Geometric interpretation of trace

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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."

Answer 4

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.

Answer 5

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'?

Answer 6

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!

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^ 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.

Answer 7

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.

Answer 8

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.

Answer 9

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.

Answer 10

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.

Answer 11

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.

Answer 12

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

Answer 13

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.

Answer 14

• 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.

Answer 15

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

Answer 16

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.

Answer 17

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?

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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.

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

Answer 18

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.

Answer 19

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.

Answer 20

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.

Answer 21

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.

Answer 22

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 $A$.

Answer 23

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.

Answer 24

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.

Answer 25

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 $f(I)$.

Answer 26

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.

Answer 27

Let $k$ be any field and $V$ an $n$-dimensional vector space over $k$. Let $A \in \operatorname{End}(V).$ Fix some basis $\{v_1, \ldots, v_n\}$ for $V$, and let $\{\phi_1, \ldots, \phi_n\}$ be a dual basis. Define the trace of the basis $\{v_1, \ldots, v_n\}$ to be vectors $\{v_1\phi_1(Av_1), \ldots, v_n \phi_n(Av_n)\}.$ Intuitively, these are the "traces" of each of the basis vectors after applying $A$. The elements $\phi_i(Av_i)$ for $1 \leq i \leq n$ are then the degrees to which each $v_i$ have been left behind by $A$, being the coefficient by which $v_i$ has been multiplied to get its "trace" after applying $A$ (this last interpretation makes most sense to me if we take $k = \mathbb{R}$).

Now, observe that $\operatorname{Tr}(A) = \sum_{i=1}^n \phi_i(A v_i)$. Thus the trace of $A$ can be interpreted as the degree to which it "leaves behind" any basis. To me this motivates the choice of the word "trace."

This is especially nice when $\{v_1,\ldots,v_n\}$ is an eigenbasis for $A$, since in that case the "trace" of each $v_i$ is exactly $A v_i$, so the degree to which $v_i$ is "left behind" by $A$ is just the eigenvalue corresponding to $v_i$.

Answer 28

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.

Answer 29

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 $\Omega$