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I am looking for general solution of the form $B = XAX^{-1}$ with the following constraints

$X^{-1}=X'$

where $X$ is unknown matrix and $A,\,B,\, X$, are $3\times 3$ matrices.

The problem appears peculiar and thus I wish to simplify it further to more practical problem X is rotation matrix ($R$), and $A$ is plane equation matrix, B is resulted homography in the sense of $B =XAX^{-1} = XAX^{T}= R(I-tn^T/d)R^{T}$ ,

where $t$ is translation vector, $n$ is plane normal vector, and $d$ is signed distance, all three are known

My dirty solution so far is as follows : Finding the Jordan Canonical form:

$MBM^{-1}=J=NAN^{-1}$,

and then $X$ is obviously

$X = N^{-1}M$

but this works only if the canonical forms are the same, otherwise cannot be utilized for solution.

The second dirty trick is to solve a system of Linear equations in terms of

$AX = BX$

but i cannot figure out how to program this in the software (e.g. Matlab linsolve). Could you help me to clarify how to write system of linear equations in the shape of 9 rows?

The trick is how to notate them on the "right side" to use in ready to use functions for solving SLE.

Thank, you

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  • $\begingroup$ well, let's simplify: X is always invertible, orthogonal and X' = X^{1} and is also 3x3 matrix $\endgroup$ – Misha Bolgarskiy Nov 14 '18 at 13:32
  • $\begingroup$ Are $A$ and $B$ symmetric then ? $\endgroup$ – nicomezi Nov 14 '18 at 13:32
  • $\begingroup$ This will not work. Take $A=0$ and $B=I$, the identity. $\endgroup$ – Dietrich Burde Nov 14 '18 at 13:33
  • $\begingroup$ No, A and B are not symmetric. For the symmetric case, there is already a valuable thread. Even I can generalize a solution of the kind MAM' = J = NBN' (finding the jordan canonical form) - if the canonical form is same, then there is solution, if not - no solution, then X = N^{-1}M $\endgroup$ – Misha Bolgarskiy Nov 14 '18 at 13:38
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Case 1. $A,B$ are not orthogonally similar; no solutions.

Case 2. $A,B$ are orthogonally similar; $A=PBP^T$, where $P\in O(n)$. Let $C(A)=\{X\in M_n(\mathbb{R});AX=XA\}$. Then, it is not difficult to show

$\textbf{Proposition}$ $\{X\in O(n);A=XBX^T\}=(C(A)\cap O(n))P$.

It remains to obtain $C(A)\cap O(n)$. When $n=3$, $C(A)$ is a vector space of dimension $3,5$ or $9$ (when $A$ is a scalar matrix).

Generically (randomly choose $A$), $A$ is cyclic

https://en.wikipedia.org/wiki/Cyclic_subspace

and $C(A)=\{aI+bA+cA^2;a,b,c\in \mathbb{R}\}$. Generically, the eigenspaces of $A$ are not orthogonal and, consequently, $C(A)\cap O(n)=\pm I_n$ and $X=\pm P$.

EDIT. $A,B$ are orthogonally similar iff the couples $(A,A^T)$ and $(B,B^T)$ are similar. Then, practically, one studies the linear system

$(S)$ $AX=XB,A^TX=XB^T$.

If $A,B$ are not orthog. similar, then the sole solution is $X=0$.

If $A,B$ are orthog. similar, then, for a generic $A$, $(S)$ admits a vector space of solutions of dimension $1$: $X=uX_0$. It remains to calculate $u$ so that $||uX_0||_2=1$.

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  • $\begingroup$ thank you, I need time to digest this.. $\endgroup$ – Misha Bolgarskiy Nov 14 '18 at 19:03
  • $\begingroup$ you are somehow, point towards similarity transformations..? $\endgroup$ – Misha Bolgarskiy Nov 14 '18 at 19:04
  • $\begingroup$ Maybe I could simplify further the problem, A is plane equation matrix in the shape of $i-pn/h$ and $X$ is a rotation matrix. $\endgroup$ – Misha Bolgarskiy Nov 15 '18 at 1:38
  • $\begingroup$ @Misha Bolgarskiy , If you want to change the problem, then open a new file. Anyway, I did the job; now it's your business. $\endgroup$ – loup blanc Nov 15 '18 at 11:08
  • $\begingroup$ I am really sorry Mr. Blanc, i am pretty newbie here $\endgroup$ – Misha Bolgarskiy Nov 15 '18 at 17:03

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