Reconstruct a matrix from its traces In my research I came across the following problem. Let $A$ be a symmetric  and $\Gamma$ be a diagonal $n\times n$ matrices. The eigenvalues of $A$ are known $\lambda_1,\ldots\lambda_n$. The traces $\mathrm{Tr}(A^k\,\Gamma)=t_k$, $k\in\mathbb{N}$ are also known, $\Gamma$ is given. Can $A$ be found based on this information? If yes, how? While I am curious about the general $n$ case, information on $n=3$ would be most valuable.  Any reference would be greatly appreciated.
P.S. 


*

*From the answer of Francesco Polizzi it is known that the reconstruction is not possible when there is an orthogonal matrix $M$ that commutes with $\Gamma$.  Fortunately, in the case of interest, this situation can be excluded. Specifically, it is known that diagonal entries of $\Gamma$ are all distinct and positive. By this answer, $\Gamma$ then commutes only with diagonal matrices.

*user44191 suggested that $B=M^{-1}AM$ ($M$-nonsingular) will have the same traces as $A$. However, apart from a trivial $\pm1$ possibility, which can be excluded by a proper sign convention, the new matrix $B$ is no longer symmetric. Thus, it leaves the question open. 

 A: I do not think you can reconstruct $A$ just from this information. 
Take $\Gamma=I_n$, let $M$ be any orthogonal $n \times n$ matrix and set $B={}^tMA M$.
Then $A$ and $B$ are two similar (symmetric) matrices, and so all their positive powers $A^k$ and $B^k$ have the same eigenvalues (and in particular the same trace). 
A: Unfortunately even the generic situation is bad. Since we know the eigenvalues, we should search for the orthonormal system of eigenvectors $v_i$ of $A$. We have ($e_i$ is the standard basis)
$$
Tr(A^k\Gamma)=\sum_i\left[\sum_j\gamma_j \langle v_i,e_j\rangle^2\right]\lambda_i^k
$$
so, in effect, you have the knowledge of $\sum_j\gamma_j \langle v_i,e_j\rangle^2$ ($i=1,2,3$). However, these three values are not independent: since the matrix $(\langle v_i,e_j\rangle^2)_{ij}$ is bistochastic, their sum is just $Tr \Gamma$, so you have only $2$ independent equations, while the orthogonal matrices form a 3D manifold. Thus in the generic case you should expect a continuous 1-parametric family of solutions.
