Let $A,B \in \mathbb{R}^{m\times n}$ be matrices such that $(A+B)X=B$ for some $X$, and $A\neq \mathbf 0$ then $B\in \mathcal{C}(A)$

Question

I came up with the following conjecture,

Let $A,B \in \mathbb{R}^{m\times n}$ be matrices such that $(A+B)X=B$ for some $X$, and $A\neq \mathbf 0$ then $B\in \mathcal{C}(A)$

where $B \in \mathcal{C}(A) $ is slight notation abuse, meaning that if $B = [ b_1, b_2 , .. , b_n] $ then $B \in \mathcal{C}(A) $ means $ b_i \in \mathcal{C}(A)$ for $i = 1..n$

Attempt at proof:

Suppose $A,B \in \mathbb{R}^{m\times n}$ are matrices in $(A+B)X=B$, and $A\neq \mathbf0 $, then we can write $AX = B(I-X)$. Assuming $(I-X)$ is invertible, we have $A X(I-X) = B$, thus $\exists P $ such that $AP=B$ showing that $ B \in \mathcal{C}(A)$. $\square$

The problem with the proof is that I assume $(I-X)$ to be invertible. However I believe to it has to be invertible otherwise $(A+B)X=B$ doesn't hold. How to be rigorous about this in the proof?

Answer 1

If that is true for $X=\begin{bmatrix} 1&0&\cdots&0\\0&0&\cdots&0\\\vdots&\vdots&\ddots&\vdots\\0&0&\cdots&0 \end{bmatrix}$, then your equality states that all the columns of $B$ are $0$ except the first one, and the first column of $A$ is $0$. It can be that the first column of $B$ is not in $\mathcal C(A)$, by setting the rest of the columns of $A$ to anything that do not span the first column of $B$.

So your statement is in general true if $I-X$ is invertible, otherwise it is not true in general.

I think if you modify the statement to have $\det(A)\neq 0$ instead it may work.