Disclaimer: This question is a follow-up question to: Symmetric matrix is positive semidefinite. In here, I tried rephrasing the original problem and made some mistakes in the process. This is the correct version of it. Thank you to all that interacted with my previous post.
Let $A=(a_{ij}) \in M_n(\mathbb R)$ be a symmetric matrix such that:
- $a_{ij} \in \mathbb N_0, \quad \forall i,j=1,...,n$
- $a_{ii} = 2n+1$,
- $0 \leq a_{ij} \leq 2n, \quad i \neq j$.
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!
Again this is not true.
$$ \det\begin{pmatrix} 9&8&7&6\\ 8&9&0&0\\ 7&0&9&0\\ 6&0&0&9 \end{pmatrix}=-5508. $$ Also, if you replace the $0$s with $1$s the determinant is $-3248$.
Edit: A counterexample when $n=3$: $$ \det \begin{pmatrix} 7&6&5\\ 6&7&1\\ 5&1&7 \end{pmatrix}=-31. $$