The linear operator $T_A(B)=AB-BA$ defined on $\mathbb M_n(\mathbb C)$ has a null determinant

Question

Let $T_A:\mathbb{M_n(C)}\rightarrow\mathbb{M_n(C)}$ be a linear operator such that $T(B)=AB-BA$, were $\mathbb{M_n(C)}$ denote the space of matrix with order $n$ and complex inputs. Then $T_A$ has null determinant.

Proof: Is sufficient to show that $T_A$ has not inverse, i.e., $T_A$ is not one-to-one. But this is inmediate because $T_A(I)=0$ so, $\ker(T_A)\neq\{0\}$. I have doubts about my proof. It is right?

Answer 1

Your proof is fine. "Sufficient" is the correect spelling, however.