# Coordinate-Free Definition of Trace.

$$\DeclareMathOperator{\tr}{trace}$$

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?

• There is actually a sketched proof on the wiki page. Have you read it? Do you need some clarification of any specific detail there? – Amitai Yuval Jul 22 '15 at 10:40
• @AmitaiYuval Somehow I was unable to notice that a proof was outlined. I kind of filtered that out as something unnecessary for my purpose. – 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 $v_1\otimes\omega_1\circ v_2\otimes \omega_2$.