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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 $X$?

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  • $\begingroup$ Assuming you're using $'$ for transpose, it's symmetric. $\endgroup$ – Robert Israel Mar 1 at 16:11
  • $\begingroup$ Yes, $\prime$ for transpose. Could there be any sufficient and necessary conditions that guarantee the existence of a solution? $\endgroup$ – yc0000 Mar 2 at 7:22
  • $\begingroup$ Did you try the case of $2\times 2$-matrices? And why do you think the question has an easy to formulate answer? $\endgroup$ – user6976 Mar 3 at 3:47
  • $\begingroup$ 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. $\endgroup$ – yc0000 Mar 4 at 9:03
  • $\begingroup$ 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). $\endgroup$ – Federico Poloni Mar 9 at 11:43
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A good resource is : Abou-Kandil, Hisham, Gerhard Freiling, Vlad Ionescu, and Gerhard Jank. Matrix Riccati equations in control and systems theory. Birkhäuser, 2012.

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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?

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  • $\begingroup$ I don't see why $X$ symmetric implies $\tilde{X}$ symmetric. A product of symmetric matrices is not symmetric. $\endgroup$ – Federico Poloni Mar 9 at 12:35
  • $\begingroup$ You are right, I just notice that and add some assumption. @Federico Poloni $\endgroup$ – yc0000 Mar 9 at 13:14
  • $\begingroup$ $X$ is symmetric because $X=D+X^{\prime}LX$ where the right-hand side is symmetric given that $L$ and $D$ is symmetric. $\endgroup$ – yc0000 Mar 9 at 13:17
  • $\begingroup$ 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. $\endgroup$ – Federico Poloni Mar 9 at 13:18
  • $\begingroup$ Yes, I definitely will. Thank you. @Federico Poloni $\endgroup$ – yc0000 Mar 9 at 13:24

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