Let \(V\) be the space of \(n \times n\) matrices over \(F\). Let \(A\) be a fixed \(n \times n\) matrix over \(F\). Let \(T\) and \(U\) be the linear operators on \(V\) defined by \[ T(B) = AB, \qquad U(B) = AB - BA. \]
True or false? If \(A\) is diagonalizable (over \(F\)), then \(T\) is diagonalizable.
True or false? If \(A\) is diagonalizable, then \(U\) is diagonalizable.
Solution
Both assertions are \(\boxed{\text{true}}\).
Let \(P\) be invertible with \[ D=P^{-1}AP=\operatorname{diag}(\lambda_1,\ldots,\lambda_n), \] and define the invertible linear map \[ \Phi:V\to V, \qquad \Phi(B)=P^{-1}BP. \]
1. Left multiplication by \(A\) is similar to left multiplication by \(D\).
For \(C\in V\), \[ \begin{aligned} (\Phi T\Phi^{-1})(C) &=\Phi\bigl(A(PCP^{-1})\bigr)\\ &=P^{-1}APC\\ &=DC. \end{aligned} \] Thus \(T\) is similar, as an operator on the \(n^2\)-dimensional space \(V\), to the operator \(L_D(C)=DC\).
For the standard matrix units \(E_{ij}\), \[ L_D(E_{ij})=DE_{ij}=\lambda_iE_{ij}. \] The \(n^2\) matrices \(E_{ij}\) form a basis of \(V\), so they form an eigenbasis for \(L_D\). Hence \(L_D\) is diagonalizable, and similarity preserves diagonalizability. Therefore \(T\) is diagonalizable [@HK71].
2. The commutator operator is similar to commutation with \(D\).
For \(C\in V\), \[ \begin{aligned} (\Phi U\Phi^{-1})(C) &=\Phi\bigl(A(PCP^{-1})-(PCP^{-1})A\bigr)\\ &=DC-CD. \end{aligned} \] Thus \(U\) is similar to \(\operatorname{ad}_D(C)=DC-CD\). Again using the matrix units, \[ DE_{ij}=\lambda_iE_{ij}, \qquad E_{ij}D=\lambda_jE_{ij}, \] and therefore \[ \operatorname{ad}_D(E_{ij})=(\lambda_i-\lambda_j)E_{ij}. \] Hence the same basis \(\{E_{ij}\}\) is an eigenbasis for \(\operatorname{ad}_D\). So \(\operatorname{ad}_D\), and consequently \(U\), is diagonalizable.