Solving for vector contained in a diagonal matrix Consider the following system of equations:
$$\mathbf{y}D_{\mathbf{x}}+\mathbf{z}D_{\mathbf{y}}=\mathbf{u}D_{\mathbf{z}}$$
where $\mathbf{x}$, $\mathbf{y}$, $\mathbf{z}$, and $\mathbf{u}$ are $1\times n$ vectors and $D_{\mathbf{x}}$, $D_{\mathbf{y}}$, and $D_{\mathbf{z}}$ are diagonal $n\times n$ matrices with $\mathbf{x}$, $\mathbf{y}$, and $\mathbf{z}$, respectively, along their diagonals (i.e., $D_{\mathbf{x}} = \mathrm{diag}(\mathbf{x})$).
My question is whether it is possible to solve for $\mathbf{y}$ here – both the $\mathbf{y}$ as a vector and the $\mathbf{y}$ along the diagonal of $D_{\mathbf{y}}$. The problem is that I do not know of operations to pull $\mathbf{y}$ out of $D_{\mathbf{y}}$. The one possibility that I have considered is to use the Hadamard product since this can be used to convert a vector into a diagonal matrix and vice versa, as discussed here. But I am not sure how this would work in this case, as it would be necessary to distribute $\mathbf{y}$ out of the resulting expression, and I don't know if this would be possible given the properties of the Hadamard product. That is, we could write
$$(\mathbf{y}e^T) \odot I_n = D_{\mathbf{y}}$$
where $\odot$ is the Hadamard product and $e^T = (1,1,\ldots)\in\mathbb R^n$. So, would it be possible to distribute out $\mathbf{y}$ and then move the remaining terms to the other side of the first equation above? For example, we couldn't do something like left-hand side since that would just be equal to $\mathbf{y}$, not $D_{\mathbf{y}}$:
$$\mathbf{y}(e^T \odot I_n) \neq D_{\mathbf{y}}$$
Edit: Oh, it seems that extracting $\mathbf{y}$ in this case would be a simple as rewriting the equation above as
$$\mathbf{y}D_{\mathbf{x}}+\mathbf{y}D_{\mathbf{z}}=\mathbf{u}D_{\mathbf{z}}$$
because rewriting the equation this way would not change the terms along the diagonal of $\mathbf{z}D_{\mathbf{y}}$. Then we can write
$$\mathbf{y}=\mathbf{u}D_{\mathbf{z}}D_{\mathbf{x+z}}^{-1}$$
But then how would solve for $\mathbf{y}$ in the following?
$$\mathbf{y}D_{\mathbf{x}}+\mathbf{y}D_{\mathbf{y}}=\mathbf{u}D_{\mathbf{z}}$$
I think that in this case, it would not be possible to solve for a single vector $\mathbf{y}$ as in the previous case. Instead, we would have a system of polynomials:
$$\mathbf{y^2} + \mathbf{y}D_{\mathbf{x}} = \mathbf{u}D_{\mathbf{z}}$$
where $\mathbf{y^2}$ is a vector where the elements are the squares of the corresponding elements of $\mathbf{y}$ – that is, $\mathbf{y^2} = \begin{pmatrix} y_{1}^2 & y_{2}^2 & \cdots & y_{n}^2 \end{pmatrix}$.
Next, what if we had an equation as follows?
$$\mathbf{y}D_{\mathbf{x}}+\mathbf{y}D_{\mathbf{yM}}=\mathbf{u}D_{\mathbf{z}}$$
where $\mathbf{M}$ is an $n \times n$ matrix. Unlike the other matrices, it is not a diagonal matrix. Thus, in each element along the diagonal of $D_{\mathbf{yM}}$, we have some linear combination.
I think in this case, the polynomial system of equations would be rather complicated. We would have the vector $\mathbf{y^2}$ again, but this time multiplied by some diagonal matrix based on the elements of $\mathbf{M}$ and $D_{\mathbf{x}}$ that is multiplied by a scalar ($n$, I believe). Then, we would have a set of vectors ($n-1$, I believe) that each have as elements different products of the elements in $\mathbf{y}$ (e.g., $y_1 y_3$) and are each multiplied by a scalar and a diagonal matrix composed of permutations of the elements in $\mathbf{M}$ and $D_{\mathbf{x}}$.
So, I have two questions here:

*

*Is my general intuition about what this equation would look like correct?

