All the coefficients of the characteristic polynomial are some traces

If $V$ is an $n$ -dimensional vector space and $A$ is an matrix over $V$ , you probably remember that in its characteristic polynomial $p_{A}(t) =\det(A-tI)$ , the coefficient of $t^{n-1}$ was minus the trace $\mathrm{Tr}(A)$ and that the constant coefficient was $\det(A)$ . Probably an absurd question might be that, does there exist a formalism and computing principle that produces the coefficient of each $t^k$ , and so $\mathrm{Tr}(A)$ and $\det(A)$ are just the two extremities of the formalism?

And despite the absurdity in the question, I like the surprise in the answer: Yes. There is a generalized notion of the trace map defined for tensors. Note that we have $\bigwedge^1A =A: V \to V$ and $\mathrm{Tr}(\bigwedge^1A)=\sum_{i=1}^n \langle A(e_i), e_i \rangle$ . Then if you let $\bigwedge^kA: \bigwedge^k V \to \bigwedge^k V$ given by $v_{1} \wedge \dots \wedge v_k \mapsto A(v_1) \wedge \dots \wedge A(v_k)$ for all $v_i \in V$ (the notation $A(e_I)$ is usually used to denote $A(e_{i_1}) \wedge \dots \wedge A(e_{i_k})$ where $I = \{ i_{1}, \dots , i_k \}$ is a multi-index of order $k$ ), the trace of $\bigwedge^kA$ is defined as

$\displaystyle \sum_{\text{All increasing } I \text{ with } \lvert I \rvert=k} \langle A(e_I), e_I \rangle$

where $\langle A(e_I), e_I \rangle$ means the evaluation of the $k$ -form $A(e_I)$ on the $k$ -vector $e_I$ .

The similarity of the last formula with that of primary trace is noticeable. It’s like that in the original trace definition, the indices are expanding in one dimension (the index was $i$ enumerating only numbers), so the space of possible pairings is $(1+1)$ -dimensional and we were only summing over in the $1$ dimensional diagonal subspace $\{ (i,i) \}_{1 \leq i \leq n}$ of the index space. Here however the indices are $k$ -dimensional so $2k$ -dimensional for the space of possible pairings; yet here as well we only sum over pairs on the $k$ -dimensional diagonal subspace $\{ (I , I) \}_{\lvert I \rvert=k}$ . For $k=n$ we have $\mathrm{Tr}(\bigwedge^nA)=\langle A (e_1) \wedge \dots \wedge A(e_n), (e_1, \dots, e_n) \rangle = \det(A)$ , and if you stare at the definitions for long enough you can easily verify that

$\displaystyle p_{A}(t)= \sum_{k=0}^n t^{n-k} \cdot (-1)^k \mathrm{Tr}\left(\bigwedge ^k A\right)$

It seems that exterior algebra is a more convenient framework for working with the combinatorial attributes of a matrix, since the alternativity properties needed to define those attributes are already built into wedge products. Let me just mention one other example. The Pfaffian $\mathrm{Pf}(A)$ when $A$ is a $2n \times 2n$ skew-symmetric matrix (not relevant or interesting in other cases) is a useful quantity in geometry and mathematical physics and it’s defined by