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Fill-in consists of new nonzero positions created by Gaussian elimination, absent from the original sparse matrix in the same ordering. For a Cholesky decomposition $PA{P}^T=C{C}^T$, compare $C$ with the lower triangle of $PA{P}^T$, not with the unpermuted input. Each new undirected edge of the elimination graph contributes one structural position in $C$. The sparse elimination of a four-leaf star gives six such positions when the center is eliminated first, and none when the leaves are eliminated first.

Symbolic factorization predicts possible nonzeros from the pattern alone. An initially zero $a_{ij}$ becomes nonzero in a single scalar step if both $a_{ik}$ and $a_{kj}$ are nonzero; however, at later steps several contributions can cancel. For an exact example, letEliminate indices $1,2$ in that order. The first Schur complement, in order $(2,3,4)$, isThe second pivot element updates the $(3,4)$ entry to $-1/3-(-1/2)(1)/(3/2)=0$. The remaining block is $\mathrm{diag}(7/3,7/3)$, so all four pivot elements are positive and $H$ is a symmetric positive-definite matrix. The elimination graph retains the structural edge $3-4$, but this exact choice of coefficients makes the numerical entry vanish. A small change to $H_{34}=H_{43}$ generally removes the cancellation.

Such an exact zero is different from discarding a small value computed with floating-point arithmetic. The symbolic factorization deliberately reserves space for possible entries without depending on special numerical identities. A perfect elimination ordering creates no structural fill-in at all; an ordering with cancellations need not have that property.