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 $t$?
Determinant and power series
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calculus
linear-algebra
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0Also, Newton's identities ["in reverse"](http://en.wikipedia.org/wiki/Newton's_identities#Application_to_the_characteristic_polynomial_of_a_matrix) 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. – 2011-09-08
2 Answers
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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 $M$.