# What is the intuition behind the trace of an endomorphism?

Looking back on my math education, I noticed that even though the trace of an endomorphism came up a lot, I'd be hard pressed to give a good description of what the trace really means. I'm looking for some good intuition about its meaning.

To elaborate on what I'm looking for: if I forgot the rigorous definition of the determinant, I could rebuild it from scratch without reference to a basis because I know that it's supposed to measure the change in volume and orientation of a parallelepiped under a linear transformation. For the same reason, I can quickly tell that it is independent of basis, multiplicative, and determines wether the endomorphism is injective or not, all without doing any calculations. I want something similar for the trace. It doesn't need to be geometric, but I want to know what the trace tells us about how the endomorphism acts.

• The trace of an endomorphism is the sum of all the eigenvalues (perhaps in some extension field) of the endomorphism, and it is an invariant of it, meaning: it doesn't depend on some choice of basis to represent the endomorphism as a matrix, for example. Also, the trace of an endomorphism in itself is a linear functional of the space of all endomorphisms to the definition field. It also has an important role distinguishing separable and non-separable extensions of fields (Fields, Galois Theory and etc.) ... Aug 11 '20 at 19:20
• @DonAntonio: Thanks, but I wasn't really looking for features and uses of the trace, but for the idea behind it. Like, if I built up linear algebra from scratch, how would I even get the idea to define the trace in the first place? Especially since it is usually defined in a coordinate dependant way, which doesn't seem natural. Aug 11 '20 at 21:03
• As many other things in science and mathematics, perhaps they "used" the idea of trace before even getting to know it and define it. I can't say, but taking a peek at the diagonals of a square matrix seems something more or less natural, and from here...And the main thing is that at some point they proved it is not coordinate dependent. Aug 11 '20 at 21:30
• There is a coordinate free definition given on wikipedia that you might find interesting. This essentially coincides with Hyperplane's explanation, but does not require an inner product and applies to arbitrary fields. Aug 12 '20 at 13:10
• mathoverflow.net/questions/13526/… might help. Aug 13 '20 at 10:04

For finite-dimensional vector spaces $$V$$, there is a canonical isomorphism of $$V$$ with its double dual $$V^{**}$$ and this makes the vector space $$V \otimes V^*$$ naturally isomorphic to its own dual space: $$(V \otimes V^{*})^* \cong V^* \otimes V^{**} \cong V^* \otimes V \cong V \otimes V^{*},$$ where the first and last isomorphisms are the natural ones involving tensor products of (finite-dimensional) vector spaces. Since $$V \otimes V^{*} \cong {\rm End}(V)$$, we get that $${\rm End}(V)$$ is naturally isomorphic as a vector space to its own dual space. If you unwrap all of these isomorphisms, the isomorphism $${\rm End}(V) \to ({\rm End}(V))^{*}$$ sends each linear operator $$A$$ on $$V$$ to the following linear functional on operators on $$V$$: $$B \mapsto {\rm Tr}(AB)$$. In particular, the identity map on $$V$$ is sent to the trace map on $${\rm End}(V)$$.

• I feel like this is a very similar approach to the one used by @Hyperplane, where they used an inner product to characterize the elements of the various tensor products in your answer. For instance, $\vert v\rangle\langle w\vert$ is an element of $V\otimes V^\ast$, if we identify it as $\operatorname{End} V$. So in a way, this should be the generalization to arbitrary fields I was asking for. Thanks! Aug 12 '20 at 13:38

One way to think about it is that the trace is the unique linear map with the property

$$\operatorname{tr}(|u\rangle\langle v|) = \langle u | v\rangle$$

In particular in the case of a rank 1 matrix $$A=\lambda|u\rangle\langle v|$$ we see that the trace in some sense measures "how much $$\ker(A)^\perp$$ is aligned with $$\operatorname{Im}(A)$$". Does this viewpoint carry over to matrices that are not rank-1?

Consider a higher rank matrix $$A=\sum_{i=1}^k \sigma_i|u_i\rangle\langle v_i|$$. Again we may assume $$u_i$$ and $$v_i$$ are normalized (you may notice that when $$k=n$$ and the $$u_i$$ and $$v_j$$ are orthogonal this is the SVD of $$A$$) then

$$\operatorname{tr}(A)= \operatorname{tr}(\sum_i \sigma_i|u_i\rangle\langle v_i|) = \sum \sigma_i\langle v_i|u_i\rangle$$

Which is a weighted sum of how much the orthogonal complements of the kernels of the rank-1 components align with their images. Due to linearity, it does not matter how we represent $$A$$ as a sum of rank-1 matrices.

Notable special cases:

• if $$A$$ has an orthogonal eigenbasis, then $$A=\sum_i \lambda_i |v_i\rangle\langle v_i|$$ and so $$\operatorname{tr}(A)=\sum_i\lambda_i\langle v_i|v_i\rangle = \sum_i \lambda_i$$. Here the orthogonal complements and the images of the rank 1 components are perfectly aligned.

