Solving matrix equation with kronecker product I'm trying to get an analytical expression for $\mathbf{X}$, a n-by-1 vector, where
$\mathbf{X}=\mathbf{A}\mathbf{X}+\mathbf{B}(\mathbf{X}\otimes \mathbf{X})+\mathbf{C}$
and $\mathbf{A}$,$\mathbf{B}$ and $\mathbf{C}$ are known. I'm guessing that it'd have two answers like a quadratic...but can't figure out how to show it.
 A: The expression has to be understood under the form of a LHS and a RHS  both  $n \times 1$ vectors. I will treat here the case $n=2$.
In order to somehow simplify this expression, I will set $D=A-I$.
Thus we have equation: 
$$DX+B (X\otimes X) +C=0$$
Clearly, the only possible interpretation is with matrices having the following dimensions:
$$\pmatrix{a&c\\b&d}\pmatrix{x\\y}+\pmatrix{e&g&i&k\\f&h&j&l}\pmatrix{x^2\\xy\\xy\\y^2}+\pmatrix{m\\n}=\pmatrix{0\\0}$$
Otherwise said, a solution $(x,y)$ is such that:
$$\tag{1}\begin{cases}ax+cy+ex^2+(g+i)xy+ky^2+m&=&0\\bx+dy+fx^2+(h+j)xy+ly^2+n&=&0\end{cases}$$
thus belongs to the intersection of 2 conical curves.
Thus, the number of intersection points may be :


*

*finite ; in this case it can be $0, 1, 2, 3$ or $4$ (the maximum possible)

*or infinite, a case that may arise if these conical curves are degenerated i.e., represent the union of two straight lines.
Two examples for the extreme cases of the finite case ; we can give values to $a, b, c \cdots n$ in such a way that (1) becomes:


*

*$\begin{cases}x^2+y^2-1&=&0\\x^2+y^2-4&=&0\end{cases}$ 


(in this case, no solution to this intersection of 2 concentric circles), or


*

*$\begin{cases}4x^2+y^2-5&=&0\\x^2+4y^2-5&=&0\end{cases}$ 


(in this case, 4 solutions $(x,y)=(1,1),(1,-1),(-1,1),(-1,-1)$ to this intersection of 2 ellipses).
Remark: In the case $n=3$, the components $(x,y,z)$ of $X$ will be interpreted as intersection points of 3 quadric surfaces, for example a paraboloid and 2 ellipsoids...
