$$V \otimes \wedge^2 V^* = V^*$$ (coming from the bilinear pairing on $V$ induced by the wedge product, namely $v \otimes w \mapsto v \wedge w$), we get an identification $$V^*\otimes V = (\operatorname{Sym}^2V \otimes \wedge^2V^*) \oplus k.$$ On the other hand, $V^* \otimes V = \operatorname{Hom}(V, \, V)$, so we get a further identification $$\operatorname{Hom}(V, \,V) = (\operatorname{Sym}^2V \otimes \wedge^2 V^*) \oplus k.$$
Proposition. Under the above identification, an endomorphism $f \colon V \to V$ satisfies $\operatorname{trace}(f)=0$ if and only if it lies in the first summand $\operatorname{Sym}^2V \otimes \wedge^2 V^*.$ In other words, $\operatorname{Sym}^2V \otimes \wedge^2 V^*$ is naturally identified with the vector space $\operatorname{Hom}_0(V, \, V)$ of trace zero endomorphisms of $V$.
