Solving two quadratic matrix equations [closed]

Question

Given $10 \times 10$ matrices $A$ and $B$, I would like to find $10 \times 10$ matrix $X$ such that

$$A = X B X^T \tag{1}$$

$$B = X A X^T \tag{2}$$

How can I solve the issue? if there is a way to solve only equation (1) or (2) that is ok also.

If anyone can already solve this and show me the way it's fine too.

Matrix $A$:

   [[0.125+0.03125i,0,0,0,0,-0.0625,-0.0625,-0.03125,0,0],      
    [0,0.0625,0,0,0,0,-0.0625,0,0,0],
    [0,0,0.0625,0,0,0,0,-0.0625,0,0],
    [0,0,0,0.15625,0,0,-0.03125i,0,-0.0625,-0.0625],
    [0,0,0,0,0.0625,0,0,0,0,-0.0625],
    [-0.0625,0,0,0,0,0.0625,0,0,0,0],
    [-0.0625,-0.0625,0,-0.03125i,0,0,0.125+0.03125i,0,0,0],
    [-0.03125,0,-0.0625,0,0,0,0,0.15625,0,0],
    [0,0,0,-0.0625,0,0,0,0,0.0625,0],
    [0,0,0,-0.0625,-0.0625,0,0,0,0,0.125]]

Matrix $B$:

   [[0.15625,0,0,0,0,-0.0625,-0.0625,-0.03125i,0,0],        
    [0,0.0625,0,0,0,0,-0.0625,0,0,0],
    [0,0,0.0625,0,0,0,0,-0.0625,0,0],
    [0,0,0,0.125+0.03125i,0,0,-0.03125,0,-0.0625,-0.0625],
    [0,0,0,0,0.0625,0,0,0,0,-0.0625],
    [-0.0625,0,0,0,0,0.0625,0,0,0,0],
    [-0.0625,-0.0625,0,-0.03125,0,0,0.125+0.03125i,0,0,0],
    [-0.03125i,0,-0.0625,0,0,0,0,0.09375,0,0],
    [0,0,0,-0.0625,0,0,0,0,0.0625,0],
    [0,0,0,-0.0625,-0.0625,0,0,0,0,0.125]]

Thanks!

Answer 1

I calculated the determinant of your two matrices:

I find $\det A=7.95808\cdot 10^{-13} \, i$ while $\det B=-4.54747\cdot 10^{-13} + 2.27374\cdot 10^{-13} i$. If your equations (1) and (2) hold, then $\det A=(\det C)^2\det B$ and $\det B=(\det C)^2\det A$, so $(\det A)/(\det B)=(\det B)/(\det A)$, which fails.

Hence there is no solution of both equations $A=XBX^{T}$ and $B=XAX^T$.

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Now for the solution of a single equation, say $A=XBX^T$. The matrix $X^T$ is the transpose of $X$, so $X_{nm}=X_{mn}$. Both matrices $A$ and $B$ are complex symmetric, $A_{nm}=A_{mn}$ and $B_{nm}=B_{mn}$. Both matrices are also invertible. Then each has a unique complex symmetric square root, $A^{1/2}$ and $B^{1/2}$, see this MO posting . A solution to $A=XBX^T$ is given by $$X=A^{1/2}B^{-1/2}.$$ The solution is not unique, any matrix $A^{1/2}OB^{-1/2}$ with $O$ real orthogonal (so $OO^T=1$) is also a solution.

I tried this for the matrices $A$ and $B$ listed above, with Mathematica. The output is very lengthy, you can download it from here and here is the evidence that it works as it should.

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_{The $10\times 10$ matrix in the OP is cumbersome, let me illustrate how this works for a $2\times 2$ matrix:}

$$a=\left( \begin{array}{cc} i & 1 \\ 1 & i \\ \end{array} \right),\;\;b=\left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \\ \end{array} \right),$$ $$a^{1/2}=\left( \begin{array}{cc} \frac{\sqrt{-1+i}}{2}+\frac{\sqrt{1+i}}{2} & \frac{\sqrt{1+i}}{2}-\frac{\sqrt{-1+i}}{2} \\ \frac{\sqrt{1+i}}{2}-\frac{\sqrt{-1+i}}{2} & \frac{\sqrt{-1+i}}{2}+\frac{\sqrt{1+i}}{2} \\ \end{array} \right),\;\;b^{1/2}=\left( \begin{array}{cc} \frac{1}{2}+\frac{i}{2} & \frac{1}{2}-\frac{i}{2} \\ \frac{1}{2}-\frac{i}{2} & \frac{1}{2}+\frac{i}{2} \\ \end{array} \right),$$ $$x=a^{1/2}b^{-1/2}=\left( \begin{array}{cc} \sqrt{\frac{1}{2}+\frac{1}{\sqrt{2}}} & i \sqrt{\frac{1}{\sqrt{2}}-\frac{1}{2}} \\ i \sqrt{\frac{1}{\sqrt{2}}-\frac{1}{2}} & \sqrt{\frac{1}{2}+\frac{1}{\sqrt{2}}} \\ \end{array}\right),$$ _{and you can readily check that $xbx^T=a$. In this particular example the matrix $x$ is symmetric, but in general it is not.}