# 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.
• It is not true for $\Gamma=I$. Since $Tr(A^k)=\sum\limits_{i=1}^n \lambda_{i}^k$, no additional information can be retrieved besides the eigenvalues (which are already given). Apr 30, 2019 at 8:45
• If $B$ is a nonsingular matrix that commutes with $\Gamma$, then $B^{-1}AB$ will have the same traces; assuming that $\Gamma$ has distinct eigenvalues, these $B$ are arbitrary diagonal matrices. It may still be possible to determine the $C_G(\Gamma)$-conjugacy class $A$ belongs to in $\mathbb{M}_n$ (though I haven't checked fully). Apr 30, 2019 at 8:45
• @user44191 Can you please elaborate about the conjugacy classes? Apr 30, 2019 at 8:57
• @JosiahPark Indeed $\Gamma=I$ is somewhat a special case. The problem may also be ill posed for $\Gamma$ in the vicinity of $I$. It is my hope, however, solutions may be found for some $\Gamma$. Apr 30, 2019 at 9:01
• @yarchik Allowing for multiple $\Gamma$'s allows for a positive answer. For instance for $n=3$ if one takes $\Gamma_{i}=e_{i}e_{i}^{T}$, $i=1,\dots,3$ one can recover $A$. Apr 30, 2019 at 9:02

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.

• Yes, this fully answers my question. Just a short clarification question, do you have any insight on what kind of 1-parameter freedom would that be? If we talk about the orthogonal matrices realizing a rotation in 3d space, would it be possible to say something like, the axis of rotation can be determined, but not the angle? Apr 30, 2019 at 14:13
• One may expect a 1-parametric family of solutions, however, that is because you allow them to be complex. $A$ is symmetric and, therefore, eigenvectors should be real. Imposing this additional condition may lead to a finite number of solutions in some cases. Do you agree with this? May 1, 2019 at 9:46
• @yarchik No. I did purely real analysis. Nothing was complex. The real orthogonal matrices are a 3D manifold and the 2 equations were real as well. May 1, 2019 at 11:18

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).

• $\Gamma=I_n$ is somewhat special. What if we do not assume from onset that gamma is the identity matrix? Apr 30, 2019 at 8:54
• In the general case, take as $M$ a (non-identical) orthogonal matrix commuting with $\Gamma$. Then $${}^tM(A^k \Gamma) M = B^k ({}^tM \Gamma M) = B^k \Gamma,$$ so $A^k \Gamma$ and $B^k \Gamma$ are similar and the same argument applies. Apr 30, 2019 at 9:06
• Can one always find a (non-identical) orthogonal matrix $M$ that commutes with $\Gamma$? For me it is not obvious, scaling and rotation typically do not commute. Apr 30, 2019 at 9:33
• Ok, let's say that if the centralizer of $\Gamma$ contains a (non-identical) orthogonal matrix, then the reconstruction of $A$ is not possible. I did not check whether this is always the case for all diagonal matrices. Apr 30, 2019 at 9:42
• @yarchik A diagonal matrix with only $\pm1$ entries is orthogonal and commutes with any diagonal matrix. For a diagonal matrix with only distinct entries, I suspect that these are the only orthogonal matrices that commute. Apr 30, 2019 at 12:03