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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?

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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.

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  • $\begingroup$ Thank you Brian 👍 $\endgroup$ – JP McCarthy Aug 22 at 14:09
  • $\begingroup$ Does this speak to discrete spectrum? $\endgroup$ – JP McCarthy Aug 22 at 14:15
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    $\begingroup$ I think so. A self-adjoint operator with a finite spectrum is a special case of self-adjoint operators with compact resolvent. You might want to look up the spectral theorem for self-adjoint operators with compact resolvent. $\endgroup$ – Brian Lins Aug 22 at 15:27
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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.

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  • $\begingroup$ Thank you Martin. You're a maestro with this stuff. $\endgroup$ – JP McCarthy Aug 22 at 16:42
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    $\begingroup$ Glad I could help! $\endgroup$ – Martin Argerami Aug 23 at 6:33

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