# How is FA=I in a linear unbiased estimator?

I am seeing the following problem statement:

$y = Ax+v$, and $\hat{x}=Fy$

• $y$ is measurmenet $y\in \mathbb{R}^m$
• $v$ is the error $v\in \mathbb{R}^m$
• $x$ is is the vector of variables to be estimated $x\in \mathbb{R}^n$
• $\hat{x}$ is our estimate for $x$
• $A$ is rank n
• $E(\hat{x})=x$

And it says somewhere in the problem statement, "since $E(\hat{x})=x$ then $FA=I$". This was not immediately obvious to me. So, I tried to prove that to myself using the information above:

$E(\hat{x}) = E(Fy) = E(F(Ax+v)) = x$

$E(FAx+Fv) = E(FAx+Fv)= E(FAx)+E(Fv)$

since $v$ is the error, and the estimator is unbiased, then $E(v)=0$

$E(FAx)+E(Fv) = E(FAx) = x$, so $FA=I$.

Is the above logic correct? Is it necessary that $A$ is full rank ( rank $n$ )? I have a feeling that I should be using that somewhere here and I'm not using it anywhere.