TITLE: Series expansion of the determinant for a matrix near the identity
QUESTION [34 upvotes]: The problem is to find the second order term in the series expansion of the expression $\mathrm{det}( I + \epsilon A)$ as a power series in $\epsilon$ for a diagonalizable matrix $A$. Formally, we will write the series as follows
$$ \det( I + \epsilon A ) = f_0(A) + \epsilon f_1(A) + \epsilon^2 f_2(A) + \cdots +\epsilon^N f_N(A), $$
In particular, we are looking for an expression in terms of the trace of the matrix $A$.

REPLY [9 votes]: The full formula for the expansion of the determinant of an nxn matrix A in a polynomial of traces of powers of A is:
$$\det(A)=\sum_{\pi\in\Pi(n)}(-1)^{|\pi|-n}\prod_{m=1}^n\frac{(Tr(A^m))^{\pi_m}}{m^{\pi_m}\pi_m!}$$
where symbols are as follows:
$\Pi(n)$ is the set of partitions of $n$ in positive integers; its elements are sequences of non negative integer multipliers, but a finite number of them different from zero, such that $$1\cdot \pi_1+2\cdot\pi_2+...n\cdot \pi_n=n.$$ E.g. the set of partitions of 3 has 3 elements, namely $$\Pi(3)=\{(3,0,\ldots),(1,1,0,\ldots), (0,0,1,0,\ldots)\},$$ meaning that the number 3 can be written as a sum of distinct positive integers with multiplicities in these ways: $$3=1\cdot3, 3=1\cdot 1+2\cdot 1, 3=3\cdot 1.$$ $|\pi|=\sum_{m=1}^{\infty}\pi_m$ is the length of the partition $\pi$, (the total number of non negative integers to be summed) e.g.$|(3,0,\ldots)|=3$, $|(1,1,0,\ldots)|=2$, etc.
The formula can be derived, without assuming that A is diagonalizable, but of  course assuming characteristic 0, by expanding $\ln(\lambda+A)=\ln\lambda +\ln(1+\frac{A}{\lambda})$ in a Taylor series, which converges for $\lambda>||A||$, then applying $\det e^B=e^{Tr(B)}$, expanding again the double series which is actually a polynomial because $\det(\lambda+A)$ is, and collecting terms of order 0 in $\lambda$.
I believe there are other ways (i.e. combinatorial, using Newton's identities or something of the kind) to derive it, but I have not seen an explicit proof.
The formula given in previous answer for a 2x2 matrix follows using the partitions of 2 $$\Pi(2)=\{(2,0,\ldots),(0,1,0,\ldots)\}$$
to compute
$$\det(A)=(-1)^{2-2}\frac{(Tr(A))^2}{1^2\cdot 2!}+(-1)^{1-2}\frac{Tr(A^2)}{2^1\cdot 1!}.$$
As another example, using the partitions of 3 given above, one would find for a 3x3 matrix:
$$\det(A)=(-1)^{3-3}\frac{(Tr(A))^3}{1^3\cdot 3!}+(-1)^{2-3}\frac{Tr(A)\cdot Tr(A^2)}{1^1\cdot1!\cdot 2^1\cdot 1!}+(-1)^{1-3}\frac{Tr(A^3)}{3^1\cdot1!}$$
or
$$\det(A)=\frac{1}{6}\left((Tr(A)^3-3Tr(A^2)\cdot Tr(A)+2Tr(A^3)\right).$$
For a 4x4 matrix one ends up with
$$\det(A)=\frac{(Tr(A))^4}{24}-\frac{(Tr(A))^2\cdot Tr(A^2)}{4}+\frac{(Tr(A^2))^2}{8}+\frac{Tr(A)\cdot Tr(A^3)}{3}-\frac{Tr(A^4)}{4}$$
and so on.
EDIT
I'm editing my post, because I had forgotten to mention, that by the same token one can obtain a general formula for all the coefficients of the characteristic polynomial of A, not just the determinant. For an nxn matrix A, by defining
$$\Delta_A(\lambda)=\det(\lambda+A)=\sum_{r=0}^{n}P_r(A)\lambda^r$$
(notice the unusual sign though) and expanding as above, one finds
$$P_r(A)=\sum_{\pi\in\Pi(n-r)}(-1)^{|\pi|-n+r}\prod_{m=1}^{n-r}\frac{(Tr(A^m))^{\pi_m}}{m^{\pi_m}\pi_m!}$$
of which $\det(A)=P_0(A)$ is a particular case.