Geometric Interpretation of Trace - MathOverflow

Geometric Interpretation of Trace

2010-01-31T01:49:07Z

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.

Answer by Rado for Geometric Interpretation of Trace

2010-01-31T01:54:42Z

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 by Yemon Choi for Geometric Interpretation of Trace

2010-01-31T02:05:24Z

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'?

Answer by Allen Knutson for Geometric Interpretation of Trace

2010-01-31T02:09:25Z

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.

Answer by Jon Yard for Geometric Interpretation of Trace

2010-01-31T02:32:26Z

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.

Answer by Vectornaut for Geometric Interpretation of Trace

2010-01-31T08:07:07Z

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:

It gets multiplied by $\lambda$ if you multiplying one of its arguments by $\lambda$.
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 by alpheccar for Geometric Interpretation of Trace

2010-01-31T20:57:39Z

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 by Sridhar Ramesh for Geometric Interpretation of Trace

2010-02-10T09:42:25Z

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 by Q.Q.J. for Geometric Interpretation of Trace

2010-02-13T13:25:59Z

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 by fpb812 for Geometric Interpretation of Trace

2010-11-17T23:01:23Z

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 by Makoto Yamashita for Geometric Interpretation of Trace

2010-11-18T02:37:05Z

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.

Answer by Sujit_Nair for Geometric Interpretation of Trace

2010-11-18T03:52:07Z

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 by Amritanshu Prasad for Geometric Interpretation of Trace

2013-01-23T08:26:50Z

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.

Answer by Jafar for Geometric Interpretation of Trace

2013-03-29T10:35:13Z

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

Answer by Paul Broussous for Geometric Interpretation of Trace

2013-03-29T12:01:37Z

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 by Bob Terrell for Geometric Interpretation of Trace

2013-04-01T23:57:57Z

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 by R salimi for Geometric Interpretation of Trace

2013-04-05T19:56:18Z

If we consider $M_n(\mathbb{R})$ like $\mathbb{R}^{n^2}$ 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($I$).