How to show $\lambda_i \in \sigma_A(x)$?

Question

Let $\sigma_A(x)$ be the spectrum of $x$ in $A$, and linear functional $\phi$ satisfying $\phi(x)\in \sigma_A(x)$ for every $x \in A$, consider $p(\lambda)=\phi((\lambda e-x)^n)$, and denote its roots by $\lambda_1$, $\lambda_2$ ...$\lambda_n$.

May I ask how to get $\lambda_i \in \sigma_A(x)$ from $0=p(\lambda_i)=\phi((\lambda_i e-x)^n) \in \sigma_A((\lambda_i e-x)^n)$?

Answer 1

If $p(\lambda) = \phi((\lambda e - x)^n) = 0$, that says $0 \in \sigma_A((\lambda e - x)^n)$. By the Spectral Mapping Theorem, $\sigma_A((\lambda e - x)^n)$ is the image of $\sigma_A(x)$ under the map $t \to (\lambda - t)^n$, i.e. there is some $t \in \sigma_A(x)$ such that $(\lambda - t)^n = 0$, but that says $t = \lambda$, i.e. $\lambda \in \sigma_A(x)$.