$\lambda\in \sigma(x)\Rightarrow \lambda^{n}\in\sigma(x^{n})$ in Banach algebra

Question

Let $\sigma(x)$ be the spectrum of an element $x$ of a unitary Banach algebra $X$ with the unity $e\in X$.

I read that, since $\lambda e-x$ divides $\lambda^n e-x^n$, we have $\lambda\in \sigma(x)\Rightarrow \lambda^{n}\in\sigma(x^{n})$.

If my calculations are correct, I understand that $\lambda^n e-x^n=(\lambda e-x)\sum_{k=0}^{n-1}\lambda^{n-k-1}x^k$ $=\sum_{k=0}^{n-1}\lambda^{n-k-1}x^k(\lambda e-x)$, but I do not understand why, if $\lambda e-x$ is not invertible, then $\lambda^n e-x^n$ is not either...

Could anyone explain the reason of this implication to me? $\infty$ thanks!

Answer 1

You have $(x-\lambda e)q(x)=(x^{n}-\lambda^{n}e)$ for a polynomial $q$. Suppose $(x^{n}-\lambda^{n}e)$ is invertible. Then $x-\lambda e$ is invertible because $$ (x-\lambda e)[q(x)(x^{n}-\lambda^{n}e)^{-1}]=e=[q(x)(x^{n}-\lambda^{n}e)^{-1}](x-\lambda e) $$ Therefore, if $(x-\lambda e)$ is not invertible, then $(x^{n}-\lambda^{n}e)$ cannot be invertible either. That is, $\lambda \in \sigma(x)\implies \lambda^{n}\in\sigma(x^{n})$.