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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 $A$).

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What do you mean by "convert"? Obviously there can be no function such that $f(Tr(A))=Det(A)$ or $g(Det(A))=Tr(A)$ for all matrices $A$, since one can prescribe values for both trace and determinant simultaniously. – Florian Eisele Feb 11 2012 at 13:55
But is there an explicit $f_{n}$ and $g_{n}$ for each $n \in \mathbb{N}$? – unknown (google) Feb 11 2012 at 14:04
Of course there can't be such an $f_n$, and if you had actually understood what Florian told you would see it. I voted to close. – Angelo Feb 11 2012 at 14:09
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 I guess the OP was searching for $f( Tr(A), Tr(A^2), \dots ) =Det(A)$ – Marc Palm Feb 11 2012 at 14:26
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 $\mathop{\rm Det} (A^i) = (\mathop{\rm Det} A)^i$. – Angelo Feb 11 2012 at 15:12
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closed as too localized by Angelo, Andreas Blass, quid, Gjergji Zaimi, Harry Gindi Feb 11 2012 at 19:51

2 Answers

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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.

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@MichaelStoll Thankyou. Is this a particular example. By this I mean if we mod out some factors can we get a formula for Trace as well? – unknown (google) Feb 11 2012 at 14:34
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 To be a bit more precise: you have $\det(A^m) = \det(A)^m$, so the determinants of powers of $A$ won't give you more information than the determinant of $A$ itself. But knowing the product of the $n$ eigenvalues (and nothing more) will never tell you what their sum is (unless $n = 1$, of course). – Michael Stoll Feb 11 2012 at 14:38
@MichaelStoll Thankyou. – unknown (google) Feb 11 2012 at 14:46
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There is a nice formula, called the Plemelj-Smithies 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...

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Man, slid that one under the wire! :-) – Todd Trimble Feb 11 2012 at 19:58
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 Wow! No kidding! :) – Jon Bannon Feb 11 2012 at 20:03

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