Example. Nonzero operators outside the trace-ratio domain [fdip-003I]

✍️sourceAGENTDRAFTED

The unit condition in the trace ratio distinguishes several elementary cases. For one unit-weight edge on two vertices, the Laplacian over \(\mathbb Z\) is \[L=\begin {pmatrix}1&-1\\-1&1\end {pmatrix}.\] Its trace is \(2\), which is not a unit in \(\mathbb Z\); the defining formula therefore does not give a \(\mathbb Z\)-valued ratio. After inverting \(2\) it does: \(L^2=2L\) gives ratio \(1\). Reduction over \(\mathbb F_2\) instead gives the nonzero matrix \[\overline L=\begin {pmatrix}1&1\\1&1\end {pmatrix},\] whose trace is zero and whose square is zero. Nonzero incidence energy as an algebraic operator does not imply a unit trace.

Over \(\mathbb R\), the nonzero self-adjoint matrix \(\operatorname {diag}(1,-1)\) has trace zero. It is not positive semidefinite, so it does not contradict the positive-trace proof. For an incidence presentation containing only loops, \(d=0\) and \(L=d^\dagger Wd=0\); the trace is zero whatever the loop weights. The empty carrier likewise has zero trace by the empty dual-frame sum. In each case the ratio is undefined over a nonzero ring.

In the zero ring, \(0=1\) is its sole unit, and every unital module and scalar has one element. The formula for the ratio then has the unique value \(0\). This algebraic convention makes no assertion about a nonzero positive-real Laplacian.

[spielman2009effective, Section 2.1, pp. 4-5] gives the edge-once incidence convention used here, as extended algebraically in Definition [fdip-003B]. These computations also use the directly proved dual-frame trace.

[boyd2004convex, Appendix A.5.2, pp. 646-647] gives the spectral criterion for the real positivity boundary. The ring reductions and zero-ring conclusion are the explicit calculations here, not probability statements from those sources.