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 $M$ very useful to guarantee convergence in Ribando's theorem at [Measuring solid angles beyond dimension three](https://doi.org/10.1007/s00454-006-1253-4). 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, even if $A$ is PSD, the matrix $M$ need not be PSD 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 to $A$ (let us call it $A_p$) that would be as nearly the same as possible to the primal matrix $A$. Is it possible?