# Determinant and power series

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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How about $$\det(I-tA)=\exp\left(-\sum_{n=1}^\infty\frac{t^n}{n}\mathrm{tr}(A^n)\right)?$$ –  anon Sep 8 '11 at 9:16
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. –  anon Sep 8 '11 at 9:26
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## 2 Answers

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}

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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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