Determining that a vector is not a least squares solution.

Question

Can I conclude that, if $Au\cdot(b-Au)≠0$, then $u$ is not a least square solution?

When is it not true that $Ax\cdot(b-Ax)≠0$ but $x$ is a least squares solution?

Answer 1

Yes, you can indeed conclude that if $Au \cdot (b - Au) \neq 0$, then $u$ is not a least squares solution. Conversely, if $Au \neq 0$ and $Au \cdot (b - Au) = 0$, then $u$ is necessarily a least squares solution.