# Geometric interpretation of trace

This afternoon I was speaking with some graduate students in the department and we came to the following quandary;

Is there a geometric interpretation of the trace of a matrix?

This question should make fair sense because trace is coordinate independent.

A few other comments. We were hoping for something like:

"determinant is the volume of the parallelepiped spanned by column vectors."

This is nice because it captures the geometry simply, and it holds for any old set of vectors over $$\mathbb{R}^n$$.

The divergence application of trace is somewhat interesting, but again, not really what we are looking for.

Also, after looking at the wiki entry, I don't get it. This then requires a matrix function, and I still don't really see the relationship.

One last thing that we came up with; the trace of a matrix is the same as the sum of the eigenvalues. Since eigenvalues can be seen as the eccentricity of ellipse, trace may correspond geometrically to this. But we could not make sense of this.

• Related question: Take the $p$-dimensional vector space over $\mathbb{F}_p$ and take the identity transformation on this space. Then the trace is $0$. What the "geometric" meaning of this, if any? – Anweshi Jan 31 '10 at 2:12
• Nice comment Anweshi! That is a very interesting question also. This is the 3rd time this week that your comments have really impressed me! – B. Bischof Jan 31 '10 at 2:31
• Your geometric description defines the determinant of a matrix just in terms of the (signed) collection of vectors that make up the rows. One reason you'll never find a totally analogous description of the trace is that it really is not a function of a collection of $n$ vectors: any reordering, and your trace is different. – Theo Johnson-Freyd Jan 31 '10 at 8:18
• Theo's comment highlights the fact that the sense in which trace is "coordinate independent" is not always the same as the sense in which the determinant is -- so perhaps underlying the original question is a more basic question about what kind of invariance property, let alone geometric property, is desired. – Yemon Choi Jan 31 '10 at 8:33
• @Anweshi the geometric meaning is that in characteritc $p$ the baricentre of a affine multiset of $p$ points is at infinity. [Equivalent projective configurations exist: Fano for $p=2$, ...]. Using the geometric interpretation of trace of a symmetric matrix (defining a quadric) of order $p$ as $p$ times the expected value for eigenvalues (medium leght of principal axes) requires the characteristic non being $p$. – user46855 Feb 15 '14 at 15:58

If your matrix is geometrically projection (algebraically $A^2=A$) then the trace is the dimension of the space that is being projected onto. This is quite important in representation theory.

• It is also important in statistics! – kjetil b halvorsen Dec 24 '12 at 0:29
• This is important everywhere in math. – Matthieu Romagny Mar 29 '13 at 17:18
• This property, together with linearity, determines the trace uniquely, and so one can view the trace as the linearised version of the dimension-counting operator. (This is basically the "noncommutative probability" way of thinking about the trace.) – Terry Tao Aug 13 '14 at 16:27

Let's use $$\det(\exp(tA)) = 1 + t\operatorname{Tr}(A) + O(t^2)$$, and think about the vector ODE $$\vec y' = A \vec y$$, solved by $$\vec y(t) = \exp(tA) \vec y(0)$$. If we take a unit parallelepiped worth of $$\vec y(0)$$, flow for short time $$t$$ under $$\vec y' = A\vec y$$, and see how its volume changes, the change will thus be $$t\operatorname{Tr}(A)$$ to first order.

Ah, Yemon Choi beat me to part of that.

• I took me a while to understand what you meant by "a unit parallelepiped worth of $\vec{y}(0)$". To clarify for future readers, you are starting with the unit parallelepiped and flowing each of its $n$ orthonormal unit vector sides $\hat{e}_{(i)}$ independently under the vector ODE $\vec{y}' = A \vec{y}$, so that the parallelepiped's time evolution is given by the $n$ different solutions of that ODE for the initial conditions $\vec{y}(0) = \vec{e}_{(i)}$. – tparker Feb 5 '18 at 3:26

V. I. Arnold sums it up very well in Section 16.3, page 113 of "Ordinary Differential Equations" (Springer Edition).

"Suppose small changes are made in the edges of a parallelepiped. Then the main contribution to the change in volume of the parallelepiped is due to the change of each edge in its own direction, changes in the direction of the other edges making only a second-order contribution to the change in volume."

I'm surprised nobody has mentioned this yet, but the trace defines a Hermitian inner product on the space of linear operators from $$\mathbb{C}^n$$ to $$\mathbb{C}^m$$: $$\langle A, B\rangle = \operatorname{Tr} A^\dagger B.$$ And every multiplicative operator on $$M_{n}(\mathbb{C})$$ which preserves the involution $$\dagger$$, must preserve this inner product. You can't get much more geometric than that.

