# the Moore-Penrose inverse of $1-ba$ when $1-ab$ is Moore-Penrose invertible

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?

• You may want to take a look at this – polfosol Jan 16 '17 at 6:44
• My 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. – Josué Tonelli-Cueto Jan 16 '17 at 10:33
• I think you mean $1+b(1-ab)^\dagger a=(1-b a)^\dagger$. – shima homayouni Jan 16 '17 at 10:50

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.
• How can I prove the second equality? Should I multiply on left by $b$ and on the right by$a$? – shima homayouni Jan 26 '17 at 17:03