Counterexample

Is this true?

Real matrix with real eigenvalues is diagonalisable

No — it is false.

The counterexample

[[1,1],[0,1]]

Why it fails

Eigenvalue 1 with geometric multiplicity 1 < algebraic 2.

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Keep practicing

Invariants don't determine

Two objects sharing an invariant were treated as the same object.

See real exam options built on this mistake Drill statements like this

Go deeper· Explained in full in Diagonalisability — the criteria card· Asked in CSIR NET December 2025· From Eigenvalues and Canonical Forms › Diagonalisability criteria

Published 23 Aug 2026. Found a mistake? Tell us — corrections are the fastest thing we act on.

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