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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.

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    $\begingroup$ Maybe this can help you math.stackexchange.com/questions/214401/…. ( I think this is not a duplicate because the matrix is not square). $\endgroup$ – Alex Silva Jan 12 '15 at 17:06
  • $\begingroup$ @Alex Silva. Unfortunately, the square case does not generalize (at least naively). Indeed in the square case $B$ is irrelevant and can be cancelled. $\endgroup$ – ziutek Jan 12 '15 at 17:11
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    $\begingroup$ Out of curiosity, pseudoinverses normally satisfy some of the four Penrose conditions. Which ones are valid for $A^{\dagger}_B$? $\endgroup$ – Algebraic Pavel Jan 12 '15 at 18:57
  • $\begingroup$ @AlgebraicPavel Thanks for pointing this out - there was a typo in the question, which I've now corrected. $\endgroup$ – ziutek Jan 13 '15 at 12:03
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    $\begingroup$ Now it makes sense, thanks. Actually the comment of @AlexSilva seems to be relevant now since you have $A^{\dagger}_BA=I$, which could lead to a "rectangular generalization" of the linked question with "inverse" replaced by "left inverse". $\endgroup$ – Algebraic Pavel Jan 13 '15 at 12:21
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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.

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