PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-16

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.

The companion matrix advances the recurrence state vector by one step

Statement

Let $a$ satisfy an order- $d$ recurrence from zero over a field, let $C$ be its row-shift companion matrix, and put $v_{n} = (a_{n}, \dots, a_{n + d - 1})^{T}$ . Then

$v_{n + 1} = C v_{n} and v_{n} = C^{n} v_{0}$

for every $n \geq 0$ .

Facts & Assumptions

Given: An order- $d$ recurrence sequence $a$ , its state vectors $v_{n}$ , and its companion matrix $C$ .

[L1]

The companion matrix has shift rows and final row $(- c_{d}, - c_{d - 1}, \dots, - c_{1})$ (The row-shift companion matrix of a linear recurrence).

[L2]

Matrix multiplication is $(A B)_{i k} = \sum_{j} a_{i j} b_{j k}$ , and $I_{d}$ is the identity matrix (Entrywise ring-matrix operations, rectangular matrix products, identity matrices and transpose).

[L3]

Matrix multiplication is associative and satisfies $I_{d} A = A = A I_{d}$ (Matrix arithmetic over a commutative ring is associative, unital and distributive, and transpose reverses products).

Proof

technique · induction

1.1givenL1L2

Multiplying $C$ by $v_{n}$ , the first $d - 1$ rows return $a_{n + 1}, \dots, a_{n + d - 1}$ and the final row returns $- c_{d} a_{n} - \dots - c_{1} a_{n + d - 1} = a_{n + d}$ , so $C v_{n} = v_{n + 1}$ .

1.2baseL2L3

At $n = 0$ , one has $C^{0} v_{0} = I_{d} v_{0} = v_{0}$ .

2.1ihstep 1.1L3

If $v_{n} = C^{n} v_{0}$ , then step 1.1 and associativity give $v_{n + 1} = C v_{n} = C^{n + 1} v_{0}$ .

3.1step 1.2step 2.1discharge-induction∎

Induction proves $v_{n} = C^{n} v_{0}$ for every $n \geq 0$ , together with the one-step identity from step 1.1.

Depends on

Used by

The Fibonacci companion matrix advances (Fₙ,Fₙ₊₁) Example

Dependency tree · two levels

9 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

M. Waldschmidt, Linear Recurrence Sequences VI, slide 17 (standard reference, not scraped)