Solve for $f_i$ in $A^Tf=A^TR_if_i$ where $R_i$ the unit matrix with a $0$ in the $i^{th}$ diagonal entry

Question

Let $m\times1$ vectors $\vec{f},\vec{f_i}$ with all entries of $ \vec{f}$ nonzero, and $m\times n$ matrix $\mathbf{A}$, with $n\leq m$. Define the $m\times m$ matrix $\mathbf{R_i}$ as the matrix with $1's$ in all of its diagonal entries but $(i,i)$ (unit matrix with $0$ on the $i^{th}$ element of the diagonal). Assuming $\vec{f},\mathbf{A}$ are the parameters, when is the equation $\mathbf{A}^T\vec{f}=\mathbf{A}^T\mathbf{R_i}\vec{f_i}$ solvable for $f_i$ and how do we derive the solution?

Answer 1

The problem is to determine if there is some $f_i$ such that $f- R_i f_i \in \ker A^T$. Without loss of generality we can assume $[f]_i = 1$, hence we can write $f = e_i + R_i f$.

The problem reduces to determining if there is some $f_i$ such that $e_i + R_i(f-f_i) \in \ker A^T$.

We note that $\{R_i(f-f_i) | f_i \text{ arbitrary} \} = \{e_i\}^\bot$, so we further reduce the problem to determining if there is some $w \bot e_i$ such that $A^T(e_i +w) = 0$.

This is equivalent to determining if the $i$th row of $A$ can be written as a linear combination of the other rows. One way of down this would be to apply the Gram Schmidt process to the other rows first and then see if the $i$th row lies in the span of the others.