In contrast, here is a definition of the trace on a finite-dimensional vector space $V$. Consider the isomorphism $ \mathrm{End}(V) \cong \mathrm{End}(V)^{\*} $$ \mathrm{End}(V) \cong \mathrm{End}(V)^{*} $ given by viewing $\mathrm{End}(V)$ as $V \otimes V^{\*} $$V \otimes V^{*} $, swapping the factors, and using the canonical isomorphism $V \cong V^{\*\*}$$V \cong V^{**}$. The trace is the image of the identity under this map. It might be said that the tensor algebra is used in an auxiliary manner, but there is (I think) a difference: only the most general, "hands-off" properties of tensor products are used. The heaviest lifting comes from the isomorphism $V^{\*\*} \cong V$$V^{**} \cong V$ (in order to show that it is actually an isomorphism, one must keep track of dimensions of a finite-dimensional vector space under dualization). So the only hard work comes on the original object of interest: the vector space $V$ itself. In contrast, the definition of the determinant required some detailed understanding of the alternating algebra, which was not originally in question.
