# Origin and use of an identity of formal power series: $\det(1 - \psi T) = \exp \left(-\sum_{s=1}^{\infty} \text{Tr}(\psi^{s})T^{s}/s\right)$

The following is a historical question, but first some background:

Let $\psi$ be a linear operator from a vector space to itself. The following two expressions, viewed as formal power series, can be shown to be equal (even for an infinite dimensional vector space):

$$\det(1 - \psi T) = \exp \left(-\sum_{s=1}^{\infty} \text{Tr}(\psi^{s})T^{s}/s\right)$$

Here is a simple example using the associated matrix in the finite dimensional case:

For more on this topic (and how to ensure the definitions make sense for infinite dimensional vector spaces) see: N. Koblitz, p-adic Numbers, p-adic Analysis, and Zeta-Functions, Springer-Verlag, New York, 1984.

In particular, see Koblitz's book for how ($p$-adic versions of) this identity can be used in Bernard Dwork's proof of the rationality of the zeta function (resolving the first of the Weil Conjectures).

I would welcome any insight as to how one thinks up, observes, or conjectures such an identity in the first place, but my question is: what is the origin of this identity and where else has it appeared (perhaps in a somewhat modified form)?

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Its discussed on the wikipedia page for the determinant. It can be viewed as a form of the identity $\exp(\mathrm{tr}(M))=\det\exp(M)$. –  Chris Godsil Jan 13 '13 at 20:10
@ChrisGodsil If you can write up something formal showing why this is the case, I will accept it as an answer. (Unless others can shed more light on where else this identity shows up and what its history is.) –  Benjamin Dickman Jan 13 '13 at 21:20
Since you ask the question in relation to its use in Dwork's proof of the first Weil conjectures... I think this kind of identity was known to Weil, who first defined the zeta-function of a variety in fairly general terms. You might look at his 1949 paper where he stated his conjectures, just to see if it provides some inspiration for you. –  KCd Jan 14 '13 at 4:23
Using $\det\exp=\exp\mathrm{tr}$, we have $$\det(I-tA) = \det \exp\log(I-tA) = \exp\mathrm{tr}\log(I-tA)$$ and as $$\log(I-tA) = -\sum_{k\ge1} \frac1k t^k A^k,$$ we have $$\exp\mathrm{tr}\log(I-tA) = \exp\left( -\sum_{k\ge1} \frac1k t^k \mathrm{tr}(A^k)\right)$$ This shows that your identity is a form of the $\det\exp$ identity.
The question about the origins is interesting and I do not have much to say (which was why I restricted myself to a comment in the first place). For complex matrices it is very easy to prove that $\det\exp(A)=\exp\mathrm{tr}(A)$, which might explain why I have never seen a name attached to it.