I've pondered this question quite a bit, because I love the geometric definition of the determinant.^ My current feeling is that, although the trace has a beautiful geometric meaning (the one given by Allen Knutson), its *raison d'être* is fundamentally algebraic:

Let $V$ be a finite-dimensional vector space over the field $F$, and let $L(V)$ be the set of linear maps from $V$ to itself. The trace is the unique (up to normalization) linear map from $L(V)$ to $F$ such that $\text{tr}(AB) = \text{tr}(BA)$ for all $A, B \in L(V)$.

This is my favorite definition to date, but I suspect that the trace has a deeper meaning: *it's what you get when a linear map eats itself*. I can't explain exactly what I mean by that, but here's some evidence in favor of it:

Because $V$ is finite-dimensional, you can think of a linear map from $V$ to itself as an element of $V^* \otimes V$. If $A = \omega_1 \otimes v_1 + \ldots + \omega_k \otimes v_k$, then $\text{tr}(A) = \omega_1(v_1) + \ldots + \omega_k(v_k)$.

In the abstract index notation used in general relativity (See Robert Wald's book for a great introduction), a vector $v$ would be written $v^a$, a linear map $A$ would be written ${A^a}_b$, and the vector $Av$ would be written ${A^a}_b v^b$. The indices show you that $v$ is being plugged into the input slot of $A$, and another vector is coming out the output slot. The trace of $A$ would be written ${A^a}_a$, which seems to represent the output of $A$ being plugged back into the input!

If someone could explain to me how the geometric, algebraic, and "self-eating" (autophagic?) meanings of the trace were related to each other, I would be very happy!

^ In fact, I love it so much that I'll repeat my favorite statement of it here! Let $V$ be a $n$-dimensional vector space over the field $F$. A *signed-volume form* on $V$ is a map from $V^n$ to $F$ with the following properties:

- It gets multiplied by $\lambda$ if you multiplying one of its arguments by $\lambda$.
- It doesn't change if you add one of its arguments to another of its arguments.

The determinant of a linear map $A \colon V \to V$ is the scalar $\det(A)$ such that $D(A v_1, \ldots, A v_n) = \det(A) D(v_1, \ldots, v_n)$ for any vectors $v_1, \ldots, v_n$ and any signed-volume form $D$.

A single number can satisfy this equation for all signed-volume forms because the signed-volume form on $V$ is unique up to normalization.