Commuting diagonalizable operators share an eigenbasis.Suggest a slogan
Let \(V\) be a finite-dimensional vector space over a field \(k\), and let \(\theset{A_i}_{i \in I}\) be a family of diagonalizable linear operators on \(V\). The \(A_i\) pairwise commute if and only if they are simultaneously diagonalizable: there is a basis of \(V\) in which every \(A_i\) is diagonal.
The implication from simultaneous diagonalizability to commuting holds for any family of operators, since diagonal matrices commute.
Commuting operators need not be simultaneously diagonalizable without the diagonalizability hypothesis: \(\begin{bmatrix} 1 & 1 \\ 0 & 1\end{bmatrix}\) commutes with itself and is not diagonalizable.