• D'oh! Yes, this is a good observation. This also crops up when one looks at (complex) representations of compact groups (cf. Schur orthogonality) – Yemon Choi Jan 31 '10 at 2:48
• As always with inner products, though, you need to check first whether you're a physicist or a mathematician so you know whether to use the formula Jon wrote or $\langle A,B\rangle = \mathrm{Tr} A B^*$. – Mark Meckes Feb 1 '10 at 14:40
• You need an inner product in order to define $\dagger$, so the trace only lets you lift this inner product from vectors to matrices. It doesn't define a totally new one. – Oscar Cunningham Sep 25 '14 at 20:03
• On the other hand, if $n = m$ then you can also define a non-degenerate symmetric bilinear form using the same formula but without the $\dagger$, and this does not need an inner product. (I'm not so clear on the significance of this.) – Robin Saunders Jun 18 '15 at 2:29

If you are just working in a finite-dimensional Euclidean space, then by using the fact that we can calculate the trace of $$A$$ as $$\sum_{j=1}^n \langle Ae_j, e_j\rangle$$ for $$any$$ choice of orthonormal basis $$e_1,\dots, e_n$$, one obtains

$$\operatorname{Tr}(A) = n\int_{x\in B} \langle Ax, x\rangle \,dm(x)$$

where $$B$$ is the Euclidean unit sphere, and $$m$$ is the uniform measure on $$B$$ normalised to have total mass $$1$$. This is perhaps not quite as geometric as you want, but perhaps seems less dependent on a choice of coordinates.

Also, the wikipedia page refers to the trace as being (related to) the derivative of the determinant -- does that not seem $$`$$geometric'?

• It should be emphasised that the trace really is a property of an operator between vector spaces, not a property of the matrix used to represent them. Again, this is not quite "geometric" -- it is really more "spectral" -- but it does I think make the trace seem more natural. – Yemon Choi Jan 31 '10 at 2:50
• This is the interpretation of trace you want to think about when proving the mean value property of a harmonic function, for example. i.e. this is saying a quadratic polynomial is harmonic if and only if it satisfies the mean value property. – Ryan Budney Jan 31 '10 at 8:12
• This interpretation is also what one uses to understand Ricci curvature and scalar curvature: very important geometrically indeed. – Benoît Kloeckner Jul 4 '13 at 6:57
• I apologize for being pedantic (in particular, 9 years after the original post), but I think the total mass of the measure $m$ should be $n$ rather than $1$ since the trace of the identity matrix is equal to $n$. – Jochen Glueck Mar 20 '19 at 18:06
• @JochenGlueck Thanks :) – Yemon Choi Mar 20 '19 at 19:59

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:

1. It gets multiplied by $\lambda$ if you multiplying one of its arguments by $\lambda$.
2. 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.

• To make tr and det even more similar, any Lie algebra map from $gl(n)$ to a commutative Lie algebra factors through trace (this is the cyclicity property you mention), whereas any multiplicative map from $gl(n)$ to a commutative monoid factors through determinant. – Theo Johnson-Freyd Jan 31 '10 at 8:16
• Slight quibble: when you say that we can regard any linear map from V to itself as an element of $V^*\otimes V$, this is assuming that V is finite-dimensional. The analogous statement is very much false in infinite dimensions – Yemon Choi Jan 31 '10 at 8:30
• I think there is a more important algebraic reason for trace to exist. Namely: the trace is the coefficient of $x^(n-1)$ in the characteristic polynomial of an $n \times n$ matrix. (Although, the properties you mention are also interesting & important.) – Ilya Grigoriev Feb 1 '10 at 3:36
• Keywords for the "self-eating" interpretation are the graphical calculus for tensor categories and string diagrams. Pointers are an article by Kate Ponto and Mike Shulman (see also accompanying slides) and a blog post by sigfpe. – Ingo Blechschmidt Feb 15 '15 at 21:26
• Sorry, I actually meant this article. "The purpose of this expository note is to describe duality and trace in a symmetric monoidal category, along with important properties (including naturality and functoriality), and to give as many examples as possible." – Ingo Blechschmidt Feb 16 '15 at 20:19

Is it bad form to answer a question twice? In my defense, I'm different now from who I was when I answered the first time...

In their answers, Allen Knutson and Jafar give a geometric characterization of the trace:

The trace is the derivative of the determinant map $\operatorname{GL}(V) \to \mathbb{R}^\times$ at the identity.

In a comment, Theo Johnson-Freyd gives an algebraic characterization:

The trace is the unique Lie algebra homomorphism $\mathfrak{gl}(V) \to \mathbb{R}$, up to scale.

These characterizations are equivalent in a very pretty way.

The determinant of a transformation in $\operatorname{GL}(V)$ is the factor by which it expands volumes. When you compose two transformations, their volume expansion factors compose as well, so the determinant is a Lie group homomorphism $\operatorname{GL}(V) \to \mathbb{R}^\times$. Therefore, its derivative at the identity is a Lie algebra homomorphism $\mathfrak{gl}(V) \to \mathbb{R}$, so it must be the trace, up to scale.

To pin down the scale, think of $\operatorname{id}_V$ as an element of $\mathfrak{gl}(V)$, and observe that $\exp(t \operatorname{id}_V) \in \operatorname{GL}(V)$ is scaling by $\exp(t)$. Therefore, $\exp(t \operatorname{id}_V)$ expands volumes by a factor of $\exp(tn)$, where $n$ is the dimension of $V$. In other words, the determinant map $\operatorname{GL} \to \mathbb{R}^\times$ sends $\exp(t \operatorname{id}_V)$ to $\exp(tn)$. Its derivative therefore sends $\operatorname{id}_V$ to $n$. The trace does the same thing, so it matches the derivative of the determinant not only up to scale, but on the nose.

