Im reading C*-algebras by Murphy, and they state and prove the following theorem.
Let $a \in A$ for $A$ a unital algebra. If $\sigma(a) \neq \emptyset$, and $p \in \mathbb{C}[z]$, then $\sigma(p(a)) = p(\sigma(a))$.
Proof: Suppose $p$ is nonconstant (trivial otherwise). If $\mu \in \mathbb{C}$, then by the Fundamental Theorem of Algebra, $$p(z)-\mu = \lambda_0 \prod_{k=1}^n (z-\lambda_k)$$ for $\lambda_0 \neq 0$ and $\lambda_1,\dots,\lambda_n \in \mathbb{C}$. Hence we have that: $$p(a)-\mu = \lambda_0 \prod_{k=1}^n (a-\lambda_k).$$ Clearly $p(a)-\mu \in \text{Inv}(A)$ if and only if $a-\lambda_k \in \text{Inv}(A)$ for all $k=1,2,\dots,n$. It follows then that $\mu \in \sigma(p(a))$ if and only if $\mu=p(\lambda)$ for some $\lambda \in \sigma(a)$.
Everything makes perfect sense up until the last sentence. If $\mu \in \sigma(p(a))$, then one of the $a-\lambda_k$'s isn't invertible, implying that there exists $j \in \{1,2,\dots,n\}$ such that $\lambda_j \in \sigma(a)$. I don't see how this implies anything about $\mu=p(\lambda)$. If someone could explain the reasoning, that would be helpful.
Thank you.
You have $$\tag1 p(z)=\mu+\lambda_0\prod_{k=1}^n(z-\lambda_k). $$ The point of $(1)$ is that $\mu=p(\lambda)$ if and only if $\lambda=\lambda_k$ for some $k$.
If $\mu\in\sigma(p(a))$, then there exists $k$ with $\lambda_k\in\sigma(a)$ and $p(\lambda_k)=\mu$ by $(1)$.
Conversely, if $\mu\not\in\sigma(p(a))$, then $p(a)-\mu$ is invertible, and so $a-\lambda_k$ is invertible for all $k$; if $p(\lambda)=\mu$, then by $(1)$ we have $\lambda=\lambda_k$, and none of them is in $\sigma(a)$. So $\mu\not\in p(\sigma(a))$.