TITLE: Equivalence of Hadamard Graph and Hadamard Matrix
QUESTION [5 upvotes]: I'm reading Distance Regular Graphs by Brouwer, Cohen, and Neumaier. In section 1.8, they explained Hadamard graphs.
Conversion from a Hadamard Matrix into a Hadamard Graph
An $n$-Hadamard graph $G$ is a graph on $4n$ vertices defined in terms of a Hadamard matrix of order $n$ $H_n = h_{ij}$ as follows:

Define $4n$ symbols $r_i^+$, $r_i^-$, $c_i^+$, and $c_i^-$, where $r$ stands for row and $c$ stands for column and take these as the vertices of the graph.
Then add two types of edges between row vertices and column vertices based on the sign of $h_{ij}$:
\begin{equation*}
\text{parallel edges $(r_i^+, c_j^+)$ and $(r_i^-, c_j^-)$ if $h_{ij} = +1$} \\
\text{crossing edges $(r_i^+, c_j^-)$ and $(r_i^-, c_j^+)$ if $h_{ij} = -1$}
\end{equation*}

Then the graph $G$ will be a bipartite graph where the set of vertices is partitioned into row vertex set of $2n$ vertices and column vertex set of $2n$ vertices. And there will be $2n^2$ edges.
Equivalence between Graph and Matrix
In theorem 1.8.1 in their book, they showed that $G$ is a distance-regular graph with an intersection array
$\{n,n-1,\frac{n}{2},1;1,\frac{n}{2},n-1,n\}$ if and only if the matrix $H$ is Hadamard matrix of order $n$. 
My Question
Their proof of this theorem seems rather brief and I have hard time in understanding the equivalence. Especially, I don't understand what role two orthogonal rows or columns in the matrix $H_n$ play in the graph $G$ so that distance-regularity is achieved. The book cited three papers for the proof, but I cannot find any of them in the Internet.
Can anyone explain what's the idea or intuition behind the proof of equivalence?

REPLY [6 votes]: First, the book by Brouwer, Cohen and Neumaier is known for a degree of terseness;
any one who uses it will have struggled with it at some point.
You describe a construction that, from an $n\times n$ Hadamard matrix, produces a
bipartite graph on $4n$ vertices that is regular of degree $n$.
This graph has diameter four; moreover for each vertex $u$ there is a unique vertex 
$u'$ at distance four from $u$. It follows that the the $4n$ vertices may be partitioned 
into $2n$ pairs.
So the way to proceed is to prove that these comments about pairs of vertices at distance four
are correct and then, using this, prove that the parameters $b_i$ and $c_i$ are well-defined.
This reduces to showing that if $v$ is at distance two from a vertex $u$, then $v$ has exactly
$n/2$ neighbours in common with $u$. (This is where the orthogonality of the rows of $H$ enters.)