I've been working on the spectral mapping theorem and I've come across a hurdle while trying to solve a problem. Let P(x) be a non-constant polynomial, T be a bounded linear operator and denote $\sigma_p(T)$ to be the eigenvalues of a linear operator $T$.
I was able to prove that $P(\sigma_p(T)) \subset \sigma_p(P(T))$. The converse is a direct consequence if the space is finite-dimensional. However, I am unable to prove the converse for infinte-dimensional spaces.
Requesting for hints or a counterexample
Let us assume we have $P(T)w=\lambda w$ with $||w||=1$. We may no define also $Q(z)=P(z)-\lambda z$, since $Q$ is some polynomial over $\mathbb C$ we can find some representation $Q(z)=\prod_{i=1}^n(z-\lambda_i)$ and therefore we have $$0=Q(T)w=\prod_{i=1}^n(T-\lambda_i)w.$$ Now $v_n:=(T-\lambda_n)w=0$ or we go iteratively $v_{n-1}:=(T-\lambda_{n-1})v_n=0$ and so on. So, there is some $k$ such that $(T-\lambda_{k})v_{n-k+1}=0$. In other words, $\lambda_k$ is an eigenvalue of $T$ with some eigenvector $u$. With that we already know that $P(\lambda_k)\in\sigma_p(P(T))$. It follows now from $0=Q(T)u=P(T)u-\lambda u$ $$P(\lambda_k)u=P(T)u=\lambda u.$$ So it holds, $\lambda=P(\lambda_k)\in P(\sigma_p(T))$, hence $\sigma_p(P(T))\subset P(\sigma_p(T))$.