# 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. –  Guess who it is. Jan 3 '12 at 4:31
Also this one. –  Guess who it is. 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

You could note that the matrix similarity \begin{align} \pmatrix{I & \mathbf{0} \\ X & -I}\pmatrix{\mathbf{0} & I \\ -C & -B} & \pmatrix{I & \mathbf{0} \\ X & -I} & \\ =\pmatrix{\mathbf{0} & I \\ C & X + B} & \pmatrix{I & \mathbf{0} \\ X & -I} & \\ =\pmatrix{ X & -I \\ X^2 + BX + C& -X - B} & \\ \end{align}

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

\begin{align} \pmatrix{G^{-1} & \mathbf{0} \\ G^{-1}M & I}\pmatrix{M & G \\ F & D} \pmatrix{G & \mathbf{0} \\ -G^{-1}MG & I} & \\ =\pmatrix{G^{-1}M & I \\ G^{-1}M^2 + F & G^{-1}MG + D} \pmatrix{G & \mathbf{0} \\ -G^{-1}MG & I} & \\ =\pmatrix{ \mathbf{0} & I \\ G^{-1}M^2G + FG - G^{-1}M^2G - DG^{-1}MG & D+G^{-1}MG} & \\ =\pmatrix{ \mathbf{0} & I \\ (F-DG^{-1}M)G & D+G^{-1}MG} & \\ =\pmatrix{0 & I \\ -C & -B} & \\ \end{align}

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.

Since diagonalization is a difficult problem without a closed form solution, it is no wonder you had such difficulty finding such a solution. Maybe this is why the equation is related to what is named the quadratic eigenvalue problem as J.M. noted.

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Thanks for the information. Currently, I am doing some studies, so I can't give this its proper due. However, I've given you a +1 and accept. :-) –  000 Jan 23 '13 at 22:37

Consider the Riccati equation (1) $XAX+XB+CX+D=0_n$ where $X\in M_n(\mathbb{C})$ is unknown and $A,B,C,D\in M_n(\mathbb{C})$ are generic given matrices. (that is, the entries of these matrices are independent commutative indeterminates OR, more simply, there are no algebraic relations with coefficients in $\mathbb{Q}$ linking the entries of $A,B,C,D$).
Then (1) can be reduced to an equation in the form (2) $X^2+AX+B=0_n$ where $A,B$ are generic matrices. This equation has exactly $\binom{2n}{n}$ solutions and solving (2) can be reduced to solve a polynomial equation $P(u)=0$ with $degree(P)=2n$ and Galois group $S_{2n}$.
Note that , in your equation (3) $AX^2+BX+C=0_n$, if $A$ is generic, then $A$ is invertible and (3) has in fact the form (2).