ExampleConstruction: AI-generatedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-16

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A companion operator with a visible cyclic vector and equal canonical polynomials

Example

For $f = x^{3} + x + 1$ , let $C = (\begin{array}{ccc} 0 & 0 & - 1 \\ 1 & 0 & - 1 \\ 0 & 1 & 0 \end{array}) .$ Then $e_{1}, C e_{1}, C^{2} e_{1}$ are $e_{1}, e_{2}, e_{3}$ , so $e_{1}$ is cyclic. The matrix is the companion matrix of $f$ , and $μ_{C} = χ_{C} = f .$

Remarks

Over $F_{2}$ , the polynomial $x^{3} + x + 1$ is irreducible and the same column orientation represents multiplication by its residue class in $F_{8} = F_{2} [x] / (x^{3} + x + 1)$ . This fixes the convention for a downstream computation of the Frobenius map of $F_{8} / F_{2}$ ; the present matrix is multiplication by the residue class, not the Frobenius operator, and no forward dependency is used here.

Facts & Assumptions

Given: The displayed companion matrix $C$ and $v = e_{1}$ .

[L1]

If $m_{T, v} = x^{d} + a_{d - 1} x^{d - 1} + \dots + a_{0}$ , then $(v, T v, \dots, T^{d - 1} v)$ is an ordered basis of $Z (v; T)$ , and in this basis $T ∣_{Z (v; T)}$ has the companion matrix with ones on the subdiagonal and last column $(- a_{0}, \dots, - a_{d - 1})$ (A vector annihilator gives a power basis and its companion matrix).

[L2]

An endomorphism of a finite-dimensional vector space has a cyclic vector if and only if $μ_{T} = χ_{T}$ (A cyclic vector exists exactly when the minimal and characteristic polynomials agree).

[L3]

For $A \in M_{n} (F)$ , the polynomial $χ_{A} (x)$ is monic of degree $n$ ( $χ_{A} (x)$ is monic of degree $n$ ; for $n \geq 1$ its $x^{n - 1}$ coefficient is $- tr (A)$ and its constant coefficient is $(- 1)^{n} \det (A)$ , while $χ_{0 \times 0} = 1$ ).

Verification

technique · computation

1.1

Matrix multiplication gives $C e_{1} = e_{2}$ and $C^{2} e_{1} = e_{3}$ , so the three power vectors are the standard basis and $e_{1}$ is cyclic.

algebra

2.1

The columns show $C e_{3} = - e_{1} - e_{2}$ , so $C^{3} e_{1} = C e_{3} = - e_{1} - e_{2} = - e_{1} - C e_{1}$ and hence $f (C) e_{1} = (C^{3} + C + I) e_{1} = 0$ . Since $f (C)$ commutes with $C$ , $f (C) C^{j} e_{1} = C^{j} f (C) e_{1} = 0$ for $j = 0, 1, 2$ , and step 1.1 makes $e_{1}, C e_{1}, C^{2} e_{1}$ a basis, so $f (C) = 0$ .

step 1.1algebra

3.1

By step 1.1 the vector $e_{1}$ is cyclic, so [L2] gives $μ_{C} = χ_{C}$ , and [L3] makes $χ_{C}$ monic of degree $3$ ; thus $μ_{C}$ is monic of degree $3$ . By step 2.1, $μ_{C}$ divides the monic degree-three polynomial $f$ , so $μ_{C} = f$ and therefore $μ_{C} = χ_{C} = f$ . Since $m_{C, e_{1}}$ divides $f$ and $(e_{1}, C e_{1}, C^{2} e_{1})$ is a basis of $Z (e_{1}; C) = F^{3}$ , $m_{C, e_{1}} = f$ , and [L1] in that basis is exactly the displayed matrix, with last column $(- a_{0}, - a_{1}, - a_{2}) = (- 1, - 1, 0)$ .

step 1.1step 2.1L1L2L3algebra∎

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