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Let $A\in M_n(K)$ be a square matrix over a field $K$. The notion of inverse matrix was generalized by Moore and Penrose for real and complex matrices (also called pseudo-inverse $A^{\dagger}$ of $A$, satisfying $AA^{\dagger}A=A, A^{\dagger}AA^{\dagger}=A^{\dagger}$ and $AA^{\dagger}$ Hermitian). This was again generalized to arbitrary fields with involutory automorphism of $K$ by Pearl, but the existence depends on additional rank conditions.

Question: What is known on the existence of a generalized inverse $A'$, which just satisfies $AA'A=A$, over an arbitrary field ? Do we also need an additional assumption, perhaps on the characteristic of $K$, or again some rank condition ?

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The condition $AA'A=A$ means that the semigroup of matrices is regular. This is really true: see Clifford - Preston, The Algebraic Theory of Semigroups, sec.2.2, ex. 6(g).

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  • $\begingroup$ Thank you, Boris. Do you have some other reference where this is proved ? (I have no access to the book). $\endgroup$ Jun 26, 2013 at 17:11
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    $\begingroup$ @ Dietrich Burde: Have you access to Semigroups and Combinatorial Applications by Gérard Lallement? If yes, then see in it Proposition 4.2. $\endgroup$ Jun 26, 2013 at 17:44
  • $\begingroup$ Yes, thank you. And the proof is really not difficult. $\endgroup$ Jun 28, 2013 at 20:27
  • $\begingroup$ @ Dietrich Burde: OK. Good luck! $\endgroup$ Jun 28, 2013 at 20:30
  • $\begingroup$ Adding on to @BorisNovikov's answer: Proposition 4.2 in Chapter 2 (p. 39). (There are also Proposition 4.2s in Chapters 6 and 8.) $\endgroup$ Sep 6, 2022 at 4:32

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