# Least Squares and Preconditioners

I've seen the preconditioned least squares objective function $$\arg\min_{x}\Vert P^{-1}Ax-P^{-1}b\Vert_2$$, where $$P^{-1}$$ is positive definite. However, I'm not sure how we show that this is the equivalent to the original $$\arg\min_x \Vert Ax-b\Vert_2$$. My attempt is that if $$\sigma$$ is the smallest eigenvalue of $$P^{-1}$$, then \begin{align} \sigma\Vert Ax-b\Vert_2 \leq \Vert P^{-1}Ax-P^{-1}b\Vert_2 \leq \Vert P^{-1}\Vert \Vert Ax-b\Vert_2 \end{align} and noting that $$\arg\min_x \sigma \Vert Ax-b\Vert_2=\arg\min_x \Vert P^{-1}\Vert \Vert Ax-b\Vert_2$$, we have \begin{align} \arg\min_x \Vert Ax-b\Vert&=\arg\min_{x}\Vert P^{-1}Ax-P^{-1}b\Vert_2 \end{align} but I'm not sure that this argument is correct.

This problem is not equivalent to the original one, essentially now you are minimizing with respect to another inner product, the one given by $$\langle x , y\rangle := xP^{-T}P^{-1}y$$. Up to an orthogonal transformation this is a weighted least squares for some weights $$w_1,\dots, w_n$$, hence this is not equivalent to the original minimization (no weights).