is there a way to solve the following tensor equation? 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.)
 A: 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 $A$.
A: 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:)
