[Conway's exercise] Suppose $A \in B(X)$, where $B(X)$ denotes the space of bounded linear operators on a Banach space $X$. Let there be a polynomial $p$ such that $p(A) \in K(X)$, with $K(X)$ representing the space of compact operators on $X$. What conclusions can be drawn about the spectrum $\sigma(A)$ of the operator $A$?
Is the following ideas correct?
For any $\lambda \in \sigma(A)$, the operator $A - \lambda I$ is not invertible. This leads to $p(A) - p(\lambda)I$ being non-invertible, and thus $p(\lambda) \in \sigma(p(A))$.
The compactness of $p(A)$ implies that $\sigma(p(A))$ is a set of isolated points, except possibly at 0. (Which theorem guarantees that?)
This discreteness of $\sigma(p(A))$ imposes similar constraints on $\sigma(A)$, specifically that it consists of isolated points under the mapping of $p$.
We will use a well known fact that $\sigma(p(A))=p(\sigma(A)).$ Assume $\deg p=N\ge 1.$ Then the set $\sigma(A)$ admits at most $N$ accumulation points. Indeed, observe that if $z$ is an accumulation point of $\sigma(A)$ then $p(z)$ is an accumulation point of $p(\sigma(A)).$ In fact if $w_n\in \sigma(A), $ $w_n\neq z$ and $\lim_nw_n=z,$ then $\lim_np(w_n)=p(z).$ Moreover $p(w_n)\neq p(z)$ for all $n$ except finitely many. Hence $p(z)$ is an accumulation point of $p(\sigma(A)), $ as claimed. Assume now that $\sigma(A)$ admits accumulation points $z_1,z_2,\ldots z_{N+1}.$ Then $p(z_n)$ is an accumulation point of $p(\sigma(A))=\sigma(p(A))$ for any $1\le n\le N+1.$ Hence $p(z_n)=0$ for all $1\le n\le N+1,$ thus $p=0.$