I'm trying to prove that for any $B\in M_n(\mathbb{C})$ the linear operator $T_B:M_n(\mathbb{C})\to M_n(\mathbb{C})$ defined by $T_B(A)=AB-BA$ has null determinant.
As a first approach i proved that given the standard basis $\mathcal{B}=\{e_{11},\cdots,e_{nn}\}$ (where $e_{ij}$ is the matrix with zeroes everywhere except in the position $ij$), the matrix $T_B(e_{ij})$ has at least $(n-1)^2$ zeroes. Hence the matrix $[T_B]_\mathcal{B}$ has at least $n^2(n-1)^2$ zeroes. Intuitively, a matrix with so many zeroes surely has null determinant, but that's no more than a simple idea; which approach should I take?
Since $T_B(B)=0$ and $B\neq0$, the map is not injective, so its determinant is zero.
(If we had $B=0$ then the map itself is zero)