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 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 $f$.

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