Spectrum of orthogonality graph (2) The orthogonality graph, $\Omega(n)$, has vertex set the set of $\pm 1$ vectors of length $n$, with orthogonal vectors being adjacent. 
I am only interested when $4|n$, since otherwise $\Omega(n)$ is empty or bipartite.  I am keen to know the spectrum of $\Omega(n)$ - the eigenvalues and their multiplicities. In particular I am seeking the inertia of $\Omega(n)$ - that is the numbers of positive, zero and negative eigenvalues. Many thanks Clive
 A: If $G$ is a Cayley graph for $\mathbb{Z}_2^n$ with connection set $C \subseteq \mathbb{Z}_2^n \setminus \{0\}$, then for each element $a \in \mathbb{Z}_2^n$ there is an eigenvector $v$ given by
$$v_x = (-1)^{x \cdot a}$$
where $x \cdot a$ is the usual inner product when $x$ and $a$ are thought of as 01-vectors (i.e., it is the number of 1's they have in common). This eigenvector has eigenvalue
\begin{align*}\sum_{c \in C} (-1)^{c\cdot a} &= |\{c \in C: c \cdot a \equiv 0 \ \text{mod} \ 2 \}| - |\{c \in C: c \cdot a \equiv 1 \ \text{mod} \ 2 \}| \\
&= |C| - 2|\{c \in C: c \cdot a \equiv 1 \ \text{mod} \ 2 \}|
\end{align*}
This gives a full orthogonal set of eigenvectors for the graph $G$.
In your case, the graph can be described as a Cayley graph for $\mathbb{Z}_2^n$ with connection set consisting of all of the elements with $n/2$ 1's in them (when written as binary strings). We can think of the binary strings as subsets of the $n$ element set $[n]$. So for each subset $S \subseteq [n]$, we get an eigenvalue of
$$\sum_{T \subseteq [n], |T| = n/2} (-1)^{|S \cap T|}.$$
If $S$ has size $k$ then this is equal to
$$\sum_{i=0}^k \sum_{T \subseteq [n], |T| = n/2, |S \cap T| = i} (-1)^i = \sum_{i=0}^k (-1)^i\binom{k}{i}\binom{n-k}{n/2 - i}.$$
I may have stolen this from the aforementioned notes of @ChrisGodsil (I hope he does not mind), but it follows from the well-known technique for computing the eigenvalues of Cayley graphs for $\mathbb{Z}_2^n$ described above.
A: To add to David E. Roberson's solution:
As already stated by Taneli Huuskonen's comment, the last sum is $0$ for odd $k$, so the eigenvalue $0$ has indeed the total multiplicity of $2^{n-1}$. For even $k=2K$ and putting $n=:2N$ (note that $N$ is still even), the sum can be simplified to $$(-1)^{k/2}\frac{(n-k)!(k)!}{(\frac n2)!(\frac {n-k}2)!(\frac k2)!} =(-1)^K\frac{\binom{2K}{ K}\binom{2(N-K)}{N-K}}{\binom NK}=:\lambda_{N,K},$$ and this eigenvalue occurs with multiplicity $2{\binom nk}=2{\binom n{2K}}$ for $K=0,...,\frac n4-1$ (by the symmetry $K\leftrightarrow N-K$) and with multiplicity ${\binom nk}$ for the "middle" one at $K=\frac n4=\frac N2.$
A: The graph $\Omega(n)$ is one of the relations of the Hamming association scheme $H(n,2)$, namely, the one corresponding to the Hamming distance $n/2$, see e.g. here. Its eigenvalues are given by the values of the Krawchuk's polynomials $K_k(n/2)$, as $0\leq k\leq n$, see details in my answer to your MO question 295493,
and multiplicities are just $\binom{n}{k}$, $0\leq k\leq n$. 
For this one can e.g. compute that if $n$ is divisible by 4 then the number of 0 eigenvalues of $\Omega(n)$ is $2^{n-1}$.
