I came across this question. Let $B\in\mathbb{C}^{m\times n}$, $A\in\mathbb{C}^{m\times n}$, $c\in\mathbb{C}^{m\times 1}$, $d\in\mathbb{C}^{1\times n}$ and $G\in\mathbb{C}^{n\times n}$. I am trying to decompose $B$ into some expression of $A$: $$B = (A + cd)G.$$ where $G$ is non-singular. $A$ and $B$ are both of full column rank.
Does anyone have any idea of determining $c$, $d$ and $G$? I do believe they exist. Thank you!
We have that $B=(A+cd)G$ implies $B-AG=cdG$. Note that by your conditions on $c$ and $d$, $\mathrm{rank}(cd)\leqslant 1$, so $\mathrm{rank}(cdG)= \mathrm{rank}(cd)\leqslant 1$ because $G$ is regular. This means that $\mathrm{rank}(B-AG)\leqslant 1$. In fact, the converse is also true, if $\mathrm{rank}(B-AG)\leq 1$, you can write $B-AG=cd'$ for appropriate column $c$ and row $d'$, and then take $d=d'G^{-1}$. So, the question is if there exists a regular matrix $G$ such that $\mathrm{rank}(B-AG)\leqslant 1$.
And it doesn't have to exist. For a counterexample take: $$B=\begin{pmatrix}1&0\\0&1\\0&0\\0&0\end{pmatrix}\ \ \mbox{ and }\ \ A=\begin{pmatrix}1&0\\0&1\\1&0\\0&1\end{pmatrix}.$$ Note that both $A$ and $B$ have full column rank, but for any regular $G=\begin{pmatrix}\alpha&\beta\\\gamma&\delta\end{pmatrix}$: $$B-AG=\begin{pmatrix}1-\alpha&-\beta\\-\gamma&1-\delta\\-\alpha&-\beta\\-\gamma&-\delta\end{pmatrix},$$ and this matrix is of rank $2$ because the determinant of the bottom $2\times 2$ part is non-zero since $G$ is regular. So, for any regular $G$, $A-BG$ cannot be written as a product of a column and a row.