# solution of $B = XAX^{-1}$ in terms of $AX = BX$ system of linear equations

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

• well, let's simplify: X is always invertible, orthogonal and X' = X^{1} and is also 3x3 matrix – Misha Bolgarskiy Nov 14 '18 at 13:32
• Are $A$ and $B$ symmetric then ? – nicomezi Nov 14 '18 at 13:32
• This will not work. Take $A=0$ and $B=I$, the identity. – Dietrich Burde Nov 14 '18 at 13:33
• 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 – Misha Bolgarskiy Nov 14 '18 at 13:38

## 1 Answer

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 $$||uX_0||_2=1$$.

• thank you, I need time to digest this.. – Misha Bolgarskiy Nov 14 '18 at 19:03
• you are somehow, point towards similarity transformations..? – Misha Bolgarskiy Nov 14 '18 at 19:04
• 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. – Misha Bolgarskiy Nov 15 '18 at 1:38
• @Misha Bolgarskiy , If you want to change the problem, then open a new file. Anyway, I did the job; now it's your business. – loup blanc Nov 15 '18 at 11:08
• I am really sorry Mr. Blanc, i am pretty newbie here – Misha Bolgarskiy Nov 15 '18 at 17:03