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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

1.1

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

baseL1L2

1.2

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 $d - 1$ .

ihL1L2

2.1

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.

step 1.2L3

2.2

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 $c_{d}$ .

step 1.2L3algebra

3.1

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.

step 2.1step 2.2algebra

4.1

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

step 1.1step 3.1discharge-induction∎

Depends on

Used by

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

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 35 results over 8 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click 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)