Let $A$ is a positive semi-definite matrix like this:

$ A = \begin{bmatrix} 1 & \alpha_{1,2} & \alpha_{1,3} & \alpha_{1,4}\\ \alpha_{1,2} & 1 & \alpha_{2,3} & \alpha_{2,4}\\ \alpha_{1,3} & \alpha_{2,3} & 1 & \alpha_{3,4}\\ \alpha_{1,4} & \alpha_{2,4} & \alpha_{3,4} & 1 \end{bmatrix}$

Then we want to check another matrix called $M$ verry useful to guarantee convergence in Ribando's theorem here https://link.springer.com/content/pdf/10.1007/s00454-006-1253-4.pdf?pdf=button. So $M$ is

$ M = \begin{bmatrix} 1 & -|\alpha_{1,2}| & -|\alpha_{1,3}| & -|\alpha_{1,4}|\\ -|\alpha_{1,2}| & 1 & -|\alpha_{2,3}| & -|\alpha_{2,4}|\\ -|\alpha_{1,3}| & -|\alpha_{2,3}| & 1 & -|\alpha_{3,4}|\\ -|\alpha_{1,4}| & -|\alpha_{2,4}| & -|\alpha_{3,4}| & 1 \end{bmatrix}$

Based on my observation, matrix $M$ is not PSD for all PSD matrix $A$ in general. My question is, what is the certain property that guarantees PSD for matrix $M$? Now, Imagine we want to compute from end to the beginning, meaning that having matrices $A$ and non-PSD $M$, modify matrix $M$ to a PSD matrix(for example removing negative eigenvalues or ...) such that results in an equivalent matrix $A$ (let call $A_p$) that would be as same as possible to the primal matrix $A$. Is it possible?