# Is there a unique solution for this quadratic matrix equation?

The quadratic matrix equation I've been looking at lately:

$$Q_{r,r}=A_{r,r}X_{r,r}^2+B_{r,r}X_{r,r}+C_{r,r}=0_{r,r}$$

Note that $A, B, C,$ and $X$ are $r \times r$ matrices. $A$ contains known elements, $B$ contains known elements, $C$ contains known elements, and $X$ contains the unknown elements that you are solving for. $0_{r,r}$ is just the $r \times r$ null matrix.

Is there any solution for $X$ in terms of $A, B,$ and $C$ (making no easy assumptions)? (e.g. $X$ is a diagonal matrix, $A=B=C$, or anything of that sort.)

I have tried to solve this and nothing has worked out. I attempted solving it generally by manipulating the matrices in variable form (i.e. actually writing out the matrices $A, B, C,$ and $X$ in variables) and finding a unique solution for all of the elements of $X$ in terms of the elements of $A, B,$ and $C$. That didn't work out beyond the case of $r=1$.

Trying to solve it by looking at $r$ at different values did not work out either; I ended up with very abysmal equations at just $r=2$. I don't know exactly how to make this appealing to the denizens of math.stackexchange, but it (as far as I know) isn't a heavily studied problem.

There is a very high possibility that I've just been doing elementary techniques and nothing of note, so I hope someone or a group of people could shed light on this.

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Your problem is finding the solvent $\mathbf X$ of the quadratic matrix equation $\mathbf A\mathbf X^2+\mathbf B\mathbf X+\mathbf C=\mathbf 0$. This is related to the more common quadratic eigenvalue problem $(\lambda^2\mathbf M+\lambda\mathbf C+\mathbf K)\mathbf x=\mathbf 0$. See this, this, this, and this for more information. –  Ｊ. Ｍ. Jan 3 '12 at 4:31
Also this one. –  Ｊ. Ｍ. Jan 3 '12 at 8:39
@Landscape, would you please have some response in meta.math.stackexchange.com/questions/9602, because I know you have anit-retagged many questions. –  doraemonpaul May 19 '13 at 23:38

gives your equation in $X$, and if the equation is solved, then the matrix $\pmatrix{0 & I \\ -C & -B}$ is block diagonalized. Also note that there is a closed form solution to bring (almost) any matrix into the form $\pmatrix{0 & I \\ -C & -B}$ through a similarity transform. I do not know what this form is called in the literature, but I like to call it the block companion form. Here is how to do it
If a solution to $X^2 + BX + C = \mathbf{0}$ were possible in a closed form here, then it could split the eigenvalue problem in half, be applied recursively and thus have a closed form solution to the eigenvalue problem. The existence of such a solution has already been dis-proven by the Abel-Ruffini Theorem.