On which conditions we have $$AA^{-1}B=B$$ A, B are nonnegative definite symmetric $n\times n-$matrices, and $A^{-1}$ is the Moore-Penrose inverse of A.
2026-03-27 04:00:02.1774584002
The inverse in the sense of MOORE-PENROSAE $AA^{-1}B=B$
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Let me write $A^+$ for the Moore-Penrose inverse.
Then $AA^+$ is the orthogonal projector onto the range of $A$.
Therefore $AA^+B=B$ iff the range of $B$ is contained in the range of $A$.