Self-adjoint bounded operator with finite spectrum implies diagonalisable?

Question

Let $T$ be a self-adjoint bounded operator on a not-necessarily finite dimensional Hilbert space.

Suppose $T$ has finite spectrum. Does it follow that the elements of the spectrum are eigenvalues, and the operator diagonlisable?

Answer 1

Yes, you can calculate the spectral projection for each eigenvalue $\lambda$ by integrating the resolvent in a small contour around $\lambda$ that avoids all of the other eigenvalues $$P_\lambda = \frac{1}{2\pi i} \int_C (T-z I)^{-1} \, dz.$$ The Hilbert space will then be the direct sum of the spectral subspaces corresponding to the spectral projections. Isolated elements of the spectrum are always eigenvalues.

Answer 2

Since this is also tagged "C$^*$-algebras", I'll answer in that setting. If $\sigma(T)=\{\lambda_1,\ldots,\lambda_n\}$, we can construct continuous functions (polynomials, even) $f_1,\ldots,f_n$ with $f_k(\lambda_j)=\delta_{kj}$. Then $\sum_k\lambda_kf_k(t)=t$, and functional calculus gives us $$ T=\sum_k\lambda_kf_k(T), $$ where $f_1(T),\ldots,f_n(T)$ are pairwise orthogonal projections.