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Given any symmetric matrix $S \in \mathbb{R}^{n \times n}$, if $S \succeq 0$, is there a way to encode $S$ into a graph such that it takes into account the positive semidefinite constraint, and ideally, also the sparsity of $S$?

For any symmetric matrix $M \in \mathbb{R}^{n\times n}$, I think we can encode it as an undirected graph by simply considering $n$ vertices, and with entries of the matrix corresponding to edge weights. However, I cannot think of any specific ways to characterize the subset of positive semidefinite matrices. Maybe we can use the determinant characterization to do some tricks, but I am not very familiar with graph theory.

The motivation is that I want to consider the minimization problem $\operatorname{min}_{X \succeq 0} \|\mathscr{A}(X) - b\|^{2}$, where $\mathscr{A}(X) = [\langle A_{1},X \rangle, \langle A_{2}, X\rangle ,\dots ,\langle A_{n}, X \rangle]$. $A_{1},\dots ,A_{n} \in \{ 0,1 \}^{n \times n}$ and are pairwise orthogonal, somewhat like a generalization of matrix completion. I just recently heard of Chordal graphs, and maybe it's related, but I certainly need to read more about it.

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    $\begingroup$ You may find Ch. 9 of this reference helpful: seas.ucla.edu/~vandenbe/publications/chordalsdp.pdf $\endgroup$ Commented Jan 23 at 4:12
  • $\begingroup$ You might want to look into the work of Miroslav Fiedler, espacially chapter 6.1 and 6.2 of his book "Matrices and Graphs in Geometry" where he assigns to graphs the Laplacian (a positive semidefinite matrix) and conversely to a simplex given by a positive definite matrix he asssigns a graph. $\endgroup$ Commented Jan 23 at 5:15
  • $\begingroup$ If you replace a solution $X$ with $\Phi(X)$ where $\Phi$ is the projection onto the coherent algebra generated by $A_1, \ldots, A_n$, then $\mathcal{A}(X)$ does not change and $\Phi$ also preserves positivity. So you can restrict your search to matrices in the coherent algebra generated by $A_1, \ldots, A_n$. If this happens to be an association scheme, then the positivity condition can be replaced by linear constraints. $\endgroup$ Commented Mar 1 at 11:49

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