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Suppose that $A$ is $n$-square matrix such that $t_r:=$ tr$(A^r), r=1, 2, \cdots, n$ are given real numbers. How shall we compute $\det(A)$ in terms of $t_r$s?

I am completely unable to do this. Please help me. Thanks in advance

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    $\begingroup$ Well the eigenvalues of $A^k$ are the $k^{th}$ power of the eigenvalues of $A$. Furthermore, the trace of matrix is the sum of its eigenvalues and the determinant of a matrix is the product of its eigenvalues. From these facts you can derive the result you want. $\endgroup$ – Calculon Sep 8 '14 at 11:20
  • $\begingroup$ @L'universo Isn't there any other simpler solution for it ? $\endgroup$ – Anjan3 Sep 8 '14 at 11:23
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You need to use Newton's identities. These express various elementary symmetric polynomials in terms of power sums.

The elementary symmetric polynomials $s_1,s_2,...,s_n$ in $n$ variables $x_1$,...,$x_n$ are

$\begin{align*} s_1 &= x_1 + x_2 + ... + x_n\\ s_2 &= \sum_{1\leq i<j\leq n} x_i x_j\\ s_3 &= \sum_{1\leq i<j<k \leq n} x_i x_j x_k\\ \vdots\\ s_n &= x_1.x_2. ... x_n.\end{align*}$

If $t_r = x_1^r + ... +x_n^r$, then (half of) Newton's identities are

$\begin{align*} s_1 &= t_1\\ 2s_2 &= s_1t_1 - t_2\\ 3s_3 &= s_2t_1 - s_1t_2 + t_3\\ 4s_4 &= s_3t_1 - s_2t_2 + s_1 t_3 -t_4\\ \vdots\\ ns_n &= s_{n-1}t_1 - s_{n-2}t_2 + ... + (-1)^{n-1} t_n\\ \end{align*}$

Now if the complex eigenvalues of your matrix $A$ are $x_1,..,x_n$, then $t_r = Tr(A^r)$ and $det(A) = x_1.....x_n = s_n$. So to compute $det(A)$ you need to use all $n$ of Newton's identities to find $s_i$ in order as $i$ goes from $1$ to $n$.

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For any matrix $B$, considering the second-to-leading term of its characteristic polynomial shows that $\mathrm tr B$ is the sum of its eigenvalues. On the other hand, diagonalizing and taking $r$th powers shows that if the eigenvalues of $A$ are (counting multiplicity)$\lambda_1, \ldots, \lambda_n$, then the eigenvalues of $A^r$ are $\lambda_1^k, \ldots, \lambda_n^r$. So, your $t_r$ is simply the sum $\sum \lambda_a^r$ of the $i$th powers of the eigenvalues of $A$.

Now, Newton's Identities relate the $t_n$ to the symmetric polynomials $$s_r := \sum_{1 \leq a_1 < \cdots < a_r \leq n} \lambda^{a_1} \cdots \lambda^{a_r}$$ in the $\lambda_a$'s (by convention we take the empty summation to give $s_0 = 1$); considering the constant term of the characteristic polynomial of $A$ gives that the determinant is just the product of all of the eigenvalues $\lambda_a$ of $A$, which by definition is just $s_n$. . The identities are usually given inductively to save room: $$r s_r = \sum_{i = 1}^r (-1)^{i - 1} s_{r-i} t_i.$$ This can be substituted to produce explicit formulas for $\det A$ in terms of the $t_r$ alone, but probably this is prohibitively messy for larger $n$.

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Something easy to implement with a computer algebra system: $\det(A)$ is the coefficient of $t^n$ in the Taylor series of $$ \exp(\ \mathrm{tr}(A)t - \frac{\mathrm{tr}(A^2)}{2} t^2 + \ldots + (-1)^{n-1} \frac{\mathrm{tr}(A^n)}{n} t^n\ )$$ at $t=0$.

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  • $\begingroup$ Can you give some explanation for how this works? $\endgroup$ – Travis Willse Sep 8 '14 at 13:35
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    $\begingroup$ Use $\log ( 1 + \lambda t) = \sum_{k \ge 1} (-1)^{k-1} \frac{\lambda^k}{k} t^k$ so $\log \prod_{\lambda} ( 1 + \lambda t)= \sum_{k \ge 1} (-1)^{k-1} \frac{\sum_{\lambda}\lambda^k}{k} t^k$. Now exponentiate and look at the coefficients of $t^n$ on both sides. Also note that on the RHS the coefficients of $t^i$, $0\le i \le n$ do not change if we truncate to order $n$, that is, we use the finite sum $\sum_{k =1}^n (-1)^{k-1} \frac{\sum_{\lambda}\lambda^k}{k}$ $\endgroup$ – Orest Bucicovschi Sep 8 '14 at 14:07

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