A star graph has no fill in leaf-first order and a leaf clique in center-first order

Write a vector as $x=(x_0,y_1,\dots ,y_m)$. The quadratic form is $x^{T}Ax=x_0^2+{\sum }_i{y}_i^2+2t{x}_0{\sum }_i{y}_i$. Since $m{\sum }_i{y}_i^2-({\sum }_i{y}_i{)}^2={\sum }_{i<j}(y_i-y_j{)}^2\ge 0$, we have $\mathrm{\mid }{\sum }_i{y}_i\mathrm{\mid }\le \sqrt{m}\mathrm{\parallel }y{\mathrm{\parallel }}_2$. Therefore $2t\mathrm{\mid }{x}_0\mathrm{\mid }\mid {\sum }_i{y}_i\mid \le 2t\sqrt{m}\mathrm{\mid }{x}_0\mathrm{\mid }\mathrm{\parallel }y{\mathrm{\parallel }}_2\le t\sqrt{m}(x_0^2+\mathrm{\parallel }y{\mathrm{\parallel }}_2^2)=\frac{1}{2}\mathrm{\parallel }x{\mathrm{\parallel }}_2^2\mathrm{.}$ It follows that $x^{T}Ax\ge \frac{1}{2}\mathrm{\parallel }x{\mathrm{\parallel }}_2^2>0$ for $x\mathrm{\ne }0$, so $A$ is positive definite.

In leaf-first order, each leaf has exactly one later neighbor, the center. By [F1] no pair of distinct remaining neighbors exists, so eliminating that leaf adds no edge. This holds for every leaf; eliminating the last center also adds no edge.

In center-first order, the center's later-neighbor set is all $m$ leaves. By [F1], the graph update inserts every pair of distinct leaves, so it has $\left(\genfrac{}{}{0px}{}{m}{2}\right)$ leaf-to-leaf fill edges not in the original star.

The pivot at the center is $1$. Its Schur complement on the leaves is $I_m-t^2{\mathbf{11}}^T$, whose off-diagonal entries are $-t^2\mathrm{\ne }0$. Thus the symbolic fill edges from step 1.3 are also numerical nonzeros for this matrix, while step 1.2 gives no fill in the other order.

Steps 1.2 and 2.1 exhibit respectively $0$ and $\left(\genfrac{}{}{0px}{}{m}{2}\right)$ fill edges for two orderings of the same positive-definite sparse matrix.