If $a,b$ are elements of a unital algebra $A$, then is there a proposition that states $1-ab$ is Moore-Penrose invertible if and only if $1-ba$ is Moore-Penrose invertible? If yes, what is the Moore-Penrose inverse of $1-ba$? How can I prove it?
the Moore-Penrose inverse of $1-ba$ when $1-ab$ is Moore-Penrose invertible
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operator-theory
banach-algebras
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0You may want to [take a look at this](http://www.emis.de/journals/ELA/ela-articles/articles/vol22_pp92-111.pdf) – 2017-01-16
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0My guess would be that $1+b(1-ab)^\dagger a=(1+ba)^\dagger$, based on power series expansions of the geometric series. However, I am only able to prove that $(1-ba)\left(1+b(1-ab)^\dagger a\right)(1-ba)=1-ba$. The rest just got messy and I don't arrive anywhere. – 2017-01-16
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0I think you mean $1+b(1-ab)^\dagger a=(1-b a)^\dagger$. – 2017-01-16
1 Answers
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Suppose that $c$ is the Moore-Penrose pseudoinverse of $1-ab$. By definition, this means that $$(1-ab)c(1-ab)=1-ab\qquad (1),$$ together with three other identities of a similar flavor.
I claim that $1+bca$ is the Moore-Penrose pseudoinverse of $1-ba$. Indeed, each of the four identities satisfied by $c$ implies the analogous identity with $1+bca$ in place of $c$ and $1-ba$ in place of $1-ab$. For example, expanding (1) gives $$c-abc-cab+abcab=1-ab.$$ Multiplying on the left by $b$ and on the right by $a$ then gives that $$bca-babca-bcaba+babcaba=ba-baba\qquad (2).$$ Expand the following expression: $$ (1-ba)(1+bca)(1-ba)=1-2ba+baba + bca-babca-bcaba+babcaba. $$ By (2), this equals $1-2ba+baba+ba-baba$, which simplifies to $1-ba$, as desired. This is the first of the four equalities needed to verify that $1+bca$ is the Moore-Penrose pseudoinverse of $1-ba$. Similar calculations establish the other three.
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0How can I prove the second equality? Should I multiply on left by $b$ and on the right by$a$? – 2017-01-26