Property of positive semi-definite

Question

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. 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?

Answer 1

I am just putting a long possible comment here. The following has many reformulations; the idea is to give a condition for which the two matrices $A$ and $M$ in the OP question are congruent by a unitary diagonal matrix. Let $A=[x_{i,j}] $ be an $n\times n$ complex hermitian matrix. Among the entries $x_{i,j}$ with $j>i$: If for every column ($2\le j\le n$) or every line ($1\le i \le n-1$) there is at most one non zero entry $x_l$ with $l=1,\ldots, k$ and $k\le n-1$, then the hermitian matrix $M=[y_{i,j}]$, where $y_{i,j}=\begin{cases}z_lx_{i,j} &\text{ for }x_{i,j}=x_l, l=1,\ldots,k;|z_l|=1\\x_{i,j}& \text{ for } j\ge i \text{ and }x_{i,j}\neq x_l, l=1,\ldots,k\end{cases}$ is unitarily congruent to $A$ $(M=U^*AU)$ by a diagonal matrix with diagonal entries $\subset\{1,z_1,\ldots,z_k\}$. The proof is a bit direct. I let you figure the relation with the original question.