Is there a pseudo inverse $X$ such that $ABX=A$?

Question

Question

The title pretty much sums it up. I need to find a matrix $X$ such that:

$A B X = A$,

with $A\in R^{n\times n}$, $\text{rank}(A)=n$, $B\in \mathbb{R}^{n\times m}$ given. The matrix $X$ should thus “cancel” the non-square $B$.

Does such a matrix $X\in\mathbb{R}^{m \times n}$ exist?

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NB:

I am aware of the Moore-Penrose pseudo inverse, but this only finds $B$ such that:

$ABA = A$.

Answer 1

If the rank of $B$ is less than $n$, then $X$ cannot exist, because the rank of $ABX$ is at most equal to the minimum rank of the factors.

Therefore …

we can assume that $\operatorname{rk}B=n$ and also $n\le m$, that is, $B$ has a right inverse $R$: $BR=I$. Thus $ABR=AI=A$ and $R$ is the requested $X$.

Alternatively, $\operatorname{rk}A=n$ means $A$ is invertible, so you equation is equivalent to

BX=I