Generalized eigenspaces of a compact operator are finite dimensional Let $T : H\rightarrow H$ be a compact operator on a Hilbert space $H$. Say that $\lambda \in \mathbb C$ is a generalized eigenvalue of $T$ if there is some $n \geq 1$ such that $(\lambda - T)^n$ is not injective. Define the generalized eigenspace corresponding to $\lambda$ to be the space $V$ of vectors $x\in H$ such that $(\lambda - T)^n x = 0$ for some $n$. I am trying to show that $V$ is necessarily finite dimensional if $\lambda\not=0$. I can show that the kernel of $(\lambda  - T)^n$ is finite dimensional for each $n$. But, I am having trouble extending this to the union of all of these kernels. Does anyone have any suggestions?
 A: This is answered in Banach Algebra Techniques in Operator Theory by Ronald G. Douglas, as part of Theorem 5.22 in the second edition.  The result you stated does not hold for $\lambda = 0$ (for instance consider the zero operator).
You already know that $\operatorname{ker}(T - \lambda)^n$ is finite dimensional for each $n$, so we need only show that the generalized eigenspace for $\lambda$ is equal to $\operatorname{ker}(T - \lambda)^n$ for some $n$. If we suppose this is not the case, then we may form an orthonormal sequence of vectors $e_k \in H$ and a strictly increasing sequence of positive integers $n_k$ such that $e_k \in \operatorname{ker}(T - \lambda)^{n_k}$ and $e_k$ is orthogonal to $\operatorname{ker}(T - \lambda)^{n_k - 1}$. In particular, $e_k$ is orthogonal to $(T - \lambda)e_k$, so
$||Te_k||^2 = ||(T - \lambda)e_k + \lambda e_k||^2 = ||(T - \lambda)e_k||^2 + |\lambda|^2 \geq |\lambda|^2.$
Because the sequence $e_k$ is orthnormal, $e_k$ tends weakly to zero, and by compactness of $T$ we have that $Te_k$ tends to zero in norm. This is a contradiction since $\lambda \not= 0$.
