TITLE: Is it possible to derive $B$ from $AB=C$ when $B$ is not invertible?
QUESTION [5 upvotes]: Let $A$ be an $m\times n$ matrix, $B$ be an $n\times k$ non-invertible matrix (where $n<k$) and $C=AB$. If we have $C$ and $A$, can we derive $B$?

REPLY [2 votes]: You are looking for a pseudo-inverse of $A$, $A^+$, since if it exists you could write:
$$C = AB \ \to \ A^+C = A^+ A B=B$$
If the rows of $A$ are independent, then the pseudo inverse is given by:
$$A^+ = A^*(A A^*)^{-1}$$ 
Where $A^*$ is the Hermitian transpose.

REPLY [2 votes]: Let $B_1$ and $B_2$ matrices such that $C=AB_1$ and $C=AB_2$. Then
$$
A(B_1-B_2)=0.
$$
Conversely, if $X$ is any $n\times k$ matrix such that $AX=0$ and $C=AB$, then
$$
A(B+X)=AB+AX=C+0=C.
$$
You can thus recover $B$ if and only if the unique solution of $AX=0$ is the zero matrix. This is true if and only if the unique solution of $Av=0$ ($v$ an $m$-row vector) is $v=0$, which is equivalent to $A$ having a left inverse, that is
$$
m\ge n=\operatorname{rank}(A)
$$