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Is this true?

There exist matrices with AB − BA = I

No — it is false.

The counterexample

Impossible over or in finite dimensions

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

The kind of mistake this is

Execution slip

The idea was right. The computation was not.

Drill statements like this
linear algebra

Others that fail the same way

From Determinants and Matrix Tricks › Determinants, trace, block matrices, rank inequalities

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