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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 $\mathrm {tr}([f]_B)$?

3 Answers3

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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!.

anon
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  • I don't understand the "another way". We have written $\mathrm{End}(V)$ as $V\otimes V^\star$, ok. What now? – Giuseppe Negro Jun 28 '14 at 19:43
  • @GiuseppeNegro ${\rm tr}:V\otimes_FV^*\to F$ is given by $v\otimes f\mapsto f(v)$ (just as it is written above). – anon Jun 28 '14 at 19:45
  • well what he is saying (in non canonical manner) is that the trace is induced from the canonical evaluation pairing $V\times V^\ast\rightarrow k$ ($k$ being canonical notation for a field), nothing more needs to be said – Zlatan der Zechpreller Jun 28 '14 at 19:46
  • I had never heard of string diagrams. They look like wigglier (or at least, rounder) versions of Feynman diagrams. –  Jun 28 '14 at 19:51
  • @ZlatanderZechpreller Perhaps you could explain how the canonical evaluation pairing induces the trace without reference to picking a basis. It's not clear to me immediately how you get from End(V) via the evaluation to an element of k. – Francis Davey Dec 22 '17 at 16:33
  • I like the minimal polynomial definition. I am afraid your second definition appears to me to be a rather more sophisticated way of saying "sum the diagonal" - in other words it isn't co-ordinate free, precisely because you have to pick a basis to do it. In other words, it doesn't really provide what is wanted. – Francis Davey Dec 22 '17 at 16:39
  • @FrancisDavey You do not need to pick a basis to do it. You need to pick a basis to show e.g. it matches the first definition. – anon Apr 18 '23 at 19:53
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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.

Muphrid
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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, $\mathrm{Com}(k,n) = \{AB - BA : A, B \in k^{n\times n}\}$.
Proof There 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 $\mathrm{End}(V) \to (\mathrm{End}(V))^*$.

GHPR
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  • This is only true if $k$ has characteristic $0$. – Charles Rezk Oct 21 '20 at 03:07
  • I thought that the proof of Shoda applied only to characteristic $0$, but Albert and Muckenhoupt (https://projecteuclid.org/download/pdf_1/euclid.mmj/1028990168) works in any characteristic? – GHPR Oct 23 '20 at 04:22