Let $V$ be the space of all $n \times n$ matrices over a field $F$ and let $B$ be a fixed $n \times n$ matrix. If $T$ is the linear operator on $V$ defined by $T(A) = AB - BA$, and if $f$ is the trace function, what is $T^t(f)$?
My attempt: $T^t:V^*\to V^*$ defined by $T^t(g)=g\circ T$, $\forall g\in V^*$. Let $A\in V$. Then $[T^t(f)](A)$ $=f\circ T(A)$ $=f(T(A))$ $=f(AB-BA)$. It’s easy to check $f=\text{tr}$ is a linear map, i.e. $f(P+Q)=f(P)+f(Q)$ and $f(c\cdot P)=c\cdot f(P)$. So $f(AB-BA)$ $=f(AB)+f(-BA)$ $=f(AB)-f(BA)$. By exercise 3 section 3.5, $f(AB)=f(BA)$. Thus $[T^t(f)](A)$ $=f(AB)-f(BA)$ $=0_F$, $\forall A\in V$. Hence $T^t(f)=0_{V^*}$. Is my proof correct?