# Sufficient conditions for positive semidefiniteness of block matrix

$\newcommand{\Re}{\mathbb{R}}$ I'm looking for sufficient conditions that may guarantee positive semidefiniteness (PSD) of a block matrix $$A = \begin{bmatrix} A_{1,1} & \cdots & A_{1,n} \\ \vdots & \ddots & \vdots \\ A_{n,1} & \cdots & A_{n,n}\end{bmatrix}$$ with consistent block dimensions, $A_{i,j} \in \Re^{d_i \times d_j}$.
I'm aware of the sufficient condition that if $A$ is block diagonally dominant (BDD) this implies PSD [1], but the BDD condition turns to be too restrictive for my case of interest (my block matrices are not BDD, yet they are PSD). I looked for some alternative but couldn't find anything in the literature. Also I'm not interested in Schur-based exact conditions, but simpler conditions that work well often (but not necessarily always).

In looking at the block diagonally dominant (BDD) condition [1] I came up with a proposal but I haven't been able to find a proof or counter-proof so far. I will briefly describe it here.
First, consider the traditional BDD condition [1] from the following perspective:

• Build the $n\times n$ matrix $M \in \Re_+^{n\times n}$ given as $$M=[m_{ij}]_{i,j=1}^n, \quad m_{ij} = \begin{cases} \inf_{x} \frac{||A_{ij}x||}{||x||} \text{ if } i=j\\ \sup_{x} \frac{||A_{ij}x||}{||x||} \text{ if } i\neq j \end{cases}$$

• Considering this condensed matrix $M$, the traditional BDD condition [1] on $A$ is equivalent to the condition that $M$ is Diagonally Dominant (DD). Thus, we can state $M$ is DD $\implies$ $A$ is PSD.

Now, my proposal/question is, might it be possible that $M$ is PSD $\implies$ $A$ is PSD? I would appreciate any direction either towards a proof or counter-proof.

PD: Usual properties can be assumed on $A$ to simplify the problem, so $A$ can be considered symmetric, and the blocks all square with $d_i=d_j=d$.

Edit: After some extensive numerical testing, I've found cases where my conjecture is contradicted. That is, $M$ is PSD $\nRightarrow$ $A$ is PSD.
Still, for the cases I found so far, even though $A$ was not PSD, it wasn't too far (the negative eigenvalue was small). So I still wonder if this condition might stand under additional constraints.

[1] Feingold, D. G., & Varga, R. S. (1962). Block diagonally dominant matrices and generalizations of the Gerschgorin circle theorem. Pacific J. Math, 12(4), 1241-1250.