Let $R$ be a fixed $S \times K$ matrix, $x$ is a $S \times 1$ column vector (of unknowns), and $q$ is a fixed $1 \times K$ row vector. Assume that $q$ satisfies the equation $q = x^TR$. Show that if $R$ has rank $S$, then $x$ is uniquely determined by $q$, that is, there is a unique solution to $q=x^TR$.
What I did was first rearrange the equation to $R^Tx=q^T$. Now consider the augmented matrix $[R^T | q^T]$. Since we are given that $R^Tx=q^T$ is consistent, it can not be the case that $rank(R^T) < rank([R^T | q^T])$ (because otherwise the system of equations would be inconsistent). So we must have $rank(R^T) = rank([R^T | q^T])$. Since we know that $rank(R^T) = rank(R) = S$, then we have $rank(R^T) = rank([R^T | q^T])=S$, so $R^Tx=q^T$ has a unique solution. Is this line of reasoning correct?
Since the rows of $R$ are indepenedent, $R^+$, the Moore Penrose inverse of $R$, satisfies $RR^+=I_S$.
Then $qR^+=x^T$...