Let \(F\) be a field, \(n\) a positive integer, and let \(V\) be the space of \(n \times n\) matrices over \(F\). If \(A\) is a fixed \(n \times n\) matrix over \(F\), let \(T_A\) be the linear operator on \(V\) defined by \(T_A(B) = AB - BA\). Consider the family of linear operators \(T_A\) obtained by letting \(A\) vary over all diagonal matrices. Prove that the operators in that family are simultaneously diagonalizable.
Solution
Let \(E_{ij}\), \(1\leq i,j\leq n\), be the standard matrix units. These \(n^2\) matrices form a basis of \(V\).
1. Every \(E_{ij}\) is an eigenvector for every operator \(T_A\) in the family.
Fix a diagonal matrix \[ A=\operatorname{diag}(a_1,\ldots,a_n). \] Left multiplication by \(A\) scales the \(i\)th row of \(E_{ij}\) by \(a_i\), while right multiplication scales its \(j\)th column by \(a_j\). Explicitly, \[ AE_{ij}=a_iE_{ij}, \qquad E_{ij}A=a_jE_{ij}. \] Therefore \[ T_A(E_{ij})=AE_{ij}-E_{ij}A=(a_i-a_j)E_{ij}. \] Thus \(E_{ij}\) is an eigenvector of \(T_A\) for every diagonal \(A\); only its eigenvalue changes with \(A\) [@HK71].
2. The matrix-unit basis diagonalizes the whole family simultaneously.
Step 1 shows that the single basis \[ \{E_{ij}:1\leq i,j\leq n\} \] consists of eigenvectors for every operator \(T_A\) with \(A\) diagonal. Hence the matrix of every such \(T_A\) in this same basis is diagonal, with diagonal entries \(a_i-a_j\) indexed by the ordered pairs \((i,j)\). Therefore the entire family is simultaneously diagonalizable.