Hermitian Schur-complement support stays inside the symmetric filled graph

At stage $0$, $S^0=A$ is Hermitian. If $S^k$ is Hermitian, then its diagonal pivot $p_k$ is real. Since it is nonzero, substituting $a=p_k,b=0$ in [F4] gives $p_k^{-1}=p_k\mathrm{/}{p}_k^2=1\mathrm{/}{p}_k\in \mathbb{R}$; the denominator is nonzero by [F3]. For remaining $i,j$, $\overline{S_{ji}^{k+1}}=\overline{S_{ji}^k-S_{j{v}_k}^k{p}_k^{-1}{S}_{v_{k}i}^k}=S_{ij}^k-S_{i{v}_k}^k{p}_k^{-1}{S}_{v_{k}j}^k=S_{ij}^{k+1}\mathrm{.}$ Thus the update preserves Hermitian symmetry.

By [F2], the graph update itself joins every pair of distinct later neighbors of $v_k$. Thus each such later-neighbor set is a clique in the next filled graph.

Induct on the elimination stage for support. Initially every nonzero upper-triangular entry is an edge by [F1]. Suppose the inclusion holds at stage $k$ and $S_{ij}^{k+1}\mathrm{\ne }0$ for $i<j$. If $S_{ij}^k\mathrm{\ne }0$, its edge is retained by [F2]. Otherwise the correction term is nonzero. Since $\mathbb{C}$ is a field by [F4], applying [F3] to $(S_{i{v}_k}^k{p}_k^{-1})S_{v_{k}j}^k$ and then to $S_{i{v}_k}^k{p}_k^{-1}$ shows that $S_{i{v}_k}^k\mathrm{\ne }0$ and $S_{v_{k}j}^k\mathrm{\ne }0$. Hermitian symmetry from step 1.1 gives the undirected edges from $i$ to $v_k$ and from $v_k$ to $j$ in the active graph. Hence both vertices are later neighbors of $v_k$, and [F2] inserts $\{i,j\}$. This proves the support inclusion at stage $k+1$.

The base case and step 2.1 prove the support inclusion at every stage; step 1.2 proves the clique property. The proof uses conjugate symmetry for Hermitian matrices and therefore supplies the complex-Hermitian extension beyond the real-symmetric supplier theorem.