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I have three known, non-singular matrices $\mathbf{A} \in \mathbb{C}^{3\times 3}$, $\mathbf{B} \in \mathbb{C}^{3\times 3}$, and $\mathbf{C} \in \mathbb{R}^{3\times 3}$ and one unknown matrix $\mathbf{X} \in \mathbb{C}^{3\times 3}$ (also non-singular). They are related to each other by the following two equations:

$\mathbf{A} = (\mathbf{C} + \mathbf{X})^H(\mathbf{C} + \mathbf{X})$

$\mathbf{B} = (\mathbf{C} + \mathbf{X})(\mathbf{C} + \mathbf{X})^H$

I am hoping to solve for $\mathbf{X}$. Obviously, in general there are infinite solutions. (E.g. if $\mathbf{A} = \mathbf{B} = \mathbf{I}$ then any $\mathbf{X}$ that makes $(\mathbf{C}+\mathbf{X})$ unitary will work.) So I will add the following constraints: $\mathbf{A}$ and $\mathbf{B}$ are not diagonal matrices. Also, $\mathbf{X}$ is not a hermitian matrix, and each of its elements is non-zero. (If it is necessary, it would also be possible for me to manipulate the situation to ensure a symmetric $\mathbf{C}$, but I am hoping that I do not have to do this.)

So, with these added constraints, is there a unique solution for $\mathbf{X}$? If the solution is not unique, are there a finite number of solutions? In either case, is there a closed form description of the solution(s)?

Any insight into this situation would be helpful. If possible, please formulate answers for non-mathematicians such as myself.

Much thanks.

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Sorry there was a typo. Fixed now.

NOTE: I think I have $A$ and $B$ reversed. Too much work to redo the math. I solved $$ A = (C+X)(C+X)^H \\ B = (C+X)^H (C+X) $$

Look at $$ A (C+X) - (C+X)B$$ This is just a linear equation in $X$ $$ AX - XB = CB-AC $$ and can be solved in any number of ways (provided $A$ and $B$ do not share any eigenvalues).

Here is a worked out example. I am using real matrices for simplicity

Let $$A=\pmatrix{-7&5&-5\cr -10&-4&9\cr -10&9&-6\cr },~~ B=\pmatrix{7&-4&-1\cr -1&-7&9\cr -4&-4&6\cr },~~ C=\pmatrix{3&-10&1\cr 2&-1&-1\cr 8&-4&5\cr }$$ Let $$X=\pmatrix{{\it x_{11}}&{\it x_{12}}&{\it x_{13}}\cr {\it x_{21}}& {\it x_{22}}&{\it x_{23}}\cr {\it x_{31}}&{\it x_{32}}&{\it x_{33}} \cr }$$ Then $$ A(C+X)-(C+X)B = \pmatrix{-\left(5\,{\it x_{31}}-5\,{\it x_{21}}-4\,{\it x_{13}}- {\it x_{12}}+14\,{\it x_{11}}+78\right)&{\cdot}&{\cdot}\cr 9\, {\it x_{31}}+4\,{\it x_{23}}+{\it x_{22}}-11\,{\it x_{21}}-10\, {\it x_{11}}+15&{\cdot}&{\cdot}\cr 4\,{\it x_{33}}+ {\it x_{32}}-13\,{\it x_{31}}+9\,{\it x_{21}}-10\,{\it x_{11}}-100& {\cdot}&{\cdot}\cr } = \pmatrix{0&0&0\cr 0&0&0\cr 0&0&0\cr }$$

I am only showing the first column These are just linear equations can be readily solved to get $$X=\pmatrix{-3&10&-1\cr -2&1&1\cr -8&4&-5\cr }$$

This shows that if there is a solution, this has to be it. Substituting in the original equation, we see that $X$ does not solve the problem. So the original problem has no solution. I am trying to find the necessary condition for a solution to exist but can't find a simple characterization yet.

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A worked out example would be very helpful! – benpro Jan 7 '14 at 20:36
See my updated answer. – user44197 Jan 7 '14 at 20:36
Also note that I had interchanged the role of $A$ and $B$. Too much work to redo the problem :) – user44197 Jan 7 '14 at 20:43
Much thanks. A couple of notes: Based on the construction of the problem, isn't it given that A and B will have the same Eigen values? Also, in my situation I am guaranteed that a solution exists. (Finding it is where I'm having trouble ;) ) (a 2x2 example would have been fine, but thanks for the extra effort). – benpro Jan 7 '14 at 20:53

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