• I've always really liked Theo Johnson-Freyd's interpretation the best, but the unification of the two in your answer is fantastic: it's a really interesting way to understand $e^{\mathrm{tr}(X)} = \det(e^X)$, a formula whose proof I have always up until now found pretty boring. – Selene Routley Apr 3 '15 at 8:56

You can think of the trace as the expected value (times the dimension of the vector space) of the eigenvalues of matrices. The notion of eigenvalue is, as you know, a geometric one because it is the ratio of distortion of length. On the other hand 'expected value' is borrowed from probability theory, but given how the trace is extensively used in the modern branches of that field, you could spare that ;-) This point of view makes it obvious that the trace is invariant under conjugation by any invertible matrix.

• This comment finds a wide extension in the notion of numerical measure of a matrix, which is supported by the numerical range. See Th. Gallay & D. S. Comm. Pure Appl. Math. 65 (2012), pp 287-336. – Denis Serre Mar 29 '13 at 15:04
• However, this answer is somehow duplicate of that by Yemon Choi. – Denis Serre Mar 29 '13 at 15:07
• Nice, but not obvious to me that expected value of eigenvalues is invariant under conjugation: if you assign different weights to values of random variable and shuffle them, then reassign what you are looking at, then kinda sorta undo that, why should the expectation remain the same? Yes, I know, trace is invariant under conjugation. No, I don't see independently of that why the average of eigenvalues should. – Michael Jun 26 '17 at 21:01

This has been lurking implicitly beneath several of the comments so far, but just to make it completely explicit why the trace of a linear operator is independent of a choice of coordinates: the multicategory of vector spaces and multilinear maps arises from a monoidal structure on the category of vector spaces and linear maps, this monoidal structure [tensor product of vector spaces] turning out to be symmetric and closed. From this, we can construct a canonical (linear) map of type $Hom(A, 1) \otimes B \rightarrow Hom(A, B)$, which, when $A$ is finite-dimensional, turns out to furthermore be an isomorphism. In particular, this gives an isomorphism between $Hom(A, 1) \otimes A$ and $Hom(A, A)$ for finite-dimensional $A$. Now, from the closed structure, we have a canonical map of type $Hom(A, 1) \otimes A \rightarrow 1$ as well. Pulling this through the aforementioned isomorphism, we obtain a map of type $Hom(A, A) \rightarrow 1$ whenever $A$ is finite-dimensional; this map is the trace operator, defined directly on abstract vector spaces and thus coordinate independent.

Phrasing this in less categorical terms, what the above reasoning demonstrates is that there is a unique linear map $Trace$ from $Hom(A, A)$ to scalars such that $Trace(x \mapsto R(x)v) = R(v)$ for all vectors $v$ in $A$ and linear maps $R$ from $A$ to scalars (assuming, as always, that $A$ is finite-dimensional). Again, since this gives an abstract definition of $Trace$, it is immediately coordinate-independent.

Whether this should count as a geometric account is in the eye of the beholder; as far as I am concerned, suitably abstract linear algebra is directly geometric, but I could certainly understand feeling otherwise.

Let $K \subset \mathbb{R}^n$ be a compact set whose boundary is a smooth manifold. Let $F:\mathbb{R}^n \rightarrow \mathbb{R}^n$ be linear map. We have that $$\int_{\partial K} F d \vec{S} = trace(F) \cdot vol(K).$$ This a consequence of the Gauss integral formula.

There is a special case where the trace has an obvious geometric interpretation. Assume that a group $G$ acts on a finite set $E$. It also acts on the vector space $F$ of functions on $E$ with values in some field $k$. Then if $g\in G$, the trace of the operator in ${\rm End}_k (F)$ attached to $g$ is the number of points in $E$ fixed by $g$. Very often in representation theory traces of operators are related to considerations on fixed point sets via Lefschetz type formulae.

For 3 by 3 matrix $A$, there is a linear vector field $v(x)=Ax$. The divergence of $v$ is the trace of $A$. In fact $Ax = {\rm curl}(-\frac{1}{3}x\times Ax)+\frac{1}{3}{\rm tr}(A)x$. So the trace determines whether $Ax$ is a curl or not.

There is an $n$ dimensional version of this expressible in differential forms. Denote by $\hat{k}$ the $(n-1)$ form obtained by deleting $dx_k$ from $dx_1\wedge\cdots\wedge dx_n$, and when $k\ne i$ denote by $\hat{ik}$ the $(n-2)$ form obtained by deleting both $dx_k$ and $dx_i$. Then $$d\left(\sum_{i< k}(x_i (Ax)_k-x_k (Ax)_i)(-1)^{i+k}\hat{ik}\right)$$ $$= n\sum_j (Ax)_j (-1)^{j-1}\hat{j}+{\rm tr}(A)\sum_j x_j (-1)^{j-1}\hat{j}$$ The trace determines whether $\sum_j (Ax)_j (-1)^{j-1}\hat{j}$ is exact or not.

• I learned this from the linear elasticity, thanks for sharing. – Shuhao Cao Mar 27 '15 at 17:58

People have almost said this but not quite:

Take any linear transformation $A$ of a finite-dimensional real vector space $V$. Let each point $v$ in $V$ start moving at the velocity $Av$. Then the volume of any set $S \subseteq A$ will start changing at a rate equal to its volume times the trace of $A$.

