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

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Powers of a Schur-stable matrix decay geometrically up to a polynomial factor

Statement

Let $M \in C^{n \times n}$ be Schur stable and let $r$ satisfy $ρ (M) < r < 1$ . For the Euclidean induced operator norm there is a finite $C_{r}$ such that $∥ M^{k} ∥_{2} \leq C_{r} r^{k} (k \geq 0) .$ For a real matrix, interpret the same statement on its complexification.

Facts & Assumptions

Given: A finite Schur-stable matrix $M$ and $r$ with $ρ (M) < r < 1$ .

[A1]

For $n \geq 1$ , the notation $ρ (M)$ means $\max {∣ λ ∣ : λ \in σ_{C} (M)}$ ; for the unique $0 \times 0$ matrix set $ρ (M) = 0$ .

[F1]

Stabilizability and detectability for continuous and discrete linear systems: Schur stable means every eigenvalue has modulus strictly less than one.

[F2]

If the characteristic polynomial of an endomorphism splits, some orthonormal basis makes its matrix upper triangular: a complex finite-dimensional endomorphism has an orthonormal basis in which its matrix is upper triangular.

[F3]

The operator norm is submultiplicative and satisfies ||T^*T|| = ||T||^2: $∥ S T ∥_{2} \leq ∥ S ∥_{2} ∥ T ∥_{2}$ for compatible finite-dimensional maps.

[F4]

The operator norm is zero on the zero domain and otherwise is max_{||v||=1} ||Tv||: $∥ T ∥_{2} = \max_{∥ v ∥_{2} = 1} ∥ T v ∥_{2}$ .

[F5]

Every nonconstant polynomial in $C [x]$ splits into linear factors: every nonconstant characteristic polynomial over $C$ splits over $C$ , so the hypothesis of [F2] holds.

[F6]

Linear isometries, and orthogonal or unitary operators on finite-dimensional inner product spaces: a unitary operator and its adjoint preserve the Euclidean norm.

[F7]

For $A \in M_{n} (F)$ , the characteristic polynomial is $χ_{A} (x) = \det (x I_{n} - A)$ when $n \geq 1$ , with $χ_{A} (x) = 1$ for the unique $0 \times 0$ matrix and The determinant of a triangular matrix is the product of its diagonal entries: for upper-triangular $T$ , $χ_{T} (z) = \det (z I - T) = \prod_{i} (z - T_{i i})$ .

[F8]

Eigenvalues, eigenvectors, eigenspaces $E_{λ} (T) = \ker (T - λ I)$ , and the spectrum $σ_{F} (T)$ of an endomorphism and For every finite-dimensional space, $σ_{F} (T)$ is exactly the set of roots in $F$ of $χ_{T}$ : the complex spectrum is the root set of the characteristic polynomial.

Proof

1.1A1F1F2F5F7F8construct

If $n = 0$ , the unique endomorphism and every power of it is the zero map on the zero domain; its operator norm is zero, so the assertion holds with $C_{r} = 0$ . Assume now $n \geq 1$ and regard a real matrix, if present, as a complex matrix. Its characteristic polynomial is a nonconstant complex polynomial, so [F5] and [F2] give $M = U T U^{*}$ with $U$ unitary and $T = D + N$ , where $D$ is diagonal and $N$ is strictly upper triangular. Similar matrices have the same eigenvalues: $M v = λ v$ is equivalent to $T (U^{*} v) = λ (U^{*} v)$ . By [F7] and [F8], the eigenvalues are the diagonal entries of $T$ ; hence $ρ (M) = \max_{i} ∣ T_{i i} ∣ = ∥ D ∥_{2}$ , where the last equality follows directly from the Euclidean induced norm. Set $ρ = ρ (M) < r$ . Each occurrence of $N$ in a product raises the row-to-column index, while $D$ preserves it, so every word containing at least $n$ copies of $N$ is zero.

2.1F3F4F6step 1.1algebra

Expand $(D + N)^{k}$ as the sum of noncommuting words. The terms with exactly $j$ copies of $N$ number $(\binom{k}{j})$ , vanish for $j \geq n$ , and for $j < n$ have norm at most $ρ^{k - j} ∥ N ∥_{2}^{j}$ by [F3] and $∥ D ∥_{2} = ρ$ . Thus $∥ T^{k} ∥_{2} \leq \sum_{j = 0}^{\min (k, n - 1)} (\binom{k}{j}) ρ^{k - j} ∥ N ∥_{2}^{j} .$ Because $U$ and $U^{*}$ preserve vector norms by [F6] and $U^{*}$ maps the unit sphere onto itself, the induced norm [F4] gives $∥ M^{k} ∥_{2} = ∥ U T^{k} U^{*} ∥_{2} = ∥ T^{k} ∥_{2}$ .

3.1step 2.1algebracases∎

If $ρ = 0$ , then $D = 0$ and $T = N$ , so $T^{k} = 0$ for $k \geq n$ ; taking $C_{r} = \max_{0 \leq k < n} ∥ T^{k} ∥_{2} r^{- k}$ proves the claim. If $0 < ρ < r$ , each fixed $j < n$ satisfies $(\binom{k}{j}) ρ^{k - j} ∥ N ∥_{2}^{j} / r^{k} \leq c_{j} k^{j} (ρ / r)^{k}$ for a finite $c_{j}$ . The ratio of consecutive terms $k^{j} (ρ / r)^{k}$ tends to $ρ / r < 1$ , so this sequence is bounded. Summing its finitely many bounds over $0 \leq j < n$ gives a finite $C_{r}$ in step 2.1.

Depends on

Used by

Dependency tree · two levels

33 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

Federico Poloni, Lecture notes on Riccati-type matrix equations (standard reference, not scraped)
A.A. Stoorvogel and A. Saberi, The discrete algebraic Riccati equation and linear matrix inequality, Memorandum COSOR 94-16 (standard reference, not scraped)