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The coefficients of lowest and next-highest degree of a linear operator's characteristic polynomial are its determinant and trace. These have well-known geometric interpretations. But what about its intermediate coefficients?

For a linear operator $f : V \to V$, we have the beautiful formula

$$\chi(f) = det(f - t) = \sum_{i=0}^n (-1)^i\ tr(\wedge^{n-i}(f))\ t^i,$$

where $\wedge^{p}(f)$ is the map induced by $f$ on grade $p$ of $V$'s exterior algebra.

While this formula is rarely mentioned (at least I haven't seen it in any of the standard textbooks), it is not too surprising if you have a good grasp of exterior algebra. It presents $\chi(f)$ as a generating function for the exterior traces of $f$.

My question is whether these traces have a simple geometric interpretation on par with $tr$ and $det$.

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This formula is certainly mentioned in more advanced books that take coordinate-free point of view. –  Victor Protsak Jul 27 '10 at 6:49
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You seem very confident that the trace has a geometric interpretation; in fact, this was the subject of a previous MO question mathoverflow.net/questions/13526/…. But if you are truly happy with the geometricity of the trace, it seems that your question comes down to asking for a geometric interpretation of "intermediate" exterior powers of a linear operator. I'm sure that some people here would be happy to speak to that... –  Pete L. Clark Jul 27 '10 at 7:02
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Pete: The exterior powers of the operator aren't mysterious to me. But if you apply the interpretation of the trace directly to this question, it will give you an answer in terms of the various $\wedge^p(V)$ vector spaces rather than $V$ itself. Anyway, I think I got it: If we take $R^3$ with a diagonal matrix $diag(a_1, a_2, a_3)$ as an example, the bivector trace is $a_1 a_2 + a_2 a_3 + a_1 a_3$. This is the second-order volume differential contributed by the edges of the unit cube, much as the vector trace $a_1 + a_2 + a_3$ is the first-order volume differential contributed by the faces. –  Per Vognsen Jul 27 '10 at 7:20
    
@PV: What you say both agrees with what I said (or meant to say, at least) and goes on to give a geometric interpretation of the sort I vaguely had in mind. Cheers. –  Pete L. Clark Jul 27 '10 at 7:31

1 Answer 1

up vote 52 down vote accepted

A rather simple response is to differentiate the characteristic polynomial and use your interpretation of the determinant.

$$det(I-tf) = {t^n}det(\frac{1}{t}I-f) = (-t)^ndet(f-\frac{1}{t}I)= {(-t)^n}\chi(f)(1/t)$$

So if we let $\chi(f)(t) = \Sigma_{i=0}^n a_it^i$, then ${(-t)^n}\chi(f)(1/t) = (-1)^n\Sigma_{i=0}^n a_it^{n-i}$

But $I-tf$ is the path through the identity matrix, and $Det(A)$ measures volume distortion of the linear transformation $A$.

$$det(I-tf)^{(k)}(t=0) = (-1)^nk!a_{n-k}$$

and a change of variables ($t\longmapsto -t$) gives (and superscript $(k)$ indicates $k$-th derivative)

$$det(I+tf)^{(k)}(t=0) = (-1)^{n+k}k!a_{n-k}$$

So the coefficients of the characteristic polynomial are measuring the various derivatives of the volume distortion, as you perturb the identity transformation in the direction of $f$.

$$a_k = \frac{det(I+tf)^{(n-k)}(t=0)}{(n-k)!}$$

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That sounds good. In my comment to Pete, I briefly outlined something similar. To put what you said in those terms, it's measuring the differential contributions to the volume distortion from the various facets of the volume element: in the R^3 case, solids, faces, edges and vertices, in that order. The alternating signs come from inclusion-exclusion counting of overlapping contributions. –  Per Vognsen Jul 27 '10 at 7:34
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+1. $ $ –  Pete L. Clark Jul 27 '10 at 7:35

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