More precisely, if $U(t) \colon V \to V$ is defined by

$$\frac{d}{dt} U(t) = A U(t)$$

$$U(0) = 1_V$$

then $U(t)$ is a smooth function of time. Take any measurable set $S \subseteq V$ and let $S_t$ be its image under $U(t)$. Then

$$\frac{d}{dt} \mathrm{vol}(S_t) = \mathrm{tr}(A)\, \mathrm{vol}(S_t) .$$

This is equivalent to Arnold's description of the trace, or the formula

$$\mathrm{det}(\exp(tA)) = \exp(\mathrm{tr}(A)),$$

since $U(t) = \exp(tA)$.

• Take $$V$$ a finite-dimensional vector space. The $$L(V)$$ is canonically isomorphic (as a vector space) to $$V\otimes V^*$$. Then you have a canonical isomorphism between $$L(V)$$ and its dual given by :

$$L(V)^* \rightarrow (V\otimes V^*)^* \rightarrow V^* \otimes V^{**} \rightarrow V\otimes V^* \rightarrow L(V).$$

Then the trace is the element sent to $$Id_V$$. I don't know if you consider this "geometrical", but it's a pretty nice characterization of the trace.

• The most geometrical statement is probably about the differential of the determinant.

• You also have this one : it's the $$n-1$$ degree coefficient of the characteristic polynomial. It can be considered important for at least two reasons :

1. The characteristic polynomial is the generic minimal polynomial of matrices (or endomorphisms), meaning that if you take a generic matrix (say the matrix $$M = (X_{ij})$$ with coefficient in $$k(X_{ij})$$), its minimal polynomial $$\mu_M$$ is the characteristic polynomial, and if you specialize $$M$$ to any matrix $$A$$ with coefficient in $$k$$, the specialization of $$\mu_M$$ gives you the characteristic polynomial $$\chi_A$$ of $$A$$.

2. If you want polynomial function on $$M_n(k)$$ that are similarity invariants (ie $$f(PAP^{-1}) = f(A)$$), then they form an algebra generated by the coefficients of the characteristic polynomial, and then the trace is the generator of the degree 1 part. Of course this amount to the already pointed out fact that $$Tr(AB) = Tr(BA)$$ characterizes the trace up to a constant.

In an attempt to provide an answer consistent with the original request, how about: "Trace is the semiperimeter of a parallelopiped as measured along its spanning column vectors."

It's important to be careful here. The original context implies an eigen problem in which a vector is mapped (perhaps with scaling) onto itself through a linear transformation (matrix multiplication). This follows from the mention of the determinant being the volume of the paralellopiped. The above answer is consistent with that. Other eigen problems should offer (require?) different interpretations of both "determinant" and "trace". -JF

I like the following perspective:

Up to scalar, trace is the only linear operator $\text{M}(n,k) \stackrel{t}{\to} k$ such that $t(AB) = t(BA)$.

If one likes vector field theory, this is the only linear operator that vanishes on commutator of vector fields. I do prefer to characterize it as the nullifier of hyperplan generated by commutators.

Trace is the last one on earth who still believes that matrices commutate. Its geometric interpretation, somehow, is its blindness.

One could look for a geometric interpretation in $k^n$, here the thesis is that trace is all about geometry of $\text{M}(n,k)$. This final consideration, I hope, also answers the comment:

Take the p-dimensional vector space over $\mathbb{F}_p$ and take the identity transformation on this space. Then the trace is $0$ What the "geometric" meaning of this, if any?

A reformulation of this observation is that trace is the only linear operator (again up to normalization) that is constant on conjugacy classes of matrices, somehow a first order approximation of Jordan normal form.

I opened a thread to investigate the content of this answer: Update.

• About the first point, I thought important to observe that "One could look for a geometric interpretation in $k^n$, here the thesis is that trace is all about geometry of $\text{M}(n,k)$". – Ivan Di Liberti Jul 13 '17 at 11:51
• Though trace is not invariant under any permutation of the matrices in a product, just cyclic permutations. – PrimeRibeyeDeal Aug 7 '18 at 2:44

It has been said before but let me rephrase it : the interpretation of the trace is not geometric but integration-theoretic (I do not say "measure-theoretic since there is no measure, see below). Of course if a matrix $A$ has itself a geometric content, its trace also will, e.g. mean curvature = $1/n$ trace(second fundamental form).

I think that the integration-theoretic content of the trace is best captured by noncommutative geometry, where one can define noncommutative integrals thanks to Dixmier traces. Hence there is a precise sense in which a trace can be viewed as an integral.

But maybe this can be viewed as far-fetched and not very illuminating by students discovering the trace for the first time. However, you can still convey the intuition that the trace is secretly an integral to undergrad students by observing that :

-when a matrix $A$ is in diagonal form, the trace is really the integral of its eigenvalues with the counting measure.

-you can extend that to functions of this matrix : the trace of $f(A)$ is the discrete integral of the function $f$ over the spectrum of $A$.

