Geometric interpretation of characteristic polynomial 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 $det$.
 A: 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 $f$.
$$a_k = \frac{det(I+tf)^{(n-k)}(t=0)}{(n-k)!}$$
A: I am reluctant to answer a question this old that already has a very nice answer, however, looking at the title the first thing that comes to my mind is something quite different from the existing answer (and maybe it will be useful to someone who comes across this question).
When $V$ is a finite dimensional vector space over $\mathbb{C}$, then the coefficients of the characteristic polynomial are global coordinates on certain moduli spaces.  I think of global coordinates as geometric since they generally give an embedding of the space in question, and sometimes they also give an explicit geometric description (as with the examples below).
First, the coefficients give a coordiante system on the moduli space of representations of $\mathbb{Z}$ into $GL(V)$.  To see this first identify $Hom(\mathbb{Z}, GL(V))$ with $GL(V)$, and then the coefficients of the characteristic polynomial, call them $c_1,....,c_n$ where $n=\dim(V)$, are conjugation invariant functions $GL(V)\to \mathbb{C}$.  As they are symmetric polynomials in the eigenvalues (on the open dense subset of diagonalizable matrices) they are algebraically independent.  As the dimension of the moduli space $Hom(\mathbb{Z}, GL(V))//GL(V)$ is $n$, we have a global coordinate system.  The moduli space is then seen to be $\mathbb{C}^{n-1}\times \mathbb{C}^*$.
Now, if one wants to allow general linear operators (still assuming $V$ is finite dimensional over $\mathbb{C}$), then similar ideas give that the coefficients are global coordinates on $End(V)//GL(V)$ which is seen to be isomorphic to $\mathbb{C}^n.$  From this point-of-view, the "geometric interpretation of the characteristic polynomial" is that its coefficients give the global geometry of the moduli space of operators themselves.
