Is the following statement true? There exist matrices with AB − BA = I

No. Impossible over ℝ or ℂ in finite dimensions trace(AB − BA) = 0 but trace(I) = n ≠ 0. (It is possible for unbounded operators — the Heisenberg relation.)

Counterexample

Is this true?

There exist matrices with AB − BA = I

No — it is false.

The counterexample

Impossible over $\mathbb{R}$ or $\mathbb{C}$ in finite dimensions

Why it fails

trace(AB − BA) = 0 but trace(I) = n ≠ 0. (It is possible for unbounded operators — the Heisenberg relation.)

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Execution slip

The idea was right. The computation was not.

Same characteristic polynomial ⇒ similar
Diagonalisable ⇒ invertible
Real matrix with real eigenvalues is diagonalisable

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

linear algebra

Go deeper· Explained in full in Determinants, trace and block matrices· Asked in CSIR NET December 2024· From Determinants and Matrix Tricks › Determinants, trace, block matrices, rank inequalities

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

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