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

Cormac Gilbert

Cormac Gilbert

Answered question

2022-02-17

I am looking for general solution of the form B=XAX1 with the following constraints
X1=X
where X is unknown matrix and A,B,X, are 3×3 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=XAX1=XAXT=R(ItnTd)RT,
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:
MBM1=J=NAN1
and then X is obviously
X=N1M
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.

Answer & Explanation

benehmenshgf

benehmenshgf

Beginner2022-02-18Added 7 answers

Case 1. A,B are not orthogonally similar; no solutions.
Case 2. A,B are orthogonally similar; A=PBPT, where PO(n). Let C(A)={XMn(R);AX=XA}. Then, it is not difficult to show
Proposition {XO(n);A=XBXT}=(C(A)O(n))P.
It remains to obtain C(A)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 and C(A)={aI+bA+cA2;a,b,cR}. Generically, the eigenspaces of A are not orthogonal and, consequently, C(A)O(n)=±In and X=±P.
EDIT. A,B are orthogonally similar iff the couples (A,AT) and (B,BT) are similar. Then, practically, one studies the linear system
(S)AX=XB,ATX=XBT.
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=uX0. It remains to calculate u so that ||uX0||2=1.

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