Of course you cannot extend this interpretation to matrices which do not commute with $A$ : if $B$ does not commute with $A$, the spectrum of $B$ is a space which bears no relation to the spectrum of $A$ (this will speak to those who have followed a course on quantum mechanics). In other words, there is no "universal spectrum" on which to define a measure. But can one define an integral without reference to measure ? You certainly want an integral to be a continuous and positive linear functional. With or without reference to Riesz representation theorem, you can go on proving that every such functional $f$ on $M_n({\mathbb C})$ is of the form $X\mapsto Tr(XM)$ for some positive matrix $M$. If you further require the normalization $f(I_n)=1$, the eigenvalues of $M$ will be non-negative numbers with sum one. Now the analogy with a probability measure should be obvious to everyone, and the requirement that the eigenvalues of $M$ be equal to mimic the uniform probability should sound natural. Hence the trace of a matrix stands out as the unique noncommutative generalization to $M_n({\mathbb C})$ of the integral of a function defined on a set of $n$ elements against the counting measure.

I had always complex matrices in mind when writing that, but you can surely extend this discussion to more general setting, though I would strongly advise against that if your aiming at undergraduate students.

Taking a broad view of the question, here are some particular geometric interpretations of the trace with respect to certain domains:

1. $\mathrm{SL}(2, \mathbb{R})$ acts by isometries on the upper half-plane $H^2$. The displacement length $\ell(g)$ of $g\in\mathrm{SL}(2, \mathbb{R})$ is the infimum of $\{d(x,gx)\ | \ x\in H^2 \}$. If $\ell(g)>0$, then $|\mathrm{tr}(g)| = 2 \mathrm{cosh}(\ell(g)/2).$
2. The trace as the Killing form is a non-degenerate bilinear form on a semisimple Lie algebra (Euclidean structure).
3. Traces of words in a finitely generated group $\Gamma$ give coordinates on the moduli space of unimodular representations of $\Gamma$.

With Example 1 in mind, in general, I intuitively think of the trace as a measure of length.

As it is the derivative of the determinant, whose absolute value measures volume, this is not unreasonable for geometric intuition (sum versus a product in the spectra). In particular, $|\mathrm{tr}(X-Y)|$ reminds one of the taxi cab metric in the spectra of $X,Y$.

With Example 3 in mind, one gets mileage from thinking of "words" as homotopy classes in a manifold and evaluating those words at representations and taking the trace as computing the length of a geodesic representative of the homotopy class. Again, this is more of "geometric intuition" than precise formulation, but there are examples where this is more precise.

• And what about the trace of a non-invertible matrix? – Yemon Choi May 14 '16 at 0:49
• Dear Sean, I was asking because the question seems to ask for a geometric interpretation of the trace of an arbitrary matrix, in a similar vein to the usual interpretation of the determinant of an arbitrary matrix – Yemon Choi May 14 '16 at 13:32
• I added a sentence to make clear that I am taking a broader view of the question, which I believe might be of interest to people who search for "geometric interpretation of trace". – Sean Lawton May 14 '16 at 14:23

Trace has a nice geometric interpretation for a rank one operator: it is the factor by which the operator scales a vector in its image. This, together with linearity, is a geometric characterisation of trace.

Traced monoidal categories are giving a nice geometrical interpretation of the trace : as a way to implement a feedback loop.

But, it is perhaps not the kind of geometrical interpretation you are interested in.

• The graphical notation for traced monoidal categories makes very explicit the "self-eating" mentioned by Vectornaut. – Tom Leinster Feb 10 '10 at 10:54

An easy calculation that may help somehow:

Any square matrix $A$ can be written as

$A = \Sigma_{i,j} u_i v_j^t$

where $u_i,v_j$ are column matrices, and there are many different choices as to how to choose {$u_i$}, {$v_j$}. Then it follows that

$Tr(A) = \Sigma_{i,j} Tr(u_i v_j^t) = \Sigma_{i,j} u_i \cdot v_j$

and now that you have a sum of dot products you may be able to make various geometric interpetations.

We have the formula $\det (e^A) = e^{\mathrm{Tr}(A)}$ and we have a good interpretation for the determinant of a matrix as the volume and then we can take the logarithm to get the trace of the matrix $A$.

• Do you have a good interpretation of logarithm of volume? – S. Carnahan Jul 4 '13 at 3:25
• The real problem is another: the exponential of a matrix. Logaritms of positive real numbers are a change of notation from multiplicative to additive (for a archimedean complete totally ordered group). Such changes of notation were already used in ancient times by some music theorist (when speaking about musical intervals), to the displeasure of phytagoric music theorists. – user46855 Feb 15 '14 at 16:07
• @S.Carnahan: A volume expansion factor is an element of the Lie group $\mathbb{R}^\times$. A trace is the rate of change of a volume expansion factor, so it lives in the Lie algebra $\mathbb{R}$. The exponential on the right side of the identity translates between $\mathbb{R}$ and $\mathbb{R}^\times$. See my newer answer for details. (I'm basically just repeating user46855's comment here, but in maybe a more concrete way.) – Vectornaut Feb 19 '15 at 0:21

What seems really odd to me is this limitation set by the original question.

The divergence application of trace is somewhat interesting, but again, not really what we are looking for.

