Let \(V\) be the vector space of \(n \times n\) matrices over the field \(F\). Let \(B\) be a fixed element of \(V\) and let \(T_B\) be the linear operator on \(V\) defined by \(T_B(A) = AB - BA\). Show that \(\det T_B = 0\).
Solution
The identity matrix \(I\) is a nonzero vector of the vector space \(V\), and \[ T_B(I)=IB-BI=B-B=0. \] Thus \(T_B\) has nonzero kernel, so it is not invertible. By the determinant criterion for invertibility of a linear operator, \[ \det T_B=0. \]