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(I tried asking that on, but did not get a satisfying answer. I am trying here as well, in case someone here will have more insight. The question was eventually abandoned there. I personally think it is a challenging question.)

Let $C(y)$ be the following function:

$C(y) = A \operatorname{diag}(A^t y) A^{-1}$

where $A$ is an invertible $m \times m$ matrix and $y$ is an $m \times 1$ vector, everything is over the reals.

Let's assume I know the values of $C(y)$ for any given $y$ (I can basically calculate $C(y)$ for any $y$). Is there a way to identify the matrix $A$ (or the set of solutions satisfying the equality above)?

Few insights that I gained through over time and through

There should be some connection between $A$ and eigendecomposition. Basically, $C(y)$ can be diagonalized for any $y$. It might mean that $A$ consists of scaled eigenvectors of $C(y)$ in certain cases.

Any hints would be greatly appreciated.

Here is the version of the question:

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Two questions: (1) does $A^t$ mean the transpose of $A$, or the $t$'th power of $A$? (2) does 'diag($B$)' refer to the vector whose elements are the diagonal elements of $B$? – Daniel Mansfield Sep 28 '11 at 23:04
I should have clarified this: $A^t$ is indeed the transpose, and $\mathrm{diag}(x)$ is an $m \times m$ matrix with the vector $x$ on the diagonal and 0 everywhere else. – dotproduct Sep 28 '11 at 23:07
thanks Douglas, good idea. I added the link. – dotproduct Sep 28 '11 at 23:11

Maybe I do not understand the question, but since the vector of eigenvalues of $C(y)$ is $A^t y$, computing the spectral decomposition for $y = e_1, \dotsc, e_m$ (the basis vectors) gives you $A.$ The problem is overdetermined, since you are also postulating that $A$ is the conjugating matrix.

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thanks! this makes sense to me, but if I understand correctly, doesn't your method imply there is a unique solution? I don't think there is a unique solution for $m=1$. I will try to think how to settle this issue. – dotproduct Sep 28 '11 at 23:05
@Igor > since the vector of eigenvalues of C(y) is Aty Isn't this true only if A is an orthogonal matrix? The question only assumes that it is invertible. I'm posting this as an answer because I'm not allowed to comment. – psd Sep 28 '11 at 23:09
I guess I am wrong and that A only needs to be invertible for this to be a vector of eigenvalues. – psd Sep 28 '11 at 23:14
oh, whoops, for $m=1$, $C(y) = ay$ not $a^2 y$. My mistake. – dotproduct Sep 28 '11 at 23:17
I guess I'm missing something: the spectral decomposition is not unique, you can permute the eigenvalues around. So without saying something more, you only get the rows of $A$ up to permutation. – euklid345 Sep 29 '11 at 3:10

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