Consider the linear least squares problem:
$${\bf x_o}=\min_{\bf x}\|{\bf Mx-b}\|_2^2$$ can be solved by normal equations: $${\bf x_o} = ({\bf M}^T{\bf M})^{-1}{\bf M}^T{\bf b}$$
Assuming $\bf M$ has no zero eigenvalues, and we modify, to solve:
$${\bf x_o}=\min_{\bf x}\|{\bf M}^k({\bf Mx-b})\|_2^2, k \in \mathbb Z^+$$
We should be sure to not change the solution $\bf x$, right?
Or do I need some extra condition on $\bf M$ for this to be true?
Or is it mostly only theoretically true and can affect solution speed and accuracy, also?
I think the reason you got downvoted is that if $$\bf{M}^k\bf{M}\bf{x}$$ is defined then $\bf{M}$ must be a square matrix but then if $\bf{M}$ has no zero eigenvalues the problem has the unique minimum of $0$ when $$\bf{x_0}=\bf{M}^{-1}\bf{b}$$