Suppose there is a Hermitian matrix $W\in\mathbb{C}^{mn\times mn}$ with size $mn\times mn$. Denote $W_{ij}$ as the submatrix with size $n\times n$ in the following position: \begin{equation} W=\begin{bmatrix} W_{11}& W_{12} &\cdots &W_{1m}\\ W_{21} &W_{22} &\cdots &W_{2m}\\ \vdots &\vdots &\ddots &\vdots \\ W_{m1}& W_{m2}& \cdots &W_{mm}\\ \end{bmatrix} \end{equation}
$W_{ij}$ is known to be Hermitian and positive semi-definite. What kind of condition can guarantee that $W$ is positive definite?
Do we have some results like if $W_{ii}\succ \sum_{j\neq i}W_{ij} $ for each $i$, then $W$ is positive definite?
Yes, this is sufficient. Let $P_i=W_{ii}-\sum_{j\ne i}W_{ij}$. Then \begin{aligned} W=\,&\pmatrix{P_1\\ &P_2\\ &&\ddots\\ &&&\ddots\\ &&&&P_m}\\ &+\pmatrix{W_{12}&W_{12}\\ W_{12}&W_{12}\\ \\ \\ &&&&} +\pmatrix{W_{13}&&W_{13}\\ \\ W_{13}&&W_{13}\\ \\ &&&&}+\cdots +\pmatrix{\\ \\ \\ &&&W_{m-1,m}&W_{m-1,m}\\ &&&W_{m-1,m}&W_{m-1,m}}. \end{aligned} The first summand by assumption is positive definite, while each of the other summands is positive semidefinite, because it is, up to a permutation of rows and columns, equal to a Kronecker product of the form $$ \pmatrix{1&1\\ 1&1\\ &&0\\ &&&\ddots\\ &&&&0}\otimes W_{ij}. $$ Hence $W$ is positive definite.