Discrete spectrum

Remarks

A normal eigenvalue is an eigenvalue, and its algebraic multiplicity, the rank of $P_{\set{\lambda}}$, is finite.

Example: if $T$ is a compact operator, then every nonzero point of $\sigma(T)$ is a normal eigenvalue. For $D(x_n) = (x_n / n)$ on $\ell^2$, each $1/n$ is isolated in $\sigma(D)$, and $P_{\set{1/n}}$ is the projection onto the span of $e_n$, of rank $1$.

Non-example: if $X$ is infinite-dimensional, then $\sigma(I) = \set{1}$ and $1$ is isolated, but $P_{\set{1}} = I$ has infinite rank, so clause 2 fails. For $D$ above, $0$ fails clause 1, since $1/n \to 0$.

For a self-adjoint operator $T$ on a Hilbert space, $\sigma_{disc}(T)$ is the set of isolated eigenvalues of finite multiplicity. Then $\sigma(T)$ is the disjoint union of $\sigma_{disc}(T)$ and the essential spectrum.