> [!theorem] > > Let $E$ be a [[Hilbert Space|Hilbert space]] and $u \in K(E, E)$ be a [[Compact Self-Adjoint Operator|compact self-adjoint]] operator and $\seq{\lambda_j}$ be the enumeration of its eigenvalues (including $0$), in descending magnitudes. Then the [[Spectral Theorem (Compact)|spectral decomposition]] > $ > E = F_n \oplus H_n > $ > where > $ > F_n = \sum_{j = 1}^n \ker(u - \lambda_j)^{r_j} \quad H_n = \text{Im}\braks{\prod_{i = 1}^n(u - \lambda_iI)^{r_i}} > $ > exhausts $E$. That is, the smallest [[Closed Set|closed]] subspace containing $\seq{F_n}$ is equal to $E$. > > *Proof*. Let $F$ be the smallest closed subspace generated by $\seq{F_n}$ and $H = \bigcap_{n \in \nat}H_n$, then $H$ is $u$-invariant, on which $u$ has no eigenvalue. Since $\pm\norm{u|_H}$ is an eigenvalue of $u|_H$, $\norm{u|_H} = 0$ and $H \subset \ker(u)$, so it can only be $\bracs{0}$. [!theorem] Let $\seq{\lambda_n}$ be an enumeration of $\sigma(u)$, including $0$, then the [[Spectral Theorem|projection-valued measure]] corresponding to $u$ is equal to $ P(E) = \sum_{\lambda \in E}\text{Proj}_{\text{Eig}(\lambda)} $ where the sum converges in the strong operator norm. *Proof*. It is clear that $P(\emptyset) = 0$, $P(\sigma(u)) = 1$, and that each $P(E)$ is an orthogonal projection. Let $E, F \subset \sigma(u)$, then $ \begin{align*} P(E \cap F) &= \sum_{\lambda \in E \cap F}\text{Proj}_{\text{Eig}(\lambda)} \\ &= \sum_{\lambda \in E}\text{Proj}_{\text{Eig}(\lambda)}\sum_{\lambda \in E \cap F}\text{Proj}_{\text{Eig}(\lambda)}\\ &= \sum_{\lambda \in E}\text{Proj}_{\text{Eig}(\lambda)}\sum_{\lambda \in F}\text{Proj}_{\text{Eig}(\lambda)} \\ &= P(E)P(F) \end{align*} $ Let $\seq{E_j}$ be a disjoint family of subsets, and $x \in \ch$. If you assume without loss of generality that they are points, then $P(\cup E_j) = \sum P(E_j)$. Therefore $P$ is a projection-valued measure. Clearly, $\int \lambda dP(\lambda) = T$ and $\int 1 dP = I$. Integration against a (any!) projection-valued measure is a $*$-homomorphism. Now let $Q$ be the projection-valued measure **obtained from the spectral theorem**. If $p$ is a polynomial in $\lambda$, then $ p(T) = \int p(\lambda) dQ(\lambda) $ by property $(1)$ from the spectral theorem. Therefore $\Gamma(p(T)) = p$. Using the fact that integration against $P$ is a $*$-homomorphism, we find that $ p(T) = \int p(\lambda) dP(\lambda) $ so integration against $P$ and $Q$ provides the same value **on all polynomials**. Since all polynomials are dense in $B(\Sigma)$, the continuity property of the integral (applies to any projection-valued measure) shows that $\int f dQ = \int f dP$ for all $f \in B(\Sigma)$. By the uniqueness part of the spectral theorem, $P = Q$.