Let $A\in M_n(\mathbb R)$ is a non-zero symmetric zero-diagonal matrix and its elements are $0$ ore $1$. What we can say about eigenvalue of $A$?

Question

Let $A\in M_n(\mathbb R)$ is a non-zero symmetric zero-diagonal matrix and its elements are $0$ ore $1$. We know that the eigenvalue of $A$ are real. I'm interesting to know the number of distinct eigenvlue of the matrix $A$. How many of them are greater than $1$ and how many of them are lower than $1$?
Since $\mathrm{trac}(A)=0$, we get that some eigenvalue of $A$ is negative and some of them are positive. In the case that $A=[a_{ij}]$ and $\Pi_{i\ne j}a_{ij}=1$ (i.e. $a_{ij}=1$ for $i\ne j$ and $a_{ii}=0$) the answer is easy and the eigenvalues are $\lambda_1=n-1$ by repeated order $1$ and $\lambda_2=-1$ by repeated order $n-1$. My question is about the case that: $\Pi_{i\ne j}a_{ij}=0$ (i.e. there is $i\ne j$ which $a_{ij}=a_{ji}=0$).

Answer 1

Like I said in the comment, it has some connection with the graph theory, since a binary symmetric matrix with zero diagonal is the adjacency matrix of a graph.

For example, the matrix with all the elements equal to 1 (except for the diagonal) is the adjacency matrix of the complete graph with $n$ nodes, and it is known that

a binary matrix $A$ with zero diagonal has exactly two distinct eigenvalues if and only if it is the the adjacency matrix of the complete graph

An other curiosity is for example that

a binary matrix $A$ with zero diagonal has the spectrum symmetric (that is, $\lambda$ is an eigenvalue if and only if $-\lambda$ is an eigenvalue) if and only if it is the the adjacency matrix of a bipartite graph

Regarding the number of distinct eigenvalue, we know an inferior bound in particular situation: Let's say that a graph $G$ is regular if every node has the same degree, and let's call $d(G)$ the diameter

Given a regular graph $G$, then $d(G)$ is always strictly less then the number of distinct eigenvalues of its adjacency matrix

Answer 2

There is basically nothing that can be said. For instance, if $$ A=\begin{bmatrix}0&1&1&\cdots&1\\ 1&0&0&\cdots&0\\ \vdots &\vdots&\ddots \\ 1&0&0&\cdots&0 \end{bmatrix}, $$ then the eigenvalues are $\sqrt{n-1}$ and $-\sqrt{n-1}$ each with multiplicity one, and $0$ with multiplicity $n-2$. Or, if $$ A=\begin{bmatrix}0&1&0&0&0&0&\cdots&0\\ 1&0&1&0&0&0&\cdots&0\\ 0&1&0&1&0&0&\cdots&0\\ 0&0&1&0&1&0&\cdots&0\\ \vdots &\vdots&&&\ddots \\ 0&0&0&0&0&1&0&1 \\ 0&0&0&0&\cdots&0&1&0 \end{bmatrix} $$ (i.e., $A_{ij}=1$ if $i=j\pm1$), then the eigenvalues are $$ 2\cos\frac{k\pi}{n+1},\ \ k=1,2,\ldots,n. $$