Maybe that is rejected because it involves a metric tensor in most textbooks about differential geometry, but the divergence requires only an affine connection, even in differential geometry. In flat Cartesian space (without a norm or inner product), it's even simpler.

First consider that matrices have two main applications, as the components of linear maps and as the components of bilinear forms. Let's ignore the bilinear forms. Linear maps are really where matrices come from because matrix multiplication corresponds to composition of linear maps.

We know that the determinant is the coefficient of the characteristic polynomial at one end of the polynomial, and the trace is at the other end, as the coefficient of the linear term. So we should think in terms of linearization and volume, or some combination of these two concepts. We know that the determinant can be interpreted as the relative volume expansion of the map $x\mapsto Ax$. So we should think in terms of maybe linearizing this in some way.

Define a velocity vector field $V(x)=Ax$ on $\mathbb{R}^n$ and integrate the flow for a short time. What happens to the volume of any region? The rate of increase of volume equals $\mathrm{Tr}(A)$. This is because the integral curves have the form $x(t)=\exp(At)x(0)$. (See Jacobi's formula.)

Thus the determinant tells you the volume multiplier for a map with coefficient matrix $A$, whereas the trace tells you the multiplier for a map whose rate of expansion has component matrix $A$.

That sounds very neat and simple to me, but only if you avoid the formulas in the DG literature which try to interpret divergence in terms of absolute volume by referring to a metric tensor or inner product.

PS. To avoid analysis, to keep it completely algebraic apart from the geometric meaning of the determinant, consider the family of transformations $x(t)=x(0)+tAx(0)$ for $t\in\mathbb{R}$ for all $x(0)\in\mathbb{R}^n$. Then the volume of a figure (such as a cube) is a polynomial function of $t$. The linear coefficient of this polynomial with respect to $t$ is $\mathrm{Tr}(A)$. There are no derivatives, integrals or exponentials here. The trace also happens to be the linear component of the characteristic polynomial. I think this is a pretty close tie-up.

PS 2. I forgot to mention that the divergence of the field $V(x)=Ax$ is $\textrm{div} V=\mathrm{Tr}(A)$. Therefore trace equals divergence. That's the geometrical significance of the trace. The function $V$ is the linear map with coefficient matrix $A$. And the trace equals its divergence if it is thought of as a vector field rather than just a linear map. You could even write $\mathrm{Tr}(A)=\mathrm{div}(A)$ if you identify the matrix with the corresponding linear map.

• (Edited from a previous comment that quoted the wrong sentence.) "The trace also happens to be the linear component of the characteristic polynomial." Almost the opposite, no? That is, it's actually the coefficient of $t^{n - 1}$, not of $t$ (assuming that's what you mean by "the linear component")? – LSpice May 22 '16 at 2:21
• At a zero of a vector field, the trace of the vector field is defined without use of any affine connection. At a point which is not a zero, there is no trace defined, by the flow box theorem. So I don't understand your mention of an affine connection. – Ben McKay Feb 5 '18 at 10:41

Was surprised not to see this here yet. Let $$V$$ be an $$n$$-dimensional real vector space with inner product.

Any linear transformation $$f:V \to V$$ can be decomposed into $$f = \left(\tfrac{\textrm{tr}(f)}{n}\right) \mathbb{I} + f^{+} + f^{-}$$ where $$f^{+}$$ is traceless-symmetric and $$f^{-}$$ is traceless-antisymmetric, and $$\mathbb{I}$$ is the identity transformation.

Each term does a different geometric operation.

• The trace term returns a vector parallel to the input.
• The antisymmetric term returns a vector orthogonal to the input.
• The symmetric term stretches and flips the input along characteristic directions, with a net scale factor of zero (it admits an eigenbasis whose eigenvalues sum to zero).

The trace of the map is the scale factor/identity map contribution. Since the trace is a statement about lengths, it makes the most sense when an inner product is present, but of course the concept is more general. This also explains the relation to determinant/volume mentioned in other answers: to first order, the change in parallelepiped volume comes from scaling the edges parallel to themselves.

• Just noticed an even better version... Some other answers have showed that the trace is the average of $v \cdot f(v)$ for $v$ on the unit sphere. So if you decompose $f \propto 1 + f_{traceless}$, then the trace (identity) term scales vectors by a constant scale factor, while the traceless term on average returns a vector orthogonal to the input. – Joe Schindler Mar 21 '19 at 3:04

If we consider $M_n(\mathbb{R})$ as $\mathbb{R}^{n^2}$ with this map [$C_1$,...,$C_n$]$\stackrel{f}\mapsto$($C_1^t$,...,$C_n^t$),$C_i$s are columns and $f$ is bijection(using this mapping,we can put topology of $\mathbb{R}^{n^2}$ on $M_n(\mathbb{R})$ and with this topology $M_n(\mathbb{R})$ is a manifold),Then for a matrix $A$ we have $f$($A$)$\in$$\mathbb{R}^{n^2}$,we consider $f(I)$=($I_1^t$,...,$I_n^t$)That $I$ is identity matrix and $I_i$s are columns of $I$, now the dot product(inner product)of $f(A)$ and $f(I)$ is trace of $A$ and trace($A$) is the length of projection of vector $\sqrt{n}f(A)$in the direction of vector $f(I)$.

