Invertibility of an element in a Banach algebra (Gelfand's formula)

Question

In Folland's A Course in Abstract Harmonic Analysis, Theorem 1.8 states that for a unital Banach algebra (with unit $e$), the spectral radius of an element $x$ is given by $\lim_{n \to \infty} \|x^n\|^{1/n}$. In the proof, Folland writes:

We have $\lambda^n e - x^n = (\lambda e -x) \sum_0^{n-1} \lambda^j x^{n-1-j}$, from which it follows that if $\lambda^n e - x^n$ is invertible then so is $\lambda e - x$.

Why is this? It seems we need to show that $\sum_0^{n-1} \lambda^j x^{n-1-j}$ is invertible when $\lambda^n e - x^n$ is, but why would that be?

Answer 1

Note the following:

If $a,b$ are so that $ab$ and $ba$ are invertible then $a$ and $b$ must both already be invertible.

It is possible to directly give the inverses:

$$a^{-1}=b(ab)^{-1}, \qquad b^{-1} = a(ba)^{-1}$$

To see that these are actually inverses you can additionally define: $$\tilde a^{-1}= (ba)^{-1} b, \qquad \tilde b^{-1} = (ab)^{-1}a$$

Note that $$aa^{-1}=1 = \tilde aa^{-1}, bb^{-1}=1 =b\tilde b^{-1}$$ ie $a^{-1}, b^{-1}$ are right inverses to $a,b$ and $\tilde a^{-1},\tilde b^{-1}$ are left inverses to $a,b$. By the standard computation: $$a^{-1}= (\tilde a^{-1}a)a^{-1}=\tilde a^{-1}(aa^{-1})=\tilde a^{-1}$$ you find that if something has both left and right inverses they are equal and the operator is already inveritble.