Decomposing a matrix into a product of two matrices and making the right-hand one irreversable Let X be any n $\times$ m matrix.
I want to decompose it like X=AB where A is n $\times$ p and B is p $\times$ m with p which is larger than both n and m.
I wondered

*

*Can we always decompose any X into A and B while we can set p freely.

*Is A inferable with X and B?

In my knowledge, B doesn't have right-hand pseudo inverse so inferring the original A from X and B is not possible. (no B' exists for $A=XB'$).

*

*Sure, you can always let $A = [X, 0_{n \times (p-n)}]$ and $B = [I_m, 0_{m \times (p-m)}]^T $, then $X = AB$ for any $p$ larger than $n$ and $m$. I guess you're not looking for such trivial solutions but then you need to tell us what $A$ or $B$ should look like.


*Not uniquely. For a given factorization $X = AB$ you can find another one $X = \bar{A} \bar{B}$ via $\bar{A} = A T$ and $\bar{B} = T^{-1} B$ for any invertible square $p$. Even if you know $B$, this does not give away $A$ since $B$ has a non-empty left null space, i.e. $y^T B = 0$ has non-trivial solutions in $y$ and hence we have $(A + A_0) + B = AB + A_0 B = X$ for any $A_0$ in the left null space of $X$. You can find one possible solution via the pseudo inverse $A_p = X \cdot B^+$, but it's not unique.
