I am reading the Wikipedia article on the trace operator. The section titled Coordinate-Free Definition defines the trace as follows.

Let $V$ be a finite dimensional vector space over a field $F$ and define a bilinear map $f:V\times V^*\to F$ as $f(v, \omega)=\omega(v)$ for all $(v, \omega)\in V\times V^*$. This maps induces a unique linear map $\tr:V\otimes V^*\to F$.

Since $\text{End}(V)$ has a canonical isomorphism with $V\otimes V^*$, we have now a notion of trace of a linear operator on $V$.

The Question: The second paragraph of the section in the article says that

This also clarifies why $\tr(AB)=\tr(BA)$.

I can't see how $\tr(AB)=\tr(BA)$ follows from this definition at all.

Can somebody give me a hint?

  • $\begingroup$ There is actually a sketched proof on the wiki page. Have you read it? Do you need some clarification of any specific detail there? $\endgroup$ – Amitai Yuval Jul 22 '15 at 10:40
  • $\begingroup$ @AmitaiYuval Somehow I was unable to notice that a proof was outlined. I kind of filtered that out as something unnecessary for my purpose. $\endgroup$ – caffeinemachine Jul 22 '15 at 16:20

To understand the trace, it is good to spell out the isomorphism between $End(V)$ and $V\otimes V^*$: $$V\otimes V^*\to End(V): v\otimes\omega \mapsto (x\mapsto \omega(x)v).$$

Under this isomorphism, composition of endomorphisms becomes $$(V\otimes V^*) \times (V\otimes V^*) \to V\otimes V^*: (v_2\otimes\omega_2, v_1\otimes \omega_1)\mapsto \omega_2(v_1)\cdot v_2\otimes \omega_1.$$

Taking the trace of this composition, one gets $\omega_2(v_1)\omega_1(v_2)$. It is then easy to see that the trace of $v_2\otimes\omega_2\circ v_1\otimes \omega_1$ is the same as that of $v_1\otimes\omega_1\circ v_2\otimes \omega_2$.


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