LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)

How statement and proof provenance work

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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The real Francis double-shift polynomial is real and annihilates its trailing block

Statement

Let $B = (\begin{array}{cc} a & b \\ c & d \end{array}) \in R^{2 \times 2}$ , set $s = tr B = a + d$ , $t = \det B = a d - b c$ , and define $p (z) = z^{2} - s z + t$ . Then $p (z) = \det (z I - B)$ and $p (B) = 0$ . If $s^{2} - 4 t < 0$ , the roots are a nonreal conjugate pair $μ, \overline{μ}$ , and for every real square matrix $H$ of positive dimension,

$p (H) = H^{2} - s H + t I = (H - μ I) (H - \overline{μ} I)$

is real. This supplies the exact polynomial used by a real double-shift step; it asserts no convergence of the iteration.

Facts & Assumptions

Given: A real $2 \times 2$ matrix $B$ and the scalars $s, t$ above.

[F1]

Real and complex Schur forms: a real $2 \times 2$ Schur block for a conjugate pair has negative characteristic-polynomial discriminant.

Proof

1.1givenalgebra

Direct expansion gives $\det (z I - B) = (z - a) (z - d) - b c = z^{2} - (a + d) z + (a d - b c) = p (z)$ .

1.2givenalgebra

Direct multiplication gives $B^{2} = (\begin{array}{cc} a^{2} + b c & b (a + d) \\ c (a + d) & d^{2} + b c \end{array})$ , so subtracting $(a + d) B$ and adding $(a d - b c) I$ makes all four entries zero. Hence $p (B) = 0$ .

2.1givenF1step 1.1algebra∎

If $s^{2} - 4 t < 0$ , the quadratic formula gives roots $μ = s / 2 + i \sqrt{4 t - s^{2}} / 2$ and $\overline{μ}$ . Their product and sum are $t$ and $s$ , so $p (H) = (H - μ I) (H - \overline{μ} I)$ . In the application where $B$ is a real Schur block for a conjugate pair, [F1] supplies exactly this negative-discriminant hypothesis; the statement's general algebraic case already assumes it. Since $s, t, H$ are real, $H^{2} - s H + t I$ is real.

Depends on

Real and complex Schur forms

Used by

Dependency tree · two levels

3 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

James W. Demmel, Applied Numerical Linear Algebra, §4.4.8 (standard reference, not scraped)
David Bindel, CS 6210 Matrix Computations, Hessenberg QR iteration (standard reference, not scraped)