TITLE: Self-adjoint bounded operator with finite spectrum implies diagonalisable?
QUESTION [5 upvotes]: 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?

REPLY [4 votes]: 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.