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

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A recurrence companion matrix has the recurrence characteristic polynomial

Statement

Let $K$ be a field, let $d \geq 1$ , and let $C$ be the row-shift companion matrix associated with

$χ (t) = t^{d} + c_{1} t^{d - 1} + \dots + c_{d} .$

Then the matrix characteristic polynomial is exactly

$χ_{C} (t) = \det (t I_{d} - C) = t^{d} + c_{1} t^{d - 1} + \dots + c_{d} .$

Facts & Assumptions

Given: A positive integer $d$ , coefficients $c_{1}, \dots, c_{d}$ , and their row-shift companion matrix $C$ .

[L1]

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

[L2]

For a field matrix $A$ , its characteristic polynomial is $χ_{A} (t) = \det (t I - A)$ (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).

[L3]

A determinant may be expanded along any row or column as the sum of entries times their cofactors (Laplace expansion computes the determinant along every row and every column over a commutative ring).

Proof

technique · induction

1.1baseL1L2

For $d = 1$ , one has $C = (- c_{1})$ and $\det (t I - C) = t + c_{1}$ .

1.2ihL1L2

Assume the formula for size $d - 1$ . In $t I_{d} - C$ , the first column has only two nonzero entries: $t$ in row $0$ and $c_{d}$ in row $d - 1$ .

2.1step 1.2L3

Expanding that column by [L3], the cofactor of $t$ is the size- $(d - 1)$ companion determinant $t^{d - 1} + c_{1} t^{d - 2} + \dots + c_{d - 1}$ by the induction hypothesis.

2.2step 1.2L3algebra

The minor of the entry $c_{d}$ is triangular with diagonal entries $- 1$ ; its determinant is $(- 1)^{d - 1}$ , and the cofactor sign is also $(- 1)^{d - 1}$ , so this contribution is $c_{d}$ .

3.1step 2.1step 2.2algebra

Therefore $\det (t I_{d} - C) = t (t^{d - 1} + c_{1} t^{d - 2} + \dots + c_{d - 1}) + c_{d}$ , which is the claimed polynomial.

4.1step 1.1step 3.1discharge-induction∎

The base case and induction step prove the formula for every $d \geq 1$ .

Depends on

Used by

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

Dependency tree · two levels

12 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)
H. Pinkham, Linear Algebra, Chapter 10 (standard reference, not scraped)