Categorification of characteristic polynomial

Yes: the action of a linear transformation on the exterior algebra. The coefficients of the characteristic polynomial are the elementary symmetric polynomials in the eigenvalues, which are nothing more than the traces of the induced action of a linear transformation $T : V \to V$ on the exterior powers of $V$. The entire characteristic polynomial can then be thought of as the graded trace of $T$ acting on $\Lambda(V)$.

Edit: Some elaboration. Let $T : V \to V$ be a linear transformation of a finite-dimensional vector space. Suppose that $T$ is diagonalizable with eigenvalues $\lambda_1, ... \lambda_n$ and attached eigenvectors $e_1, ... e_n$. Then

$$\text{tr}(T) = \sum_{i=1}^n \lambda_i.$$

Moreover, $\Lambda^k(V)$ is spanned by wedge products of increasing sequences of $k$ distinct eigenvectors $e_i$; these are all eigenvectors for the induced action $\Lambda^k(T) : \Lambda^k(V) \to \Lambda^k(V)$ with eigenvalue the corresponding product of the eigenvalues $\lambda_i$. It follows that

$$\text{tr}(\Lambda^k(T)) = e_k(\lambda_1, ... \lambda_n)$$

is the $k^{th}$ elementary symmetric polynomial in the eigenvalues. Since this is a coefficient of the characteristic polynomial, it is in particular a polynomial function of the entries of $T$ as a matrix, and since the diagonalizable matrices are Zariski-dense in the space of all matrices, the above relation holds identically.

Now for the graded trace. A graded vector space is a vector space $V$ with a canonical direct sum decomposition

$$V = \bigoplus_{i=0}^{\infty} V_i$$

where the $V_i$ are the homogeneous components of degree $i$. A morphism $T$ of graded vector spaces is a linear transformation which respects the grading. If all of the $V_i$ are finite-dimensional, then for every endomorphism $T : V \to V$, isolate its homogeneous components $T_i : V_i \to V_i$, and then we can form the graded trace

$$\text{tr}(T) = \sum_{i=0}^{\infty} \text{tr}(T_i) z^i.$$

This is a formal power series in $z$ rather than a scalar, but it has all of the same properties as the usual trace: it behaves well under addition, direct sum, tensor product, etc.

The exterior algebra is naturally a graded vector space with homogeneous components $T_i = \Lambda^i(V)$, but it turns out that the components with $i$ odd should be regarded as having negative dimension for reasons which are too involved to get in to here, so the corresponding graded trace should actually be

$$\text{tr}(T) = \sum_{i=0}^{\infty} \text{tr}(T_i) (-1)^i z^i.$$

Taking the graded trace of the induced action of a linear transformation on its exterior algebra, we then almost recover the characteristic polynomial of $T$, except that the coefficients are reversed.

A conceptual explanation for why this works is to again restrict to the case that $T$ is diagonalizable and use the following two facts:

• Let $T : V \to V$ and $S : W \to W$ be two endomorphisms of two graded vector spaces, and let $T \otimes S : V \otimes W \to V \otimes W$ be their tensor product. Then $\text{tr}(T \otimes S) = \text{tr}(T) \text{tr}(S)$.
• There is a canonical isomorphism $\Lambda(V \oplus W) \cong \Lambda(V) \otimes \Lambda(W)$.

Hence if $T$ is diagonalizable, we can compute the graded trace of its induced action on the exterior algebra by computing it for each of its eigenspaces (easy), then multiplying the results together. Again by Zariski-density, our results hold for all $T$.

For reference, the tensor product of two graded vector spaces $V, W$ with homogeneous components $V_i, W_i$ is the graded vector space with homogeneous components

$$(V \otimes W)_n = \bigoplus_{i+j=n} V_i \otimes W_j.$$