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?