Im reading through the proof of the following theorem and I'm stuck on a single step.
Let $u$ be a compact operator on a Banach space $X$. Then $\sigma(u)$ is countable, and each nonzero point of $\sigma(u)$ is an eigenvalue of $u$ and an isolated point of $\sigma(u)$.
Proof: If $\lambda \in \sigma(u) \setminus \{0\}$, then by the Fredholm alternative, $u-\lambda I$ is not injective, so $\lambda$ is a eigenvalue of $u$. The operator $u-\lambda I$ has finite ascent $p$, so one can write: $$X = Y \bigoplus Z$$ where $Y = \text{ker}(u-\lambda I)^p$ and $Z = (u-\lambda I)^p(X)$.
The spaces $Y,Z$ are closed and $u$-invariant. Hence $u-\lambda I = (u_{|Y}-\lambda I_{|Y}) \bigoplus (u_{|Z}-\lambda I_{|Z})$. Since $(u-\lambda I)^p=0$, it follows that $\sigma(u_{|Y}) = \{\lambda\}$.
Why does $(u_{|Y}-\lambda I_{|Y})^p=0$ imply the spectrum is a singleton set $\{\lambda\}$? I get that $\lambda \in \sigma(u_{|Y})$, but why is it the only point in the spectrum?
Any insight would be appreciated. Thank you.