The trace of a linear operator $f$ can be defined as the trace of the matrix $A$ representing $f$ with respect to some basis $B$. However the trace does not depend on the basis chosen. This suggests to me that there is some definition of the trace of $f$ independent of matrices (and thus coordinate-independent). Any suggestions as to how I could define $\mathrm {tr}(f) $ without defining it as $\mathrm {tr}([f]_B)$?
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$\begingroup$Another common definition is the sum of eigenvalues (in the algebraic closure), counted with multiplicity. Since the eigenvalues are algebraic (they satisfy $A$'s minimal polynomial) we can invoke Galois theory to show the sum, being invariant, is in the original scalar field. By tensoring the space against the algebraic closure of the scalar field ("extension of scalars") and then writing it as a sum of generalized eigenspaces, corresponding to Jordan blocks of $A$, we can show that the sum of eigenvalues is equal to the sum of diagonal entries of $A$ in some basis (granted, after we extend the scalars), which we can then show is invariant under basis-change.
Another way is if we write ${\rm tr}:{\rm End}(V)\cong V\otimes_F V^*\to F$, where $v\otimes f\mapsto f(v)$ in the obvious way, and the expression $V^*$ denotes the dual vector space (i.e. $\hom_F(V,F)$). By choosing an ordered basis we can show this is the same as summing the diagonal entries in that basis. This fits into the perspective of string diagrams, a nice visual language for describing tensor facts, including currying and ${\rm tr}(AB)={\rm tr}(BA)$. Proving identities means wiggling strings around, yay!.
$\endgroup$ 6 $\begingroup$Another easy way to define the trace is through calculus: the trace of a linear operator is is the divergence of that linear operator with respect to its linear argument.
Indeed, this calculus-based approach also yields another characteristic object: the (generalized) curl of a linear operator is a bivector, corresponding to the antisymmetric part of that linear operator.
$\endgroup$ 1 $\begingroup$Several nice answers here. There is another way, also, that uses the following lemma. Let $k$ be a field, and let $k^{n\times n}$ denote the space of $n \times n$ matrices with coefficients in $k$.
Lemma The kernel of the trace operator, regarded as a linear map $k^{n\times n} \to k$, is the space of commutators, $\mathrm{Com}(k,n) = \{AB - BA : A, B \in k^{n\times n}\}$.
ProofThere are several proofs for this; a nice one by Kahan is here.
Definition The trace operator $\mathrm{tr}:k^{n\times n} \to k$ is the unique linear map such that $\mathrm{Ker}(\mathrm{tr}) = \mathrm{Com}(k,n)$ and $\mathrm{tr}(I) = n$, where $I$ denotes the $n \times n$ identity matrix.
There are also some nice definitions here 1. These include viewing trace as the unique operator such that $\mathrm{tr}(|u \rangle \langle v |) = \langle u | v \rangle$, and as the image of the identity map $I$ under a certain canonical isomorphism $\mathrm{End}(V) \to (\mathrm{End}(V))^*$.
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