If normal linear regression finds $A$ such that $AX \sim Y$, then how do I solve $BAX \sim Y$?

Question

If normal linear regression finds $A$ such that $$AX \sim Y$$ then how should I solve $$BAX \sim Y$$ where $B$, $X$ and $Y$ are given (non-invertible) matrices? I could of course derive the solution by hand, but I need a numerically stable solution, and would rather make a call to a pre-written library function.

Answer 1

I recommend using a $QR$-decomposition of $B$ which is quite stable.

$B = Q_B R_B$ which results in:

$$AX \approxeq R_B^{-1} Q_B^t Y =: Z$$

Then you can use your prefered method for finding $A$. (I'd do it with a QR-decomposition of $X$ too.)

So just for convenience I rewrite this as

$$ X^t A^t = Z^t$$

Then we calculate the QR-decomposition of $X^t$, that is $Q_x R_x = X^t$ and we get

$$ A^t = R_x^{-1} Q_x^t Z^t = R_x^{-1} Q_x^t Y^t Q_B R_B^{-t}$$

or if you prefer

$$ A = R_B^{-1} Q_B^t Y Q_x R_x^{-t}$$