Conceptual approach to the formula $\sum_{i=1}^n \det(v_1, \ldots, Av_i, \ldots, v_n)=\text{tr}(A)\det(v_1, \ldots, v_n)$ I answered this question earlier showing that 
$$\sum_{i=1}^n \det(v_1, \ldots, Av_i, \ldots, v_n)=\text{tr}(A)\det(v_1, \ldots, v_n),$$
and while I am happy with my answer, I feel like there should be a high level approach to understanding what is going on.
My first thought is that a linear map $A:V\to V$ extends to a degree $0$ derivation of $\bigwedge V$, and this map from the question is just the top degree part of this map.  However, I do not see a way to exploit this fact.
My second thought is that, since $\det(Av_1,Av_2,\ldots,Av_n)=\det(A)\det(v_1,\ldots,v_2)$, there might be a formula for the various coefficients of the characteristic polynomial where sums are taken over $k$ element subsets of $\{1,2,.\ldots, n\}$ and summands are determinants applied to the vectors where $k$ of them have had $A$ applied to them.  However, even if a formula of this sort was true, I would like a conceptual (as opposed to computational) approach to it.  For these purposes, checking conjugation invariance and continuity and then evaluating on diagonal matrices would not be acceptable.
Another thought I had was that the formula reminds me of the fact that the derivative of the determinant near the identity is trace.  However, I am unsure of exactly how to relate this idea to the formula.
So how do you conceptually explain the formula?  Is there a good way to fit it into a larger context?
 A: Here's a hint on your third thought. How does one differentiate a determinant? One row/ column at a time, holding the others fixed. 
Let $V = [v_1,v_2,\cdots, v_n]$ where $v_i'$ are columns. Then, look at the matrix $e^{tA}V.$ Can you differentiate this function at $t=0$ in two ways that correspond to the two sides? 
A: Usually this comes up in the context of Cayley-Hamilton.  See that the left side is exactly the $n-1$ order term in $\lambda$ when one starts expanding $\det ( A - \lambda I)$.
But Cayley-Hamilton can be derived not in terms of determinants but traces of linear operators on the exterior algebra.  In particular, let $A_{(k)}$ denote the natural analogue of $A$ that is $\bigwedge^k V \mapsto \bigwedge^k V$.  Explicitly, this is
$$A_{(k)}(v_1 \wedge v_2 \wedge \ldots \wedge v_k) = A(v_1) \wedge A(v_2) \wedge \ldots \wedge A(v_k)$$
Then the Cayley-Hamilton theorem can be written in terms of traces of the $A_{(k)}$:
$$p(A) = 0 = A^n + \sum_{i=1}^{n-1} (-\lambda)^{n-i} (\mathrm{Tr} \, A_{(n-i)}) A^i$$
See that the $i=n-1$ term gives us the trace of $A_{(1)} = A$.
This way of looking at it, proved in Hestenes and Sobczyk, has an appealing geometric interpretation: the traces tell us about eigenvalues, after all, and even for the $A_{(k)}$, we're looking at the sum total of whether directions are dilated or shrunk, and then we're extending that to planes, volumes, and generalized such objects.
