False statementConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-08-16

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FALSE: The characteristic polynomial determines whether an endomorphism is diagonalisable

Statement

False claim. Two endomorphisms with the same characteristic polynomial are either both diagonalisable or both non-diagonalisable.

Facts & Assumptions

Given: The matrices $I_{2}$ and $J = (\begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array})$ .

[L1]

A block-triangular characteristic polynomial is the product of those of its diagonal blocks (The characteristic polynomial of a block upper- or lower-triangular matrix is the product of the characteristic polynomials of its diagonal blocks).

[L2]

Diagonalisability is controlled by the minimal polynomial being a product of distinct linear factors (An endomorphism is diagonalisable if and only if its minimal polynomial is a product of distinct linear factors).

Refutation

technique · direct

1.1L1

By [L1], both matrices have characteristic polynomial $(x - 1)^{2}$ .

2.1L2algebra∎

The identity is diagonal. For $J$ , $(J - I)^{2} = 0$ but $J - I \neq 0$ , so $μ_{J} = (x - 1)^{2}$ and [L2] says $J$ is not diagonalisable. Thus the shared characteristic polynomial does not decide diagonalisability.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources

Keith Conrad, The Minimal Polynomial and Some Applications, Example 4.2 and Theorem 4.11 (standard reference, not scraped)