Show that the trace of the operator $S \wedge T$ is zero

Question

I have some difficulties with the following problem:

Let $V$ be a finite dimensional vector space over $\mathbb{K}$.

Let $S,T \in L(V,V)$.

Show that the trace of the operator $S \wedge T$ is zero.

One way to prove this is showing that the operator $S \wedge T$ has skew-symmetirc matrix, but it seems complicated to proceed this way.

I know that:

\begin{split} (S \wedge T)(x \otimes y) & = S(x) \wedge T(y) \\ & = \frac{1}{2!} \sum_{\sigma \in S_2} sgn(\sigma) P(\sigma)(S(x) \otimes T(y)) \\ & = \frac{1}{2} (S(x) \otimes T(y) - T(y) \otimes S(x)) \end{split}

Thanks a lot!

Answer 1

By definition, we have that $$S \wedge T = \frac{1}{2}(S \otimes T - T \otimes S).$$

Then, as the trace is a linear form, $$tr(S \wedge T) = \frac{1}{2}(tr(S \otimes T) - tr(T \otimes S)).$$

As $$tr(S \otimes T) = tr(S) \cdot tr(T)$$ we have $$tr(S \wedge T) = \frac{1}{2}(tr(S) \cdot tr(T) - tr(T) \cdot tr(S)) = \frac{1}{2} \cdot 0 = 0.$$