How to solve for square matrix $y$ given $x = A'yA$ for $A'A = I$

Question

I am trying to find a solution to $x = A'yA$ for the square matrix $y\in \mathbb{R}^{M\times M}$, where $A\in \mathbb{R}^{M\times N}$ ,$A'A = I^{M\times M}$ is the identity matrix and $x\in \mathbb{R}^{N\times N}$ is a square matrix. Can you please help me? I am not sure if it is possible to find an exact solution. Is there a numerical solver I can use (in MATLAB) to solve for the matrix $y$ given I know $A$ and $x$? Thank you for your help and/or suggestions.

Answer 1

You can, starting with $x=A^{-1}yA$, left multiply by $A$ and right multiply by $A^{-1}$ to get $$ y=AxA^{-1} $$

MATLAB has a function to invert a matrix and another to multiply two matrices, so givin $A$ and $x$ that equation lets you determine $y$.

Answer 2

By inspection $Y = AXA^T$ is a solution since

$$ A^TYA = (A^TA)X(A^TA) = IXI = X. $$

More generally, if $M > N$, then any matrix of the form $Y = AXA^T + Z$ is a solution, where $Z$ is an $M\times M$ matrix whose rows belong to the left nullspace of $A$.