Consider a matrix $A \in \mathbb{R}^{n \times m}, n > m$ with independent columns and non-negative entries.
Consider the oblique pseudo-inverse of $A$, i.e. the matrix $A^\dagger_B = (B
^\top A)^{-1} B^\top$ for some $B \in \mathbb{R}^{n \times m}$ such that the inverse $(B^\top A)^{-1}$ exists.
Characterize the class of $B$s for which the matrix $A^\dagger_B$ has non-negative entries.
Edit:
There was a typo (now corrected) in the previous version of this post in the definition of $A^\dagger_B$. $A^\dagger_B$ fulfills the following Moore-Penrose conditions for any choice of $B$ such that the inverse $(B^\top A)^{-1}$ exists.
- (1) $A A^\dagger_B A = A$
- (2) $A^\dagger_B A A^\dagger_B = A^\dagger_B$
- (4) $A^\dagger_B A = I$ is symmetric
The condition
- (3) $AA^\dagger_B$ is symmetric
is only true for $B=A$, when $A^\dagger_B$ becomes the standard Moore-Penrose pseudoinverse.
I am aware of the following result for the special case of $B = A$, i.e. the Moore-Penrose pseudoinverse: given a non-negative $A$, $A^\dagger$ is non-negative if and only if rows of $A$ are (up to a permutation) composed of rank-1 blocks orthogonal to each other ("Nonnegative Matrices in the Mathematical Sciences", A. Berman, R. J. Plemmons, Theorem 5.2).
I am wondering whether the class of matrices having the desired property is richer in the oblique case.
Best Answer
It turns out the condition $A^\dagger_B A = I$ together with the non-negativity implies that you cannot have more than one non-zero entry in any column of $A^\dagger_B$. Hence the answer for the oblique case is the same as for the usual Moore-Penrose pesudo-inverse.