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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$\begingroup$ 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. $\endgroup$– Florian EiseleCommented Feb 11, 2012 at 13:55
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$\begingroup$ But is there an explicit $f_{n}$ and $g_{n}$ for each $n \in \mathbb{N}$? $\endgroup$– TurboCommented Feb 11, 2012 at 14:04
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1$\begingroup$ 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. $\endgroup$– AngeloCommented Feb 11, 2012 at 14:09
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3$\begingroup$ I guess the OP was searching for $f( Tr(A), Tr(A^2), \dots ) =Det(A)$ $\endgroup$– Marc PalmCommented Feb 11, 2012 at 14:26
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2$\begingroup$ $\mathop{\rm Det} (A^i) = (\mathop{\rm Det} A)^i$. $\endgroup$– AngeloCommented Feb 11, 2012 at 15:12
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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.
answered Feb 11, 2012 at 14:22
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$\begingroup$ @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? $\endgroup$– TurboCommented Feb 11, 2012 at 14:34
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2$\begingroup$ 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). $\endgroup$ Commented Feb 11, 2012 at 14:38
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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
Gohberg, Goldberg and Krupnik Traces and Determinants of Linear Operators, Integral Equations and Operator Theory 1996, Volume 26, Issue 2, pp 136-187.
This paper was eventually extended into a very nice book.
answered Feb 11, 2012 at 18:38
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$\begingroup$ Man, slid that one under the wire! :-) $\endgroup$ Commented Feb 11, 2012 at 19:58
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$\begingroup$ That link currently requires a password—is it freely available somewhere? $\endgroup$– LSpiceCommented Mar 11, 2015 at 1:02
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1$\begingroup$ @LSpice: It seems not (that's annoying). I included a reference to the paper, but it seems just getting the book via ILL is probably the best bet. $\endgroup$ Commented Mar 11, 2015 at 12:23