The system is undetermined and consistent and $x$ is a solution of the system.
This is a proof I need as part of a larger problem, but I'm kinda stuck. I don't know if I need to use projections or the idea is simpler and I'm missing something.
Basically, I need to show that $Ax^\perp=\textbf{0}$ is not true (except for $x^\perp=0$). Now, this is clear for $x$ because it is a solution and has to be $Ax=b$. But how can I use the orthogonal relationship between $x$ and $x^\perp$ to prove it for $x^\perp$?
This is false. For example consider a vector in $\mathbb{R}^2$ and $A$ the projection onto the first coordinate. Take any vector with a non-zero first coordinate and zeroes everywhere else. This vector will satisfy $Ax=b$ for some non-zero $b$ but every vector orthogonal to it is in the kernel of $A$.