For non-square non-negative matrix $A$, under what conditions does a non-negative matrix $B$ exist such that $AB = I$? For non-negative matrix $A$ with dimensionality $m \times n$ where $m < n$, I am trying to find all possible non-negative matrices $B$ that are its pseudoinverses:
$AB = I$
We can find $B = A^T(AA^T)^{-1}$ that sometimes satisfies this condition. But is there only one unique $B$ that satisfies $AB = I$ for a given $A$? And is that $B$ non-negative? How do I go about rigorously approaching this question? How do I enforce the non-negativity constraint on $B$?
 A: You want the columns $b_1, \ldots, b_n$ of $B$ to satisfy
$$
AB = (Ab_1 \mid \cdots \mid Ab_n) = I.
$$
That's the same as looking for solutions to
$$
Ab_i = e_i. \tag{1}
$$
Assuming such a solution $b_i$ for $(1)$ exists, all solutions are given by $b_i+x$ with $Ax = 0$. Hence all possible matrices would be of the form
$$
B = (b_1+x_1\mid \cdots \mid b_n+x_n)
$$
where $Ab_i = e_i$ and $Ax_i = 0$ for all $i$.
By size considerations, you matrix $A$ is not injective (meaning there are non-zero $x$ for which $Ax = 0$). So, if you have such a matrix $B$, there will be infinitely many. And the existence can be phrased as: are $e_1,\ldots,e_n$ in the image of $A$? In turn, this means for $A$ to have full rank.
The coordinate free approach would be as follows: you matrix a defines a lineal map $f \colon \Bbbk^m \to \Bbbk^n$, you want $g$ such that $fg = \mathrm {id}$. This is the same as having $fg(e_i) = e_i$ for all $i$. Also, we can define $g$ on a basis, so the problem reduces to picking $y_i \in f^{-1}(e_i)$ for each $i$ and defining $g(e_i) := y_i$.
Of course, such an element need not be unique and the non-uniqueness is measured by $\ker f$, which by dimension considerations is a non trivial subspace.
On the other hand, existence is equivalent to having $\mathrm{im} f = \langle e_1, \ldots, e_n\rangle = \Bbbk^n$, i.e. to $f$ being surjective.
