Suppose $y=Ax+v$, where $x\in {\mathbf R}^n$ is some set of parameters you wish to estimate, $y\in {\mathbf R}^m$ is a set of measurements, and $v$ represents a noise. We assume $m>n$, and $A$ is full rank. Consider an estimator of the form $\hat x=By$.
Choosing $B$ to be any left inverse of $A$ yields $\hat x =x$, no matter what $x$ is, provided $v=0$.