TITLE: Determinant and power series
QUESTION [8 upvotes]: 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 $t$?

REPLY [10 votes]: To expand on anon's comment:
$$\begin{align}\det (I-A) &= \exp\log\det (I-A)  \\&=\exp\text{Tr}\log (I-A) \\ &=\exp\text{Tr}\left(-\sum_{n=1}^\infty\frac{A^n}n\right) \\&=\exp\left(-\sum_{n=1}^\infty\frac{\text{Tr}A^n}n\right)\;.\end{align}$$

REPLY [5 votes]: 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 $M$.