(I tried asking that on math.stackexchange.com, 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 math.stackexchange.com:

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 math.stackexchange.com version of the question: http://math.stackexchange.com/questions/67282/is-there-a-way-to-solve-the-following-tensor-equation