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I have the following tensor (takes a vector of length $m$ and returns a matrix $m \times m$):

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

for some invertible matrix $A$ of size $m \times m$ ($y$ is of size $m \times 1$).

Let's say I have $C(y)$ (a way to compute it, etc.) for any $y$ I want - is there a way to solve for $A$? (i.e. identify what $A$ is.)

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    (in any case, I suspect that there is no single solution for this. for example, if we consider $m=1$, then we have $C(y) = a^2 y$. In this case, it seems like there is no single solution! if $a = c$ is$a$solution, then so is $a = -c$. I am not sure how it generalizes to arbitrary $m$.)2011-09-25

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Posting my comment: I assume that both $y$ and $C(y)$ are givens. If we decompose $C=Q\Lambda Q^{-1}$ (with $Q$ normalized) and convert $\Lambda$ to the vector $\lambda$ then we have $\lambda=A^Ty$. Note that the matrix in the eigendecomposition doesn't have to be normalized so we have $A=Q\mathrm{diag}(v)$ for some vector of scaling factors $v$. Then $\lambda=\mathrm{diag}(v)Q^Ty$, hence $v$ can be solved for componentwise by dividing the components of $\lambda$ by those of $Q^Ty$, thus in effect solving for $A$.

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    Right, you have to pick a reasonable $y$. Picking $y$ with distinct non-zero entries should work. You can see it as $C(y)$ have eigenvalues $A^T y$ and the eigenvectors being the columns of $A$ - but when we get an eigenvalue decomposition of $C(y)$ the eigenvectors we find will be mutliples of the columns of $A$, hence $A = Q \operatorname{diag}(v)$.2012-02-23
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I think you may need Hadamard product.

Here is an incomplete answer.

Denote the $z\in\mathbb{R}^m$ with $z_i=[C(y)]_{ii}$. Using Hadamard product, $C(y)=A\mathrm{diag}(A^Ty)A^{-1}$ can be rewritten as $z=[A\circ(A^{-1})^T]A^Ty$ Since $y$ and $z$ is known, you need solve $B=[A\circ(A^{-1})^T]A^T$ from $By=z$. $B$ can be solved as $B=B_1+B_0$ where $B_1=(y^Ty)^{-1}zy^T\in\mathbb{R}^{m\times m}$ and $B_0y=0$. You may need SVD to construct $B_0$. Obviously the solution to $B$ is not unique.

After obtaining $B$, $A$ might be able to be extracted from $B$. But I haven't figure out this step yet:)

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    You are correct. But I mean the solution using the above method is not unique. The solution to the original problem at least is contained in the infinite solutions to the above problem. I'm not sure if the above is a proper method because extracting $A$ from $B$ still is unsolved.2011-09-25