# when is $B^{-1} = A(BA)^+$?

For $C$ a matrix, let $C^+$ be the pseudo inverse (Moore-Penrose inverse).

Let $B$ a square matrix ($n \times n$) with an inverse, and $A$ some matrix ($n \times m$). Under what conditions is it true that

$B^{-1} = A(BA)^+$ ?

(maybe some conditions on the rank of $A$ or something similar?)

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$B^{-1}=A(BA)^+$ if and only if $BA(BA)^+=I$. Hence the necessary and sufficient condition is $\textrm{rank}(BA)=n\le m$, i.e. $A$ is a full-rank "wide" matrix.