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Is this true?

Commuting matrices are simultaneously diagonalisable

No — it is false.

The counterexample

A = B = [[0,1],[0,0]]

They commute but neither is diagonalisable. Need each to be diagonalisable first.

The kind of mistake this is

Converse assumed

The theorem runs one way. You used it in the other.

Drill statements like this
linear algebra

Others that fail the same way

From Eigenvalues and Canonical Forms › Diagonalisability criteria

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