• For a projection matrix $$P$$ (i.e. $$P^2=P$$) we have $$\operatorname{tr}(P)=\dim \operatorname{Im}(P)$$. Since $$P$$ acts like an identity on the subspace is projects onto, again the orthogonal complements of the kernel and the image are perfectly aligned.

• For a nilpotent matrix $$N$$, the trace is zero. In fact, we can write $$N$$ as a sum of rank-1 matrices where the orthogonal complement of the kernel is orthogonal to the image for each of them. (can be proven for example via Schur decomposition)

• Thanks, this looks good! Can this line of thinking be extended to vector spaces over arbitrary fields? After all, inner products are usually only defined for real or complex vector spaces. Aug 11 '20 at 23:06
• You ask if this approach works over other fields, but in your post the description of the determinant without calculations is not meaningful (volume, orientation) over arbitrary fields. Does that concern you?
– KCd
Aug 11 '20 at 23:43
• @KCd: just a little, but not really. I can define alternating multilinear forms on arbitrary fields, resulting in an abstracted version of a signed "volume" which retains all the algebraic features at least. I was wondering wether such a thing is possible with the algebraic properties discussed above. For instance, if we replaced the inner product by a non-degenerate, symmetric bilinear form or something like that. Aug 12 '20 at 8:52
• An inner product is not intrinsic to an abstract vector space, so imposing one to “characterize” the trace would only tell you about a characterization of the trace of a linear operator on vector spaces equipped with an inner product (or nondeg. symm. bilinear form). In Rotman’s Advanced Modern Algebra there is a characterization of the trace of a linear operator on a finite-dim. vector space $V$ using a canonical graded derivation on the exterior algebra of $V$ built from the operator; its effect on the top exterior power of $V$ is scaling by the trace. See Prop. 9.166 (page 771).
– KCd
Aug 12 '20 at 13:01

Here's a cute geometric interpretation: the trace is the derivative of the identity at the origin. That is, we have

$$\det (1 + At) = 1 + \text{tr}(A) t + O(|t|^2)$$

So if you think of the determinant geometrically in terms of volumes, the trace is telling you something about how a matrix very close to the identity changes volumes. Similarly we have the identity

$$\det \exp(At) = \exp(\text{tr}(A) t).$$

This identity explains, among other things, why the Lie algebra of the special linear group $$SL_n$$ is the Lie algebra $$\mathfrak{sl}_n$$ of matrices with zero trace.

The argument in KCd's answer can be substantially generalized and you can get some pretty pictures out of it. There is a way of defining the trace using what are called string diagrams which, among other things, makes it immediately clear why the trace satisfies the cyclicity property $$\text{tr}(AB) = \text{tr}(BA)$$ (note that this is at least apparently slightly stronger than being conjugation-invariant): see this blog post and this blog post. As a teaser, once the appropriate notation has been introduced and the appropriate lemmas proven, here is a complete proof of cyclicity:

There are many great answers here. Here is one more, very elementary one. 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; K. Shoda (1936) Japan J. Math. 13 361-5 gave an argument that works for fields of finite characteristic. A.A. Albert and B. Muckenhoupt (1957) Michigan Math. J. 4 1-3 establishes the claim for any characteristic. Kahan gives a nice argument, though I'm not sure this was vetted in peer review.

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.

Let $$f:U\subset \mathbb R^n\to\mathbb R^n$$ be a $$C^1$$ vector field, then the trace of differential of $$f$$ (or trace of jacobian matrix), is equal to, the divergence of $$f$$. Explicitly, $$\operatorname{trace}\big(df|_p\big)=\operatorname{trace}\left(\array{\frac{\partial f^1}{\partial x^1} & \cdots & \frac{\partial f^1}{\partial x^n}\\ \vdots & \ddots & \vdots \\ \frac{\partial f^n}{\partial x^1}& \cdots & \frac{\partial f^n}{\partial x^n}}\right)|_p=\sum_{i=1}^n \frac{\partial f^i}{\partial x^i}|_p=\operatorname{div} f|_p$$ And divergence is coordinate free in the sense of Euclid inner space $$\mathbb R^n$$ or $$\operatorname{SO}(n)$$, because $$\operatorname{div} f|_p=\lim_{r\to 0}\frac{1}{\operatorname{Vol}(U_r)}\int_{\partial U_r} \langle f,\mathbf{n} \rangle(x)\, dx$$ where (1) $$U_r$$ can be a $$r$$-radius ball or rectangle that center at $$p$$, (2) $$\mathbf{n}$$ is outer unit normal vector of surface $$\partial U_r$$ at $$x$$, (3) $$\langle f,\mathbf{n} \rangle(x)$$ is the inner product at $$x\in \partial U_r$$, (4) surface integral.