Kunze Linear Algebra · Problem 3.7.3

The adjoint operator $T^t$ reverses the action of $T$ in the sense of the duality pairing between $V$ and its dual space $V^*$. Since $f$ is the trace function (a linear functional on $V$), you need to find which linear functional $T^t f$ is by using the definition: $(T^t f)(A) = f(T(A))$ for all $A \in V$. The key insight is that the trace has a special property—$\text{tr}(XY) = \text{tr}(YX)$—which makes commutators vanish under the trace, so $f(T(A)) = \text{tr}(AB - BA) = 0$ for every matrix $A$.

The adjoint operator $T^t$ reverses the action of $T$ in the sense of the duality pairing between $V$ and its dual space $V^*$. Since $f$ is the trace function (a linear functional on $V$), you need to find which linear functional $T^t f$ is by using the definition: $(T^t f)(A) = f(T(A))$ for all $A \in V$. The key insight is that the trace has a special property—$\text{tr}(XY) = \text{tr}(YX)$—which makes commutators vanish under the trace, so $f(T(A)) = \text{tr}(AB - BA) = 0$ for every matrix $A$.