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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?

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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$~V$.

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