Symmetric matrix is positive semidefinite

Question

Let $A=(a_{ij}) \in M_n(\mathbb R)$ be a symmetric matrix such that:

1. $a_{ij} \in \mathbb Q, \quad \forall i,j=1,...,n$
2. $a_{ii} = 1$,
3. $0 \leq a_{ij} \leq \frac{n-1}{n}, \quad i \neq j$.

I want to prove that such a matrix is positive semidefinite. It is well known that since $A$ is symmetric, this amounts to show that all eigenvalues of $A$ are non-negative. For that, one has to find a lower bound for the eigenvalues of $A$ that is greater or equal to zero. I tried using Gershgorin theorem, but the bound is not strong enough, the same happens with the one on this paper: https://www.math.uwaterloo.ca/~hwolkowi/henry/reports/bndseigs80.pdf

Does anyone one know stronger bounds on eigenvalues of matrices of this kind? Or to prove directly that this matrix is positive semidefinite?

Thanks in advance!

Answer 1

This is not true. Random counterexample: $$ A=\pmatrix{1&\frac34&0&\frac12\\ \frac34&1&\frac34&0\\ 0&\frac34&1&\frac12\\ \frac12&0&\frac12&1}. $$ $A$ isn’t positive semidefinite because $\det(A)=-\frac58<0$.