Is there something like a power series expansion for a determinant? I mean the following thing: if $k$ is a field (of characteristic zero) and $M$ and $N$ are two square matrices of the same size over $k$, can we then express $\det(M + tN)$ in a nice way as a power series - which should actually be a polynomial - in $t$?

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    $\begingroup$ How about $$\det(I-tA)=\exp\left(-\sum_{n=1}^\infty\frac{t^n}{n}\mathrm{tr}(A^n)\right)?$$ $\endgroup$
    – anon
    Sep 8 '11 at 9:16
  • $\begingroup$ Also, Newton's identities "in reverse" allow one to write the characteristic polynomial $\det(tI-A)$ as a polynomial in $\mathrm{tr}(A^k)$, for $k=1,2,\dots,d$, and this can be generalized to what you want. The coefficients will be defined recursively, so it's arguable how nice of an expression it is. Also, the case when $\det M = 0$ might cause technical issues. $\endgroup$
    – anon
    Sep 8 '11 at 9:26

To expand on anon's comment:

$$\begin{align}\det (I-A) &= \exp\log\det (I-A) \\&=\exp\text{Tr}\log (I-A) \\ &=\exp\text{Tr}\left(-\sum_{n=1}^\infty\frac{A^n}n\right) \\&=\exp\left(-\sum_{n=1}^\infty\frac{\text{Tr}A^n}n\right)\;.\end{align}$$

  • $\begingroup$ Is the formula $\log\det=\text{tr}\log$ true also when the matrix $I-A$ does not admit a logarithm, so it is just a formal power series? Because I found proofs of this using the relation $\det\exp(A)=\exp\text{tr}(A)$ and writing $\log(B)=A$, but if $\log(B)$ is not defined I don't know how to work. $\endgroup$
    – Losten
    May 21 '19 at 12:17

Well, if a row of a matrix is a sum of two rows then the determinant is equal to the sum of the corresponding determinants. So $\det(M + tN)$ can be represented as $2^k$ summands: $$ \det(M + tN)=\sum_{m=0}^k t^m\sum \det A_{i_1,\ldots,i_m}^{i_{m+1},\ldots,i_{k}} $$ where $A_{i_1,\ldots,i_m}^{i_{m+1},\ldots,i_{k}}\;$ is a matrix which has rows $i_1,\ldots,i_m$ of the matrix $N$ and the rows $i_{m+1},\ldots,i_{k}$ from the matrix $M$.


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