# Self-adjoint bounded operator with finite spectrum implies diagonalisable?

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?

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.
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.