Let $A$ be an $n \times n$ matrix. Are there formulas that convert linear combinations of traces of powers of $A$ to determinant of $A$ and vice versa from linear combinations of determinants of powers of $A$ to trace of $A$? (I am mainly looking for the latter  conversion of determinants to trace of $A$).
closed as too localized by Angelo, Andreas Blass, quid, Gjergji Zaimi, Harry Gindi Feb 11 '12 at 19:51This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question. 


You can find the determinant of a matrix $A$ of size $n$ in terms of the traces of $A^m$, for $m = 1, \ldots, n$. The trace of $A^m$ is the sum of the $m$th powers of the eigenvalues of $A$, and you can express the elementary symmetric polynomials (so in particular the product of the eigenvalues, which is the determinant) in terms of the power sums, see for example here. The converse is not possible. For example, the determinants of all powers of $\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ and of $\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ are 1, but the traces are different. 


There is a nice formula, called the PlemeljSmithies formula that allows the computation of the determinant of certain operators in terms of the moments (Tr(A^n)) of A. I'm including this as an answer because the formula organizes the information nicely for all finite dimensions, and holds also in infinite dimensions. For the formula, see section 1.2 of the paper here. This is a paper by Gohberg, Goldberg and Krupnik that eventually they extended into a very nice book. Later on, I'll try to come back and type in the formula... 