We show here how any interpretation of $$\operatorname{Tr} A$$ when $$A : V \to V$$ is an isomorphism can be extended to an interpretation of the trace of an arbitrary endomorphism by showing that $$\operatorname{Tr} A$$ actually only depends on a special induced sub-vector space of $$V$$.

To begin, let $$V^{(0)} = \operatorname{domain}(A) = V,\;\; V^{(i+1)} = A\left(V^{(i)}\right),\; \textrm{ and }d^i = \dim V^{(i)}$$ so that $$V^{(1)} = \operatorname{Im} A = A\left(V^{(0)}\right)$$, $$V^{(i+1)} \subseteq V^{(i)}$$, and $$d^{i+1} \leq d^i$$. Let $$N \geq 0$$ be the smallest integer s.t. $$d^{N+1} = d^N$$ and denote this common value by $$d$$. Let $$W := V^{(N)}$$.

We prove below that the restriction $$A\big\vert_W : W \to W$$ of $$A$$ onto $$W := V^{(N)}$$ is an isomorphism. Furthermore, $$\operatorname{Tr}(A) = \operatorname{Tr}\left(A\big\vert_W\right)$$ and it will be clear that $$W$$ is the unique largest vector subspace $$S$$ of $$V$$ on which $$A$$ restricts to an isomorphism $$A\big\vert_S : S \to S$$. All of this allows us to conclude that to geometrically interpret $$\operatorname{Tr}(A)$$, one may restrict their focus to geometrically interpreting the trace of the isomorphism $$A\big\vert_W : W \to W$$ rather than $$A : V \to V$$ itself.

This isn't entirely surprising since just as the trace of a matrix does not depend on the "elements off the diagonal", so too does the geometric interpretation of trace not depend on the "space off of $$W$$." Note that this also gives some geometric intuition about how the trace of a matrix can simultaneously depend only on its diagonal elements while also equaling quantities that non-trivially depend on the whole matrix (such as the sum of its eigenvalues). $$\rule{17cm}{0.4pt}$$

Proof: We now prove the above claim. Inductively construct a basis $$\left(e_1, \dots, e_{\dim V}\right)$$ for $$V$$ such that for all $$i \geq 0$$, $$\left(e_1, \dots, e_{d^i}\right)$$ is a basis for $$V^{(i)}$$. Let $$\left(\varepsilon^1,\dots, \varepsilon^{\dim V}\right)$$ be the dual basis of $$e_{\bullet}$$ and note in particular that: $$\textrm{(1) whenever }d^{i + 1} < l \leq d^i\textrm{ then }\varepsilon^l\textrm{ vanishes on }V^{(i + 1)}.$$

Since $$(e_1, \dots, e_{d^1})$$ is a basis for the range of $$A$$ we may, for any $$v \in V^{(0)},$$ write $$A(v) = \varepsilon^1(A(v)) e_1 + \cdots + \varepsilon^{d^1}(A(v)) e_{d^1}$$ so that $$A = (\varepsilon^l \circ A) \otimes e_l$$ (the sum ranging over $$l = 1, \dots, d^1$$) and hence $$\operatorname{Tr}(A) = (\varepsilon^l \circ A)(e_l) = \varepsilon^1(A(e_1)) + \cdots + \varepsilon^{d^1}\left(A\left( e_{d^1} \right)\right)$$ which shows that $$\operatorname{Tr}(A)$$ actually depends only on the range of $$A$$ (i.e. $$V^{(1)}$$). Now since $$e_1, \dots, e_{d^1}$$ are (by definition) in $$V^{(1)}$$, all of $$A\left(e_1\right), \dots, A\left(e_{d^1}\right)$$ belong to $$A\left(V^{(1)}\right) = V^{(2)}$$ so that from $$(1)$$ it follows that $$\operatorname{Tr}(A) = \varepsilon^1\left(A\left(e_1\right)\right) + \cdots + \varepsilon^{d^2}\left(A\left( e_{d^2} \right)\right)$$

Continuing this inductively $$N \leq \dim V$$ times shows that $$\operatorname{Tr}(A) = \varepsilon^1\left(A\left(e_1\right)\right) + \cdots + \varepsilon^{d}\left(A\left(e_d\right)\right)$$ so that $$\operatorname{Tr}(A)$$ depends only on $$W = V^{(N)}$$. Since by definition of $$N$$, the map $$A\big\vert_W : W \to W$$ is surjective, it is an isomorphism and furthermore, it should be clear that $$W$$ is the unique largest subspace of $$V$$ on which $$A$$ restricts to an isomorphism. $$\blacksquare$$

As described elsewhere, if you view $$A : V \to V$$ as a vector field on $$V$$ in the canonical way then the trace of $$A$$ is the same as its divergence so in the case where $$A$$ is an isomorphism there is a pleasing geometric interpretation readily available, which I'll assume that you're comfortable with. However, this interpretation is not satisfactory when $$A$$ is not surjective, as shown by the example below.

