Symmetry of the product of the inverse and the transpose of a matrix Let $D = 
\begin{pmatrix}
1 & 0 & 0 \\
-1 & 1 & 0 \\
-2 & -1 & 1
\end{pmatrix}$
and let $\Phi = -D^{-1}D^t$. Then $\Phi$ is a Coxeter transformation (which is probably not important for this question.) However, we have
$$\Phi = \begin{pmatrix}
-1 & 1 & 2 \\
-1 & 0 & 3 \\
-3 & 2 & 6
\end{pmatrix}, \text{ and } \Phi^{-1} = 
\begin{pmatrix}
6 & 2 & -3 \\
3 & 0 & -1 \\
2 & 1 & -1
\end{pmatrix},$$
so $\Phi^{-1}$ is just a point reflection of $\Phi$, with reflection center being the middle. Do you know why this holds? Probably one can use that Coxeter transformations can be written as a product of reflections and use that the reflections are their own inverse and then do something with permutation matrices. But I would like to know if there is some more elementary proof? Probably we also need that $D + D^t$ does not only have the property that it is symmetric, but also the property that it is point-symmetric (since this seems to go in in my other proof idea).
 A: A direct computation shows that any lower-unitriangular $3\times 3$-matrix
$$
D=\begin{pmatrix} 1 & 0 & 0 \cr a & 1 & 0 \cr b & c & 1 \end{pmatrix}
$$
satisfies this with $\Phi=-D^{-1}D^t$ if and only if $a=c$. The computation is elementary, too.
A: I want to give my original proof idea (which works) to indicate more directly why it is true.
Let D be a lower unitriangular matrix which has only negative integer entries below the diagonal and such that  $A = D + D^t$ is a point-symmetric matrix. It is always a generalized Cartan matrix. Therefore $\Phi = -D^{-1}D^t$ can be written as a product of reflections $\Phi = R_n \cdots R_1$ with $R_i(e(j)) = e(j) - A_{ji}e(i)$, see this paper by Ladkani (I changed some conventions a little bit). In his notation, this can be written as $\Phi = \Phi(A, \text{id})$. Let $\pi: \{1, \dots, n\} \to \{1, \dots, n\}$ be the permutation $i \mapsto n+1-i$. Let $P^\pi$ be the corresponding permutation matrix (i.e. $e(i) \mapsto e(\pi(i))$). For a matrix $B$, define $B^{\pi} = (P^{\pi})^{-1} B P^{\pi}$. Then since $A$ is point-symmetric we have $A^{\pi} = A$. It follows, since the reflections $R_i$ are self-inverse:
$$\Phi^{-1} = \Phi(A, \text{id})^{-1} = R_1 \cdots R_n = \Phi(A, \pi) = P^{\pi} \Phi(A^{\pi}, \text{id})(P^{\pi})^{-1} = \Phi(A, \text{id})^{\pi},$$
which is precisely the statement that $\Phi^{-1}$ is the point-reflection of $\Phi$.
