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If $A$ is a matrix of rank $r$, is its Moore-Penrose pseudoinverse also of rank $r$?

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closed as off-topic by M. Vinay, MathOverview, Ramiro, Leucippus, 6005 Jun 16 '16 at 18:58

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Yes. This is a direct consequence of singular value decomposition. More specifically, let $A=USV^\ast$ be a SVD, where $U$ and $V$ are unitary and $S=\operatorname{diag}(s_1,s_2,\ldots,s_r,0,\ldots,0)$ with $s_1,\ldots,s_r>0$. Then $A^+=VS^+U^\ast$, where $S^+=\operatorname{diag}(\frac1{s_1},\frac1{s_2},\ldots,\frac1{s_r},0,\ldots,0)$.

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