Spectral theorem for bounded self-adjoint opertaor

Spectral Theorem

Lemma

Suppose $T$ is a bounded self-adjoint linear operator on a Hilbert space \H. Then any eigenvalue of $T$ is real, and if $f_1$ and $f_2$ are eigenvectors corresponding to two distinct eigenvalues, then $f_1$ and $f_2$ are orthogonal.

Spectral Theorem for Compact Self-Adjoint Operator

Suppose $T$ is a compact self-adjoint operator on a separable Hilbert space \H. Then there exits an orthonormal basis $\{e_i{\}}_{i=1}^{\infty }$ of \H, that consists of eigenvectors of $T$. Moreover, if $T{e}_k={\lambda }_k{e}_k,$then ${\lambda }_k\in \mathbb{R}$ and ${\lambda }_k\to 0$ as $k\to \infty$. Conversely, every operator of the above form is compact and self-adjoint. The collection $\{{\lambda }_k\}$ is called the spectrum of $T$.

Spectral Theorem for Compact Normal Operator

A linear operator on $\mathcal{H}$ is normal if $T{T}^{\ast }=T^{\ast }T$. If $T$ is normal and compact, then $T$ can be diagonalized.

Proof

If $T_1$ and $T_2$ are two linear symmetric and compact operators on $\mathcal{H}$ that commute (that is, $T_1{T}_2=T_2{T}_1$), show that they can be diagonalized simultaneously. In other words, there exists an orthonormal basis for $\mathcal{H}$ which consists of eigenvectors for both $T_1$ and $T_2$.

Corollary

If $U$ is unitary, and $U=\lambda I-T$, where $T$ is compact, then $U$ can be diagonalized.