# Is there any general reference on matrix quadratic equations?

I am studying a problem where a quadratic matrix equation emerges. The equation is as follow (all capital letters are n by n matrices)

$$(I-X^{\prime}L)X=D$$

where $$L$$ and $$D$$ are both symmetric and positive definite. How much can I say about a solution $$X$$?

• Assuming you're using $'$ for transpose, it's symmetric. – Robert Israel Mar 1 at 16:11
• Yes, $\prime$ for transpose. Could there be any sufficient and necessary conditions that guarantee the existence of a solution? – yc0000 Mar 2 at 7:22
• Did you try the case of $2\times 2$-matrices? And why do you think the question has an easy to formulate answer? – user6976 Mar 3 at 3:47
• I don't presume this would be easy but wonder if there could be any results about the existence of a solution. I am not familiar with matrix equations. In the case of 2 by 2, this would be a system of quadratic equations and may be reduced to a quartic equation. – yc0000 Mar 4 at 9:03
• You write in your answer that "in the original equation $X$ must be symmetric", but this is not stated in your question, and on the contrary you used $X'$ there which suggests that there is a difference between the two. Is $X$ symmetric or not? If the answer is yes, then this is equation is well studied (algebraic Riccati equation; see here). – Federico Poloni Mar 9 at 11:43

A good resource is : Abou-Kandil, Hisham, Gerhard Freiling, Vlad Ionescu, and Gerhard Jank. Matrix Riccati equations in control and systems theory. Birkhäuser, 2012.

I figure out something and would like to share and check whether it is correct.

It is perfectly analogous to real quadratic equations.

Since $$L$$ is symmetric and positive definite, they can be decomposed:

$$L=U_L^{\prime} \Lambda_L U_L$$ where $$U_L$$ is orthonormal, and $$\Lambda_L$$ is diagonal with positive entries.

Let $$Q=U_L^{\prime} \Lambda_L^{-1/2} U_L$$, which is the inverse of the square root of $$L$$, and it is symmetric.

Then the above equation is equivalent to

$$\tilde{X}^{\prime} \tilde{X}-Q\tilde{X}=D$$ where $$\tilde{X}=Q^{-1}X$$

Suppose, in addition, $$D$$ and $$L$$ are simultaneously diagonalizable, I conjecture that $$\tilde{X}$$ is symmetric. Therefore, the above is equivalent to

$$(\tilde{X}-\frac{1}{2}Q)^{\prime}(\tilde{X}-\frac{1}{2}Q)=\frac{1}{4}L^{-1}-D$$

Note that the right-hand side is symmetric; hence, there exists a real solution of $$\tilde{X}$$ if and only if the right-hand side is positive semi-definite.

And after a bit of algebra, if solutions exist, it would be

$$X=\frac{1}{2}L^{-1}[I \pm (I-4LD)^{1/2}]$$ where $$(I-4LD)^{1/2}$$ is the real p.s.d square root of $$(I-4LD)$$, which exists because $$(I-4LD)^{1/2}$$ is p.s.d. and symmetric.

Hence, there are at most two solutions.

Are all of these arguments correct?

• I don't see why $X$ symmetric implies $\tilde{X}$ symmetric. A product of symmetric matrices is not symmetric. – Federico Poloni Mar 9 at 12:35
• You are right, I just notice that and add some assumption. @Federico Poloni – yc0000 Mar 9 at 13:14
• $X$ is symmetric because $X=D+X^{\prime}LX$ where the right-hand side is symmetric given that $L$ and $D$ is symmetric. – yc0000 Mar 9 at 13:17
• I see, thanks. Then I suggest you to take a look at the question I linked in the comment above. That should solve your problem. – Federico Poloni Mar 9 at 13:18
• Yes, I definitely will. Thank you. @Federico Poloni – yc0000 Mar 9 at 13:24