Solve $AX^2 + BX + C = 0$

Question

How does one solve equations of the form $AX^2+BX+C=0$ where $A,B,C$ are square matrices and $X$ is a matrix to be solved for?

More generally how does one solve equations of the form $AX^2B+CXD+E=0$?

Even more generally how does one solve higher order equations of this form such as $AX^3B+CX^2D+EXF+G=0$

In all cases, I could simply express the entries of the matrix $X$ as variables $x_1,x_2,\dots$ and multiply out the matrices and derive equations corresponding to each entry of the matrices however this would result in (for the first and most simple case at least) a system of 4 quadratic equations in 4 variables which I have no experience with. Is this the best approach or can the fact that we are dealing with matrices help to simplify the problem?

Thanks

Answer 1

For (i): $AX^2+BX+C=0$; if $A$ is invertible, then $X^2+A^{-1}BX+A^{-1}C=0$, that is a particular case of the Riccati equation. To solve this type of equation (for every $n$), you can see my post in

Find solution to matrix equation

For (ii): $AX^2B+CXD+E=0$. If $A,B$ are invertible, then we can write the equation in the form $X^2+BXC+D=0$, that is a non-unilateral equation ($X$ is between $B,C$). Solving this equation is feasible for $n=2$ and is not for $n>2$ (except numerically).

See my paper https://arxiv.org/pdf/1304.2506.pdf

For (iii): $AX^3B+\cdots=0$. Unfeasible (except numerically)...

Answer 2

This is only a guess at best, but it might help.

Consider $\overrightarrow v$ to be an eigenvector of your solution matrix, $X$, with it's corresponding eigenvalue $\lambda$. Further assume $\overrightarrow v \ne \overrightarrow 0$. Hence, we post multiply this to our equation and get $$AXX\overrightarrow v + BX\overrightarrow v + C \overrightarrow v = \overrightarrow 0$$ $$\implies (\lambda^2A + B\lambda + C)\overrightarrow v = \overrightarrow 0$$ $$\implies \lambda^2A + \lambda B + C = 0$$ Here, $\lambda$ is a scalar. I'm guessing this is easier to solve. Once all solutions to the above are found, $X$ would be such that it's eigenvalue decomposition would give us these and only these $\lambda$