Identifiability of arbitrary matrix after matrix multiplication

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Let $B \in \mathbb{R}^{n \times m}$ be an arbitrary real matrix and let $A \in \mathbb{R}^{n \times n}$ and $C \in \mathbb{R}^{m \times m}$ be invertible square matrices. Consider the matrices $AB$, $BC$ and $ABC$.

First question: It is clear that $B = A^{-1}ABC\,C^{-1}$. Is it possible to proof that $B$ cannot be "recovered" from $ABC$, if one does not know $A$ and $C$? (I'm aware that there is no clear notion of "recoverability", but I assume you have an intuition for what I mean. Formal definitions are also appreciated!)

Second question: Is there a way of "recovering" B, if one knows $ABC$ as well as $AB$ and $BC$? For example, one could try to calculate $C = (AB)^{-} ABC$ (where $X^{-}$ is the Moore-Penrose pseudoinverse of $X$) and then $B = BC \, C^{-1}$ (or the other way around). Under what conditions does this work?

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First question: Yes, you can prove that by a simple counterexample.

Let $B \in \mathbb{R}^{n \times m}$ be arbitrary, $A = I_n$ and $C = I_m$. Then, $ABC = B$. Now let $B' = 2B$, $A' = \frac{1}{2}I_n$ and $C' = I_m$. Then, $A'B'C' = B = ABC$. So, just from the knowledge of $ABC$ alone, you could never recover an arbitrary $B$ as you can't distinguish between the 2 cases.

Second question: Your procedure for recovering $B$ does indeed work, with the condition that the $AB$ has a left-inverse (note that if you can find $C$, you don't need any additional conditions for your last equation, since it's already assumed that $C$ is invertible).

The question then becomes: when does $AB \in \mathbb{R}^{n \times m}$ have a left-inverse? You need $AB$ to have full column rank (linearly independent columns). Since $A$ is square and full rank, $\text{rank}(AB) = \text{rank}(B)$. In other words, your procedure works when $B$ has full column rank. Note that this implies $n \geq m$.

Of course, for $n \leq m$, you could do the other way around as you pointed out, using the right inverse of $BC$. In that case, the condition would be that $B$ has full row rank.

So, in summary, you could recover $B$ as long as $B$ is full rank (either by row or columns).