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 $0=p(\lambda\_i)=\phi((\lambda\_i e-x)^n) \in \sigma\_A((\lambda\_i e-x)^n)$?



How to show $\lambda_i \in \sigma_A(x)$?
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. $\lambda \in \sigma\_A(x)$.


