Suppose we have invertible $N\times N$ square matrix $A$ (these are overly strong hypotheses for what follows). Suppose the eigenvalues of $A$ are $\lambda_1, \ldots, \lambda_2$.

It is well known that, for whatever definition of $\text{det}$ is used, that

$$ \text{det}(A) = \prod_{i=1}^N \lambda_i = \sum_{\sigma \in S_N} \text{sgn}(\sigma) \prod A_{i, \sigma(i)}. $$

Here $S_N$ is the symmetric group of size $N$, $\sigma$ is a permutation, $\text{sgn}(\sigma)$ is the parity of that permutation and if $\sigma(i)=j$ then $A_{i, \sigma(i)}$ is the element at the $i^{\text{th}}$ row and $j^{\text{th}}$ column of $A$.

My question is asking for intuition about the second equality. Why should we expect the product of eigenvalues of a matrix to be related to this particular antisymmetric combination of elements of the matrix?

Some straws I've tried to grasp at:

  • The simplest answer is to say just calculate and don't ask for intuition. In this case we take some definition of the determinant (for example unique multilinear, antisymmetric, and normalized function of $N$ vectors) and prove that it is equal to the expressions I've given above. This of course doesn't satisfy what I'm trying to understand.
  • Above I speak about matrices. Perhaps I could get further by thinking more generally about more general linear transformations, but I'm not sure my expression at the far right really has coordinate-free meaning so I think we'd just come back to coordinates.
  • This question was motivated by some study in differential geometry. I know there are some facts there that feel relevant, but I think any discussion of differential geometry would be putting the cart before the horse. For example, as far as I can tell, volume forms have the alternating form that they do precisely because of the equalities I've given above. That is, volumes are alternating because determinants are alternating, and it makes sense for volumes to be like determinants because determinants give volumes of parallelepiped. (this last fact is closely related in my mind to the produce of eigenvalues characterization of the determinant).

All in all, however, I have not been able to come up with nice intuition for why the second equality should hold and would appreciate any insight others can provide.

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    We could connect the right-hand-side of the equation to the volume-based interpretation as is [done here](https://math.stackexchange.com/q/668/81360), and from there make the connection to eigenvalues. For me, that is enough to bridge the gap here. – Ben Grossmann Nov 29 '21 at 16:23
  • *"Why should we expect the product of eigenvalues of a matrix to be related to this particular antisymmetric combination of elements of the matrix?"* Suppose $A$ is upper triangular, then the equality follows by a simple calculation. Moreover, assume your field is $\mathbb C$ or some algebraically closed field: since the determinant is a homomorphism form $GL_n(\mathbb F)\longrightarrow \mathbb F-\{0\}$ and $A$ is necessarily conjugate to an upper triangular matrix, the equality follows – user8675309 Nov 29 '21 at 20:16

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