 Sparse elimination of a four-leaf star

Take a star graph $G$ with center $c$ and four leaves $1,2,3,4$. Let $L_G$ be its Laplacian matrix and $I$ the identity matrix. In the order $(c,1,2,3,4)$, the original example isFor a nonzero vector $u$, its quadratic form isThus $A$ is a symmetric positive-definite matrix and either ordering below admits a Cholesky decomposition. Changing the ordering changes the work, rather than the mathematical solvability.

Center first. Eliminating $c$ with pivot element $5$ leaves the Schur complementEvery pair of leaves now has coupling $-1/5$: six new edges, hence six undirected fill-in positions. The four remaining vertices form a clique. In the Cholesky decomposition, the first column has $C_{cc}=\sqrt{5}$ and $C_{ic}=-1/\sqrt{5}$. The rest of the lower triangular matrix is dense. To check that later cancellation does not remove its entries, after $r$ leaves have also been eliminated, the active leaf block isIndeed a leaf step sends $\alpha$ to $\alpha +{\alpha }^2/(2-\alpha )$. All active off-diagonal entries remain negative. The successive pivot elements in ordinary symmetric Gaussian elimination, equivalently the squared diagonals in the Cholesky decomposition, areThe Cholesky decomposition therefore stores $5+4+3+2+1=15$ nonzero entries, including its diagonal.

Leaves first. Permute to $(1,2,3,4,c)$. Each leaf has only one remaining neighbor, so its elimination changes only the center's diagonal, reducing it by $(-1{)}^2/2=1/2$. The successive pivot elements are $2,2,2,2,3$, and the exact Cholesky decomposition isThere is no fill-in, and only $5+4=9$ nonzero entries in $C$. This is a concrete instance of leaf elimination of a tree-pattern positive-definite matrix; its general proof explains why the same idea works on every tree pattern. The minimum degree algorithm also chooses a leaf at every nontrivial stage of this particular example, but that success gives no general optimality guarantee.

The input has 13 nonzero matrix elements when both symmetric halves are counted, or nine in its lower triangle. The six added undirected edges correspond to six added positions in one triangular factor, or 12 positions if both symmetric halves of the filled pattern are displayed. Comparing all 13 input entries directly with the 15 entries of $C$ would mix counting conventions.