The following question is from my assignment in spectral theory and I am not able to prove the assertions I am asked to prove.
Question: Let X be a Complex Banach space, and let $A,B \in L(X)$ . It is assumed that $AB=I$ and $BA\neq I$. Consider any $\lambda \in \mathbb{C}$ such that $I-\lambda B$ is invertible. Show that $(A- \lambda I) B (I-\lambda B)^{-1} =I $ and $B ( I-\lambda B)^{-1} ( A-\lambda I) \neq I$. Also deduce that $A-\lambda I$ is not invertible (provided by that $I-\lambda B$ is invertible).
Very unfortunately, I am not able to prove any of the assertions asked. For spectral theory, I have been following my class notes only. The problem I think I am facing is that I am not sure which exact proposition could be helpful in the question.
I have studied the following concepts in this section: K-algebra, Normed algebra, Spectrum, Spectral radius, unitary Banach algebras, invertible elements.
Can you please outline which results I should use to prove the asked assertions. No need to give complete proofs.
I shall be grateful!
There is no section to check here, what you need to do is just a computation.
The equalities assume that $I-\lambda B$ is invertible. Since $AB=I$, \begin{align}\tag1 (A- \lambda I) B (I-\lambda B)^{-1}& =(AB-\lambda B)(I-\lambda B)^{-1}\\[0.3cm] &=(I-\lambda B)(I-\lambda B)^{-1}=I. \end{align} For the second part it is useful to note that $B(I-\lambda B)^{-1}=(I-\lambda B)^{-1}B$. This can be trivially checked from the fact that $(I-\lambda B)B=B(I-\lambda B)$, and multiplying on the left and on the right by $(I-\lambda B)^{-1}$. We then have $$\tag2 B ( I-\lambda B)^{-1} ( A-\lambda I) = ( I-\lambda B)^{-1} ( BA-\lambda B) \ne I. $$ If this were to equal $I$, we would have $BA-\lambda B=I-\lambda B$ and thus $BA=I$, a contradiction.
Finally, if $A-\lambda I$ is invertible, we get from $(1)$ that $B(I-\lambda B)^{-1}=(A-\lambda I)^{-1}$; but this contradicts $(2)$.