{dp}(\alpha)$ is the length of a shortest word $w$ such that $w\cdot \alpha$ is a negative root.
Assuming that any reflection has length achievable by a symmetric word, this formula is rather easy to check, but I couldn't find a proof for just this fact.
In X Fu's thesis, Lemma 1.3.19, the formula is proven but I'm looking for a more elementary proof of this fact:
Question: Is the length of a reflection in a Coxeter group achievable by a symmetric word?
 A: Let $\beta$ be a positive root. Let $u$ be an element of length $\mathrm{dp}(\beta)$ such that $u(\beta) <0$. Then $u(\beta)=-\alpha$ for some simple root $\alpha$ (because otherwise we could multiply $u$ by a left descent to get an element of shorter length inverting $\beta$), so $u^{-1}(\alpha)=-\beta$ and hence $\ell(s_\alpha u) <\ell(u)$.
Now since $u^{-1}s_\alpha(\alpha) = \beta$, we have that $(s_\alpha u)^{-1}s_\alpha (s_\alpha u)=s_\beta$. Thus $s_\beta$ has a symmetric word of length $2\mathrm{dp}(\beta)-1$.
This proves that $\ell(s_\beta) \leq 2\mathrm{dp}(\beta)-1$. For the opposite inequality, let $(s_1,\ldots,s_k)$ be a reduced word for $s_\beta$. Let $k+1-i$ be the maximal index such that $s_{k+1-i}\cdots s_k(\beta) =-\alpha<0$. Then $k+1-(k+1-i)= i\geq\mathrm{dp}(\beta)$. Now $s_1\cdots s_{k-i }s_{k+1-i}(\alpha)=-\beta$, so $s_{k+1-i}\cdots s_{2}s_{1}(\beta)=-\alpha$, so $k+1-i\geq\mathrm{dp}(\beta)$. Hence $k+1\geq 2\mathrm{dp}(\beta)$, and the result follows. 
A: In response to the reference request:  The existence of a symmetric reduced word is Exercise 10 (page 22) in Chapter 1 of the Björner-Brenti book.
