Simultaneous diagonalisation

Statement

Let $F$ be a field and let $A$ and $B$ be diagonalisable $n \times n$ matrices over $F$. Then there is one invertible $n \times n$ matrix $P$ such that both $P^{-1} A P$ and $P^{-1} B P$ are diagonal if and only if $AB = BA$.

Remarks

Diagonal matrices commute, so the condition $AB = BA$ is necessary. For sufficiency, $B$ maps each eigenspace of $A$ to itself, and it is diagonalisable on each of them. The same holds for each family of pairwise commuting diagonalisable matrices.

Explanations

L1 human

Two diagonalisable matrices can be diagonalised by the same change of basis exactly when they commute. Then they share a basis of eigenvectors. This fact is important in quantum mechanics, where commuting observables can be measured together.