1
$\begingroup$

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?

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

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.