I have a question about a proof.
In the proof $\sigma(T)$ is $\{\lambda \in\mathbb{C}: T-\lambda I\text{ is not invertible}\}$.
In the proof they use this lemma:
Here is the proof, my problem is with the red equivalence:
By Lemma 4.35, the $\leftarrow$ is clear. But how do they get the $\rightarrow$ ? How can he say that if the product is invertible, then each $T-\mu_jI$ is invertible?
As I have understood with we have to show that it is bounded, 1-1, and onto. It is easy to show that it is bounded by $\|T\|+|\mu_j|$. However, from general function theory, if $f_1f_2$ is bijective, we can only say that $f_2$ is 1-1, and $f_1$ is onto. So the only info I have is that all $T-\mu_jI$ is bounded(and linear). $T-\mu_nI$ is 1-1, $T-\mu_1I$ is onto, hence I am not able to show that even one of the $T-\mu_jI$ is invertible.
Can someone please help me with the $\rightarrow$?
If one of the members of the product was NOT invertible,it would map a sequence of vectors, that doesn't converge to zero, to a sequence converging to zero.But then the product would also do that, making the product non-invertible.