ExampleConstruction: Literature-sourcedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-24

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The real quarter-turn has rational canonical form but no real Jordan form

Example

The real quarter-turn

$A = (\begin{array}{cc} 0 & - 1 \\ 1 & 0 \end{array})$

is the companion matrix of $x^{2} + 1$ and hence is in rational canonical form over $R$ , but it has no Jordan canonical form over $R$ . Over a field containing $i$ , its Jordan form is $diag (i, - i)$ .

Facts & Assumptions

Given: Rational canonical form from Existence and uniqueness of rational canonical form and the module construction of Jordan form after splitting (Jordan canonical form from the elementary divisors of $V_{T}$ ).

[L1]

An endomorphism has a Jordan canonical form over its base field if and only if its characteristic polynomial splits into linear factors there (Jordan form over the base field exists exactly when the characteristic polynomial splits).

Verification

technique · direct

1.1

With $e_{1} = (1, 0)$ one has $A e_{1} = e_{2}$ and $A^{2} e_{1} = - e_{1}$ , so $e_{1}, A e_{1}$ is a cyclic basis with annihilator $x^{2} + 1$ . The displayed matrix is exactly $C (x^{2} + 1)$ and is its one-block rational canonical form.

givenalgebra

2.1

The polynomial $x^{2} + 1$ has no real root and does not split over $R$ . By [L1], the real operator has no real Jordan form.

step 1.1L1

3.1

After adjoining $i$ , the polynomial splits as $(x - i) (x + i)$ with distinct roots. The module Jordan theorem gives the two one-dimensional blocks $i$ and $- i$ , whose dimensions sum to two.

step 2.1given∎

Depends on

Used by

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Sources

A. Apisa, Wisconsin Math 542, Remark 39 (standard reference, not scraped)
M. Brussel, Finitely Generated Modules over a PID, canonical-form discussion (standard reference, not scraped)