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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.

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    $\begingroup$ 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? $\endgroup$
    – Anweshi
    Commented Jan 31, 2010 at 2:12
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    $\begingroup$ 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. $\endgroup$
    – Theo Johnson-Freyd
    Commented Jan 31, 2010 at 8:18
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    $\begingroup$ 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. $\endgroup$
    – Yemon Choi
    Commented Jan 31, 2010 at 8:33
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    $\begingroup$ @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 $p$. $\endgroup$
    – user46855
    Commented Feb 15, 2014 at 15:58
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    $\begingroup$ I want to bump up Sujit Nair's answer, which is how I think about it. Trace is the derivative of |1+tA|. This is the lie theoretical interpretation. $\endgroup$
    – Dmitry Vaintrob
    Commented Jan 12, 2016 at 23:30

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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 $\Omega$

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