How to get determinant of $A$ in terms of tr$(A^k)$? 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
 A: 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 $n$.
A: 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 $n$.
A: 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 $t=0$. 
