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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.
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    $\begingroup$ 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). $\endgroup$
    – Josiah Park
    Commented Apr 30, 2019 at 8:45
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    $\begingroup$ 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). $\endgroup$
    – user44191
    Commented Apr 30, 2019 at 8:45
  • $\begingroup$ @user44191 Can you please elaborate about the conjugacy classes? $\endgroup$
    – yarchik
    Commented Apr 30, 2019 at 8:57
  • $\begingroup$ @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 $\Gamma$. $\endgroup$
    – yarchik
    Commented Apr 30, 2019 at 9:01
  • $\begingroup$ @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 $A$. $\endgroup$
    – Josiah Park
    Commented Apr 30, 2019 at 9:02

2 Answers 2

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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.

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    $\begingroup$ 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? $\endgroup$
    – yarchik
    Commented Apr 30, 2019 at 14:13
  • $\begingroup$ 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? $\endgroup$
    – yarchik
    Commented May 1, 2019 at 9:46
  • $\begingroup$ @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. $\endgroup$
    – fedja
    Commented May 1, 2019 at 11:18
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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).

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  • $\begingroup$ $\Gamma=I_n$ is somewhat special. What if we do not assume from onset that gamma is the identity matrix? $\endgroup$
    – yarchik
    Commented Apr 30, 2019 at 8:54
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    $\begingroup$ 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. $\endgroup$
    – Francesco Polizzi
    Commented Apr 30, 2019 at 9:06
  • $\begingroup$ 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. $\endgroup$
    – yarchik
    Commented Apr 30, 2019 at 9:33
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    $\begingroup$ 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. $\endgroup$
    – Francesco Polizzi
    Commented Apr 30, 2019 at 9:42
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    $\begingroup$ @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. $\endgroup$
    – M. Winter
    Commented Apr 30, 2019 at 12:03

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