Example: If $$A : \mathbb{R}^3 \to \mathbb{R}^3$$ is such that $$\operatorname{Im} A$$ is $$2$$-dimensional and $$A$$ is volume-increasing (i.e. $$\operatorname{div}(A) = \operatorname{Tr}(A) > 0$$) then $$A$$ takes a $$3$$-d bucket of water (e.g. $$\mathbb{R}^3$$) and ''compresses it down'' into a $$2$$-d ''paper'' (e.g. into the 2-d set $$\operatorname{Im} A$$); but it's not clear (at least to me) how anyone could be expected to say that $$A$$ has increased the volume of this 3-d space (since $$A$$ is, after all, "volume-increasing") simply because the quantity $$\operatorname{div}(A) = \operatorname{Tr}(A)$$ happens to be positive! $$\blacksquare$$

Nevertheless, the equality $$\operatorname{div}(A) = \operatorname{Tr}(A)$$ is our best bet at finding a geometric interpretation of trace since it establishes a direct simple relationship between the trace and a readily interpretable quantity: $$\operatorname{div}(A)$$.

To summarize, going back to example above, in the case where $$A : V \to V$$ is an arbitrary linear map we can imagine being given some initial ''bucket of water'' $$V = V^{(0)}$$ and then imagine $$A$$ as repeatedly (and eternally) deforming this same water until eventually (i.e. after $$N$$ iterations) $$A$$ would have ''pushed'' or ''compressed'' all of $$V$$ onto some vector subspace $$W = V^{(N)}$$ (which is also the unique the largest subspace $$W$$ of $$V$$ that $$A$$ maps back onto itself) i.e. all of $$V$$ would eventually ''flow into'' $$W$$. It is at this point that $$A$$ no longer ''compresses'' this water down by some dimension(s) so that $$A$$ does nothing more than bijectively move this $$d = \dim W$$-dimensional water around. It now makes sense to ask by how much the isomorphism $$A\big\vert_W : W \to W$$ is increasing or decreasing this $$d$$-dimensional volume, which is what $$\operatorname{Tr}(A) = \operatorname{Tr}\left(A\big\vert_W\right) = \operatorname{div}\left(A\big\vert_W\right)$$ represents.

Remark: This may not really answer your question since you stated that "The divergence application of trace is somewhat interesting, but again, not really what we are looking for." Nevertheless, whatever alternative non-divergence based interpretation of the trace of an isomorphism you choose, I hope that this will help you to extend it to the case where the map isn't an isomorphism.

• If $A$ is not an isomorphism, what does that have to do with the vector field? For example, if $A$ is the $3 \times 3$ diagonal matrix with diagonal $(2,0,0)$, then the divergence is $2$, this "fluid" is expanding in the positive $x$ direction at every point... it's not compressing the fluid to a plane or line. – Zach Teitler Jun 23 '17 at 18:56
• @Zach Teitler The image of your diagonal matrix is the $x$-axis and when your diagonal matrix $A$ is restricted to the $x$-axis then it becomes the $1 \times 1$ matrix $(2)$, which "expands the $1$-dimensional fluid" that is the $x$-axis. The vector field off the $x$-axis is irrelevant just as the coefficients of $A$ off the diagonal are irrelevant when computing the trace. In short, after finding $W$ we identify $A$ with the canonical vector field that it induces and interpret the trace of $A$ as the divergence of this vector field's restriction to $W$. – Matthew K. Jun 23 '17 at 22:55
• I have no comment on the merits of this answer, but would it not be possible to write this first of all in a text editor and check it before posting, rather than make almost 10 edits in less than 2 hrs? – Yemon Choi Dec 1 '19 at 5:03
• @Yemon Cho. My apologies if my editing has somehow disrupted your life. – Matthew K. Dec 1 '19 at 5:05
• Matthew, I was merely trying to suggest ways to behave in a way that seemed more appropriate to this site, since your only MO activity seems to have been with this answer. Editing bumps questions to the top of the "stack". By the way, it would be appreciated if you took the time to get my name right :) – Yemon Choi Dec 1 '19 at 14:49

Here's more on the semi-perimeter, generalized to $$n$$ dimensions. The trace is the sum of the signed edge lengths of the rectangular parallelepiped whose first edge length = the first entry of row 1, the second edge length = the second element of row 2, and so on. We could have also used columns instead.

Here, the edge lengths can have non-positive values. While the determinant is a product of signed lengths, the trace is a sum of signed lengths.

Permuting rows/columns can drastically the trace. But in a way, taking the trace of different permutations of $$\mathbf{A}$$ tells you more information about $$\mathbf{A}$$. You don't get more information from taking the determinant of those permutations.

In the $$2 \times2$$ case, the trace is equal to half the signed perimeter of the rectangle created by the rows/columns, in the process described above. Here, it's a semi-perimeter.

According to the interesting answer of Rado, "trace" is an algebraic way to represent the dimension of fibres of a vector bundle on a compact Hausdorff space $X$. Every vector bundle on $X$ corresponds to an idempotent matrix valued function $A(x),\;x\in X$ . The dimension of fibre of a vector bundle is equal to $tr(A(x))$. Assuming $X$ is connected, this quantity is fix along $X$. I learned this from "Very basic noncommutative geometry" By Masoud Khalkhali (Can be found in arxiv).

The terminology "Trace" is also used in PDE as an operator which restricts functions in sobolov space $H^{s}(\Omega)$ to the boundary of $\Omega$