Let $E$ be a [[Banach Space|Banach space]] and $u \in K(E, E)$ be a [[Compactness|compact]] operator. > [!theorem] > > If $\lambda \ne 0$, then $\lambda \in \sigma_p(u)$ is an eigenvalue. > > *Proof*. If $u - \lambda I$ is injective, then it is an isomorphism[^1]. > [!theorem] > > Let $\lambda \in \sigma(u)$ with $\lambda \not\in 0$, then there exists $n \ge 0$ such that > $ > \ker(u - \lambda I)^r = \ker(u - \lambda I)^n > $ > The smallest of such $n$ is known as the **exponent** of $\lambda$. > > *Proof*. Assume without loss of generality that $\lambda = 1$. If $\ker(u - I)^r = \ker(u - I)^{r + 1}$ for some $r \ge 0$, then this will hold for all $r' > r$. > > Suppose for contradiction that the above does not hold, then there exists a strictly increasing sequence > $ > \ker(I - u) \subsetneq \ker(I - u)^2 \subsetneq \cdots > $ > Since each space is closed, there exists a sequence $\seq{x_n}$ such that $x_n \in \ker(I - u)$, $\norm{x_n} = 1$, and the [[Quotient Norm|quotient norm]] $\norm{x_n}_{/\ker(I - u)^{n-1}} > 1/2$. For any $k < n$, > $ > \norm{ux_n - ux_k} = \norm{x_n - Tx_n - x_k + Tx_k} \ge 1/2 > $ > where $-Tx_n - x_k + Tx_k \in \ker(I - u)^{n - 1}$. This contradicts the fact that $u$ is compact. > [!theorem] > > Let $\lambda \in \sigma(u)$ with $\lambda \ne 0$ and $r$ be its exponent, then there is a decomposition > $ > E = \ker(u - \lambda I)^r \oplus (u - \lambda I)^r(E) > $ > into [[Closed Set|closed]], $u$-invariant subspaces. If $\mu \in \sigma(u)$ with $\mu \ne 0, \lambda$, and $s$ is its exponent, then > $ > \ker(u - \mu I)^s \subset (u - \lambda I)^r(E) > $ > *Proof*. Let $T = (u - \lambda I)^r$, then $T$ is Fredholm of index $0$ with $\ker{(T)}$ and $T(E)$ being $u$-invariant closed subspaces, with $\ker{(T)} \cap T(E) = \emptyset$. Thus we can split $E = \ker(u - \lambda I)^r \oplus (u - \lambda I)^r(E)$. > > Now let $\mu \in \sigma(u)$ with $\mu \ne 0, \lambda$ of exponent $s$, and $S = (u - \mu I)^s$. Since $ST = TS$, $\ker(T)$ and $T(E)$ are $S$-invariant subspaces as well. Let $x \in \ker(S)$ and split $x = y + z \in \ker(T) \oplus T(E)$. Since $\ker(T)$ is the eigenspace of $\lambda$, $y = 0$, and $x = z \in T(E)$. > [!theorem] > > Let $\seq{\lambda_j}$ be an enumeration of $\sigma(u)$ with decreasing norms, and $r_j$ be the exponent of $\lambda_j$, then there exists subspaces > $ > F_n = \sum_{i = 1}^{n}\ker(u - \lambda_i)^{r_i} > $ > such that $F_n \subset F_{n + 1}$ for all $n \in \nat$, such that each $F_n$ admits a complementary closed subspace > $ > H_n = \text{Im}\braks{\prod_{i = 1}^n(u - \lambda_iI)^{r_i}} > $ > such that $H_n \supset H_{n + 1}$ for all $n \in \nat$ which allows inductively constructing > $ > E = F_n \oplus H_n > $ > known as the **spectral theorem** for compact operators. [^1]: See [[Fredholm Operator|Fredholm operator]].