# Simultaneously diagonalizable linear operators on a space of square matrices [duplicate]

Let $$V$$ be the space of $$n×n$$ matrices over the field $$F$$. For a fixed $$n×n$$ matrix $$A$$ over $$F$$, let $$T_A$$ be the linear operator on $$V$$ defined by $$T_A(B) = AB −BA$$. Consider the family of linear operators obtained by letting $$A$$ vary over all diagonal matrices. Prove that the operators in this family are simultaneously diagonalizable.

Consider any two linear operators $$T_{A_1}$$ and $$T_{A_2}$$, then $$T_{A_2}\left( T_{A_1}(B) \right) = T_{A_2}(A_1B-BA_1) =$$ $$A_2(A_1B-BA_1)-(A_1B-BA_1)A_2 = A_2A_1B-A_2BA_1-A_1BA_2+BA_1A_2 =$$ $$A_1(A_2B-BA_2)-(A_2B-BA_2)A_1 = T_{A_1}\left( T_{A_2}(B) \right)$$. Hence $$T_{A_2}$$ and $$T_{A_1}$$ commute. Now it is enough to show that $$T_{A_2}$$ and $$T_{A_1}$$ are diagonalizable, since operators are simultaneously diagonalizable if and only if they commute and are diagonalizable. But how do i show that $$T_{A_2}$$ and $$T_{A_1}$$ are diagonalizable?

The effect of the (additive) commutator $$T_A$$ with a diagonal matrix$$~A$$ on an elementary matrix $$E_{k,l}=(\delta_{i,k}\delta_{j,l})_{i,j=1}^n$$ is scalar multiplication by $$a_i-a_j$$, where $$a_i$$ is the diagonal entry of$$~A$$ at position $$i,i$$. Therefore all such operators $$T_A$$ diagonalise on the basis $$\{E_{k,l}\mid k=1,\ldots,n; l=1,\ldots,n\,\}$$ of$$~V$$.