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

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.

Thanks in advance.

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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
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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
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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
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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
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Voting to close as "no longer relevant", because new answers to this are really very submodular... –  Suvrit Apr 2 '13 at 0:49
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17 Answers

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.

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It is also important in statistics! –  Kjetil B Halvorsen Dec 24 '12 at 0:29
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This is important everywhere in math. –  Matthieu Romagny Mar 29 '13 at 17:18
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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

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

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'?

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the problem is twofold. First, that means that the determinant is changing, so the matrix is a function of some parameter t. I was hoping for something related to any matrix. The second problem, I don't entirely understand what is meant in that wiki entry. This answer however is getting much closer to what we were hoping for. –  B. Bischof Jan 31 '10 at 2:29
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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
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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
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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
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Let's use $\det(\exp(tA)) = 1 + t\ Tr(A) + O(t^2)$, and think about the vector ODE $\vec y' = T \vec y$, 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 $\vec y' = T\vec y$, and see how its volume changes, the change will thus be $t\ Tr(A)$ to first order.

Ah, Yemon Choi beat me to part of that.

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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 $\mathbb{C}^m$: $$\langle A, B\rangle = \text{Tr}\ A^\dagger B.$$ You can't get much more geometric than that.

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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
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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
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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.

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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
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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
    
@Yemon Choi: Thanks for pointing out the finite-dimensionality thing! I hadn't known that before. @Theo Johnson-Freyd: Interesting! I never understood why the cyclicity property was a natural thing to ask for; maybe this explains it. What you said isn't obvious to me, though, so I'll have to take some time to think about it... –  Vectornaut Jan 31 '10 at 21:17
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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
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By the way, I'd also be very interested in understanding the "self-eating" interpretation of the trace - it's extremely important for tensors, but I never found an explanation of how to think of it, and why it works so well. –  Ilya Grigoriev Feb 1 '10 at 3:37
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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.

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

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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.

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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

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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 thing because it is the ratio of distortion of length. On the other hand 'expected value' is bollowed 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.

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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
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However, this answer is somehow duplicate of that by Yemon Choi. –  Denis Serre Mar 29 '13 at 15:07
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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."

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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 characterization of trace.

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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 $A$.

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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 at 16:07
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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.

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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.

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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 $f(I)$.

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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.

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