Determine Commutativity

The linear transformation $T_A$ $(A\in\mathcal V)$ on vector space $M_n(\mathbb C),$ defined by $T_A(X)=AX-XA,$ is diagonalizable since $A$ is diagonalizable. (See this problem .) So is $T_A|_{\mathcal V},$ the restriction of $T_A$ to the invariant subspace $\mathcal V.$ (Consider their minimal polynomials respectively.)

We have to show that $T_A|_{\mathcal V}=O$ for all $A$ in $\mathcal V.$ Since $T_A|_{\mathcal V}$ is diagonalizable, this amounts to showing that $T_A|_{\mathcal V}$ has no nonzero eigenvalues. Suppose, on the contrary, that $T_A|_{\mathcal V}(B)=aB\, (a\ne 0)$ for nonzero $B$ in $\mathcal V.$ Then $T_B|_{\mathcal V}(A)=-aB$ is itself an eigenvector of $T_B|_{\mathcal V},$ of eigenvalue $0.$ On the other hand, we can write $A$ as linear combination of eigenvectors of $T_B|_{\mathcal V}$ ( $B$ being diagonalizable also); after applying $T_B|_{\mathcal V}$ to $A,$ all that is left is a combination of eigenvectors which belong to nonzero eigenvalues, if any. This contradicts the preceding conclusion.