LemmaStatement: Literature-sourcedProof: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

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AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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Schur reduction preserves strict unit-disk root location

Statement

Let $p (z) = z^{n} + a_{n - 1} z^{n - 1} + \dots + a_{0}$ be a real monic polynomial of degree $n \geq 1$ with $∣ a_{0} ∣ < 1$ . Define the reversed polynomial $p^{#} (z) = z^{n} p (1 / z)$ and $q (z) = \frac{p (z) - a_{0} p^{#} (z)}{z (1 - a_{0}^{2})} .$ If every zero of $p$ lies strictly inside the unit disk, then every zero of $q$ does too.

Facts & Assumptions

Given: A real monic polynomial $p$ of degree at least one, $∣ a_{0} ∣ < 1$ , and the assumption that all its zeros lie in $∣ z ∣ < 1$ .

[A1]

The real polynomial is monic of degree at least one, $∣ p (0) ∣ < 1$ , and every root of $p$ is strictly inside the unit disk.

[F1]

Rouche's theorem in the classical strict-inequality form: if $∣ g (z) ∣ < ∣ f (z) ∣$ on a simple closed contour, then $f$ and $f + g$ have the same number of zeros inside, counted with multiplicity.

Proof

1.1A1givenF1algebra

On $∣ z ∣ = 1$ , real coefficients give $p^{#} (z) = z^{n} \overline{p (z)}$ , so $∣ p^{#} (z) ∣ = ∣ p (z) ∣$ . The strict root hypothesis makes $p$ nonzero on the unit circle, and $∣ a_{0} p^{#} (z) ∣ < ∣ p (z) ∣$ . Apply [F1] to $p$ and $- a_{0} p^{#}$ ; $p - a_{0} p^{#}$ has the same $n$ zeros in the disk as $p$ .

2.1A1givenstep 1.1algebra∎

The polynomial $p - a_{0} p^{#}$ has constant term zero and leading coefficient $1 - a_{0}^{2} \neq 0$ , so it is $z$ times a degree- $(n - 1)$ polynomial. One of its $n$ disk zeros is the factor zero at $z = 0$ ; the remaining $n - 1$ zeros are exactly those of the quotient. Dividing by the nonzero scalar $1 - a_{0}^{2}$ changes no roots. Hence all zeros of $q$ lie strictly inside the disk.

Depends on

Rouche's theorem in the classical strict-inequality form

Used by

The direct three-block ADMM extension can diverge Counterexample

Dependency tree · two levels

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Sources

R. W. Howell and J. H. Mathews, Complex Analysis, §8.7, Corollary 8.7.12 and Remark 8.7.13 (Rouché theorem used in the proof) (standard reference, not scraped)