Hyperbolic quadratic — the series
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Every eigenvalue real, and a test that says so
A quadratic eigenvalue problem has no reason to have real eigenvalues. One class does, as a property rather than an outcome, and the proof is a Cholesky that completes. The boundary of the class has a closed form, and at the boundary the arithmetic loses half its digits with nothing ill conditioned anywhere.
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A class a longer chain takes away
Symmetry survives a bigger problem. Hyperbolicity does not. The damping that certifies a chain of seven masses is refused by a chain of eight, the damping the class demands grows like the length without bound, and the certificate has to be earned again at every size — which costs one Cholesky, and the alternative is a proof quietly inherited from a smaller problem.
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A guarantee paid for in digits
A hyperbolic quadratic has a real spectrum, but the route that computed it so far, a companion linearisation and a nonsymmetric Schur form, only happens to return one. The theory offers a route on which it cannot do otherwise: a symmetric linearisation that is a definite pencil, reduced by a Cholesky to a symmetric eigenproblem. Near the class's boundary that guarantee is the expensive route. On a damped chain at damping β*(1 + ε) the definite pencil's Crawford number — what its Cholesky divides by — vanishes like 0.228ε, every combination of the pencil shrinking with it, while the colliding pair the companion route has to resolve separates like 0.98√ε. At ε = 10⁻¹² the definite route keeps two or three digits and the companion route eight or nine. And on twenty rotated chains the companion route returns no complex pair at any margin from 10⁻⁴ to 10⁻¹², only on the boundary itself.