*Are there techniques to solve for the elements of $\mathbf{y}$ in this system?

 A: I am not sure what the notation $D_\mathbf{x}$, $D_\mathbf{y}$, and $D_\mathbf{z}$ represent and if the importance is significant, but assuming $D_{\mathbf{x}}$ is invertible, you may solve for $\mathbf{y}$ as follows given the original equation:
$$ \mathbf{y}D_{\mathbf{x}}+\mathbf{z}D_{\mathbf{y}}=\mathbf{u}D_{\mathbf{z}} $$
$$ \mathbf{y}D_{\mathbf{x}}=\mathbf{u}D_{\mathbf{z}}-\mathbf{z}D_{\mathbf{y}} $$
$$ \mathbf{y}=\bigr( \mathbf{u}D_{\mathbf{z}}-\mathbf{z}D_{\mathbf{y}} \bigr) D_{\mathbf{x}}^{-1} $$
Note: I can't verify that your edit is correct because I don't understand what $D_{\mathbf{x} + \mathbf{z}}$ represents.
Edit 1:
Thank you for the response as I didn't realize the meaning of the matrices $D_{\mathbf{x}}$, $D_{\mathbf{y}}$, and $D_{\mathbf{z}}$. Yes, your edit is correct, but perhaps, I can provide how I would work the problem if that is of use to you.
Since $D_{\mathbf{x}} = \text{diag}(\mathbf{x})$ and similarly for the other matrices, we have
$$
\begin{bmatrix} y_1 & \cdots & y_n \end{bmatrix}  \begin{bmatrix} x_1 &  &   \\  & \ddots &  \\  &  & x_n\end{bmatrix} + \begin{bmatrix} z_1 & \cdots & z_n \end{bmatrix}  \begin{bmatrix} y_1 &  &   \\  & \ddots &  \\  &  & y_n\end{bmatrix} = \mathbf{u}D_{\mathbf{z}}
$$
and multiplying through we have
$$
\begin{align}
\begin{bmatrix} y_1 x_1 & \cdots & y_n x_n \end{bmatrix} + \begin{bmatrix} y_1 z_1 & \cdots & y_n z_n \end{bmatrix} &= \mathbf{u}D_{\mathbf{z}}\\
\begin{bmatrix} y_1 x_1 + y_1 z_1 & \cdots & y_n x_n + y_n z_n \end{bmatrix} &= \mathbf{u}D_{\mathbf{z}} \\
\begin{bmatrix} y_1 (x_1 + z_1) & \cdots & y_n(x_n + z_n) \end{bmatrix} &= \mathbf{u}D_{\mathbf{z}} \\
\end{align}
$$
Therefore, this can be written in matrix form as
$$
\begin{bmatrix} y_1 & \cdots & y_n \end{bmatrix}  \begin{bmatrix} x_1 + z_1 &  &   \\  & \ddots &  \\  &  & x_n + z_n \end{bmatrix} = \mathbf{u}D_{\mathbf{z}}
$$
or more concisely as
$$ \mathbf{y} (D_{\mathbf{x}} + D_{\mathbf{z}} ) = \mathbf{u} D_{\mathbf{z}} $$
which is exactly what is given in your edit:
$$ \mathbf{y} = \mathbf{u} D_{\mathbf{z}} (D_{\mathbf{x}} + D_{\mathbf{z}} ) ^{-1}$$
Edit 2:
As for the case where you have $\mathbf{y}^2 + \mathbf{y} D_{\mathbf{x}} = \mathbf{u} D_{\mathbf{z}}$ where $\mathbf{y}^2 = \begin{bmatrix} y_1^2 & \cdots & y_n^2 \end{bmatrix}$, you would not be able to solve for $\mathbf{y}$ as far as I can tell... To see this, multiply together the matrices, which would give the following result (skipping intermediate steps):
$$
\begin{bmatrix} x_1 & \cdots & x_n \end{bmatrix}  \begin{bmatrix} y_1^2 + y_1 &  &   \\  & \ddots &  \\  &  & y_n^2 + y_n \end{bmatrix} = \mathbf{u}D_{\mathbf{z}}
$$
where we cannot solve for the matrix containing the $y$ variables because we cannot eliminate $\mathbf{x}$ from the left side of the equation.
