Set up: Assuming we are working with real numbers, let
- $W$: $q\times K$ with $q<K$ and full row rank
- $c$: $q\times 1$ vector
- $F$: $K\times K$ matrix
- $\beta,d$: $K\times 1$ vectors
Claim: Given $W\beta= c$, if $F\beta=d$ then there is some $K\times q$ matrix $G$ such that $F=GW$ and $d=Gc$.
Attempt: I tried playing around with some $(K-q)\times K$ matrix $T$ such that $A\equiv\begin{pmatrix} W\\T\end{pmatrix}$ is nonsingular and $WT'=0$. I also looked at $(WW')^{-1}$ e.g. $$ d=F\beta\implies d=F(WW')^{-1}WW'\beta $$ but then I'm getting $W'\beta$ instead of $W\beta$.
Could someone help me please?
p.s. The reverse claim holds and is easily verifiable.
p.p.s. This appears without proof in Amemiya (1985). We can think of $\beta$ as the vector of interested parameters and $W\beta=c$ is a linear constraint imposed on these parameters.
Note that if $F=GW$ then $d=F\beta=GW\beta=Gc$ is automatic. So it remains to show that there exists $G$ such that $F=GW$. Now one always has $\def\rk{\operatorname{rk}}\rk GW\leq\rk W=q$, so if $\rk F>q$ there is no such$~G$. There is nothing in your hypotheses that prevents this, for instance one could take $F=I$ and $d=\beta$, in which case $\rk F=K>q$. Your claim is not true.