Two definitions for the trace of a linear operator

Question

The trace of a linear map $f:V\rightarrow V$, $\dim V=n$ finite, can be defined in two ways.

One uses the induced linear map on $\bigwedge^nV^*$ given by \begin{align*} T\mapsto\sum_{i=1}^nT(v_1,\dots,f(v_i),\dots,v_n) \end{align*} (or on $\bigwedge^nV$ with the analogous map) and so from $\dim\bigwedge^nV^*=\binom{n}{n}=1$ we can write $\sum\limits_{i=1}^nT(v_1,\dots,f(v_i),\dots,v_n)=(\text{tr} f)T(v_1,\dots,v_n)$.

Another definition can be given by identifying $\text{End}(V)$ with $V^*\otimes V$, and requiring $\text{tr}:V^*\otimes V\rightarrow\mathbb{F}$ be linear and $\text{tr}(\phi\otimes v)=\phi(v)$.

I am trying to reconcile these two definitions. One way to do this is to choose a matrix representation and show these produce the same formula with the entries of the matrix. I am wondering if there is another way (possible coordinate-free, possibly not) which is more intuitive, since the fact that the two definitions (especially the latter) just happen to yield the sum of the diagonals is really unclear a priori.

Why should I expect these definitions to coincide? How do I show they do?

I'm not really sure where to get started on this. Honestly, I'm still pretty new to "coordinate-free" linear algebra so I'm finding the isomorphism between $\text{End}(V)$ and $V^*\otimes V$ unwieldy. I think part of my difficulty as well is the first definition I gave I'm having trouble really understanding - algebraically it's quite simple, but what $f$ is "really doing" to $\bigwedge^nV^*$ is not obvious. Clarification on any of these points is appreciated.

Answer 1

As a first comment, you are giving two distinct roles to $f$, which doesn't help (it is an endomorphism in your first definition, and a linear functional in the second one).

Second comment: I might be wrong, but I don't think there is an obvious connection between the two definitions. There are many ways of characterizing the trace, and the connection between them isn't obvious either. For example the trace of $f\in \operatorname{End}(V)$ is

• the sum of the diagonal entries of $f$ is some basis
• the sum of the diagonal entries of $f$ in any other basis
• the sum of the eigenvalues of $f$ (in the algebraic completion of $\mathbb F$)
• the only linear map $\varphi:\operatorname{End}(V)\to\mathbb F$ such that $\varphi(I)=n$ and $\varphi(f\circ g)=\varphi(g\circ f)$ for all $f,g\in\operatorname{End}(V)$.

It is not hard to prove that all these definitions define the same linear functional, but I wouldn't say it's obvious. And some of the implications require at least some theory, like for example to go to/from the eigenvalue definition you need something like the Jordan Form, Schur Diagonalization, or the Spectral Theorem.

In light of the above, I wouldn't expect an obvious abstract connection between your two definitions. Not saying that there isn't one, though.

The most direct way, for me, to see the connection between the two definitions is the following. It requires specifying a basis, as otherwise there is no way to express what the elements of $\bigwedge^nV^*$ are .

So fix a basis $e_1,\ldots,e_n$ of $V$, and consider the dual basis $e_1^*,\ldots,e_n^*$ of $V^*$. Then $\bigwedge^nV^*=\mathbb F\,e_1^*\land\cdots\land e_n^*$. It is easy to see that $\operatorname{End}(V)$ is the span of the rank-one operatos $\{f_{kj}\}$, where $f_{kj}(v)=e_j^*(v)\,e_k$. Now \begin{align} \sum_{s=1}^n(e_1^*(v_1)\land\cdots\land e_s^*(f_{kj}(v_s))\land\cdots\land e_n^*(v_n)) &=\sum_{s=1}^n\,e^*_s(e_k)\,(e_1^*(v_1)\land\cdots\land e_j^*(v)\land\cdots\land e_n^*(v_n))\\[0.3cm] &=\delta_{kj}\,(e_1^*\land\cdots\land e_n^*)(v_1,\ldots,v_n). \end{align} So $\operatorname{tr}(f_{kj})=\delta_{kj}$. By linearity, for $f=\sum_{k,j}\beta_{kj}\,f_{kj}$, we get $\operatorname{tr}(f)=\sum_{k}f_{kk}$.

With the tensor approach, the $f_{kj}$ from above are $e_j^*\otimes e_k$, so $$ \operatorname{tr}(e_j^*\otimes e_k)=e_j^*(e_k)=\delta_{kj}. $$