Let $A \in \mathbb{C}^{n\times n}$, is a hermitian matrix with guaranteed real and positive diagonal entries. Also $\det(A)>0$. Is $A$ positive-definite?
Positive-definite matrices have real and positive diagonal entries, but I have not found any resource whether the converse is true or not. I know that this is true in case of eigenvalues, but I wish not to calculate eigenvalues. This question solves it for a real matrix and I was wondering if the same can be done for complex matrix?
The linked question only solves it for diagonally dominant matrices, and the same argument made there could be made for diagonally dominant complex matrices.
In general, the answer to your question is no, even for real matrices. For example, the matrix $$\begin{bmatrix}1&-3 &&-3\\ -3 &1&&2\\-3&&2&&1\end{bmatrix}$$ has two negative eigenvalues and one positive eigenvalue.
Note, of course, that the matrix I provide is not diagonally dominant.