Solving matrix equation $XA=AY$ with known $X$ and $Y$

Question

I am having problem in solving set of matrix multiplication. There are three matrices $A,X$ and $Y$, all are non-singular $2\times 2$ matrices. Where matrix $X$ and $Y$ are known and $A$ is unknown. $$ X = \begin{bmatrix} x_{11} & x_{12} \\ x_{21} & x_{22} \end{bmatrix} $$ $$ Y = \begin{bmatrix} y_{11} & y_{12} \\ y_{21} & y_{22} \end{bmatrix} $$ $$ A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix} $$

and multiplication is as follows

$$ X\cdot A=A\cdot Y $$

I expanded it and tried to solve it as such I would be able to get elements of matrix $A$ at the end but it end up in homogenous linear equation having trivial solution zero. That makes all calculation meaningless.

$$ X\cdot A - A\cdot Y = 0 $$

What is the better way to compute it as such I can find result of matrix $A$ (in terms of elements of matrics $X$ and $Y$) at the end.

Answer 1

See The theory of matrices, chapter VIII.

Answer 2

In attempting to give you my complicated general formula to your question, a curious thing happened. All the zero terms made everything dissapear. This makes sense as $\mathbf{0}$ is obviously a solution. For a non-zero solution, $X$ and $Y$ must be similar matrices $$XA=AY \Rightarrow A^{-1}XA = Y$$

Mainly for reference purposes, and to illustrate the subtle complexity of the equation, here is the $2 \times 2$ general form of the equation along with the solution (I used my own variables as I did not want to translate what I have, and the zeroes are the terms I mentioned): \begin{array}{rcl} \pmatrix{e & f \\ g & h}\pmatrix{q & r \\ s & t}=\pmatrix{q & r \\ s & t}\pmatrix{a & b \\ c & d} \\ \Rightarrow \pmatrix{q & r \\ s & t} &= & \pmatrix{\hat{q}\over {\Delta} & \hat{r}\over {\Delta} \\ \hat{s}\over {\Delta} & \hat{t}\over {\Delta}} \\ \end{array}

with

$$\Delta = a^2\left(d^2 -de -dh + eh -fg \right) + a\left(-2bcd + bce + bch -d^2e -d^2h + de^2 + 2deh + dh^2 -e^2h + efg -eh^2 + fgh \right) + bc\left( bc + de + dh -e^2 -2fg -h^2 \right) + d\left(deh -dfg -e^2h + efg -eh^2 + fgh \right) + e^2h^2 -2efgh + f^2g^2 $$ $$\hat{q} = 0 $$ $$\hat{r} = 0 $$ $$\hat{s} = 0 $$ $$\hat{t} = 0 $$

The formula for $\Delta$ here is a test for equal eigenvalues - it is zero if $X=\pmatrix{e & f \\ g & h}$ and $Y=\pmatrix{a & b \\ c & d}$ share at least one eigenvalue(not necessarily all eigenvalues).

Therefore, if $\Delta$ is non-zero, there is not a solution to your equation.

tl;dr: If you attempt to solve the equation by looking at the individual terms and getting simultaneous equations, you will be re-inventing a wheel called the Kronecker product. It is the tool you want to learn for this equation.

Answer 3

I expanded it and tried to solve it as such I would be able to get elements of matrix A at the end but it end up in homogenous linear equation having trivial solution zero. That makes all calculation meaningless.

Why? That's a valid solution, if only trivial.

To investigate if there are additional non-trivial solutions, we see that $X A−A Y=0$ is a particular (homogeneus) case of the Sylvester equations. In the link it's explained that the equation can be rewriten as a linear (in our case homogeneus) with a matrix equation of size $n^2 \times n^2 $ ($4 \times 4$ ), which will have non-trivial solutions if its singular -which is equivalent to matrices X and Y having at least one common eigenvalue. Our solution, in vectorial form, is the full null space of that matrix.

One example in Octave/Matlab

>>> X=[1,2;0,3]  % triangular, eigenvalues: 1 , 3
X=
   1   2
   0   3
>>>  Z=[5,6,7,8]; % arbitrary
>>>    Y=Z*[3,5;0,4]*inv(Z)  % eigenvalues : 3, 4
Y =
   111.500   -77.500
   150.500  -104.500
>>>  D=kron(eye(2),X)-kron(Y',eye(2))  % kronecker products
D =
  -110.50000     2.00000  -150.50000     0.00000
     0.00000  -108.50000     0.00000  -150.50000
    77.50000     0.00000   105.50000     2.00000
     0.00000    77.50000     0.00000   107.50000
>>> a=null(D)   % D is singular, it has in this case a dim-1 null space
a =
   0.57359
   0.57359
  -0.41352
  -0.41352
>>> A=[x(1),x(3);x(2),x(4)];
A =
   0.57359  -0.41352
   0.57359  -0.41352
>>> X*A-A*Y  %  zero, save for rounding errors
ans =
  5.1958e-014  -2.4425e-014
  -2.1094e-014  1.9318e-014

Of course, $\alpha A$ is also solution