title: Discrete spectrum (mathematics)
text: # Discrete spectrum (mathematics)
  
  In mathematics, specifically in spectral theory, a discrete spectrum of a closed linear operator is defined as the set of isolated points of its spectrum such that the rank of the corresponding Riesz projector is finite.
  
  ## Definition
  
  A point $\lambda \in \mathbb {C}$ in the spectrum $\sigma (A)$ of a closed linear operator $A:\,{\mathfrak {B}}\to {\mathfrak {B}}$ in the Banach space ${\mathfrak {B}}$ with domain ${\mathfrak {D}}(A)\subset {\mathfrak {B}}$ is said to belong to discrete spectrum $\sigma _{\mathrm {disc} }(A)$ of $A$ if the following two conditions are satisfied:
  
  * $\lambda$ is an isolated point in $\sigma (A)$;
  * The rank of the corresponding Riesz projector $P_{\lambda }={\frac {-1}{2\pi \mathrm {i} }}\oint _{\Gamma }(A-zI_{\mathfrak {B}})^{-1}\,dz$ is finite.
  
  Here $I_{\mathfrak {B}}$ is the identity operator in the Banach space ${\mathfrak {B}}$ and $\Gamma \subset \mathbb {C}$ is a smooth simple closed counterclockwise-oriented curve bounding an open region $\Omega \subset \mathbb {C}$ such that $\lambda$ is the only point of the spectrum of $A$ in the closure of $\Omega$; that is, $\sigma (A)\cap {\overline {\Omega }}=\{\lambda \}.$
  
  ## Relation to normal eigenvalues
  
  The discrete spectrum $\sigma _{\mathrm {disc} }(A)$ coincides with the set of normal eigenvalues of $A$:
  
      $\sigma _{\mathrm {disc} }(A)=\{{\mbox{normal eigenvalues of }}A\}.$
  
  ## Relation to isolated eigenvalues of finite algebraic multiplicity
  
  In general, the rank of the Riesz projector can be larger than the dimension of the root lineal ${\mathfrak {L}}_{\lambda }$ of the corresponding eigenvalue, and in particular it is possible to have $\mathrm {dim} \,{\mathfrak {L}}_{\lambda }<\infty$, $\mathrm {rank} \,P_{\lambda }=\infty$. So, there is the following inclusion:
  
      $\sigma _{\mathrm {disc} }(A)\subset \{{\mbox{isolated points of the spectrum of }}A{\mbox{ with finite algebraic multiplicity}}\}.$
  
  In particular, for a quasinilpotent operator
  
      $Q:\,l^{2}(\mathbb {N} )\to l^{2}(\mathbb {N} ),\qquad Q:\,(a_{1},a_{2},a_{3},\dots )\mapsto (0,a_{1}/2,a_{2}/2^{2},a_{3}/2^{3},\dots ),$
  
  one has ${\mathfrak {L}}_{\lambda }(Q)=\{0\}$, $\mathrm {rank} \,P_{\lambda }=\infty$, $\sigma (Q)=\{0\}$, $\sigma _{\mathrm {disc} }(Q)=\emptyset$.
  
  ## Relation to the point spectrum
  
  The discrete spectrum $\sigma _{\mathrm {disc} }(A)$ of an operator $A$ is not to be confused with the point spectrum $\sigma _{\mathrm {p} }(A)$, which is defined as the set of eigenvalues of $A$. While each point of the discrete spectrum belongs to the point spectrum,
  
      $\sigma _{\mathrm {disc} }(A)\subset \sigma _{\mathrm {p} }(A),$
  
  the converse is not necessarily true: the point spectrum does not necessarily consist of isolated points of the spectrum, as one can see from the example of the left shift operator, $L:\,l^{2}(\mathbb {N} )\to l^{2}(\mathbb {N} ),\quad L:\,(a_{1},a_{2},a_{3},\dots )\mapsto (a_{2},a_{3},a_{4},\dots ).$ For this operator, the point spectrum is the unit disc of the complex plane, the spectrum is the closure of the unit disc, while the discrete spectrum is empty:
  
      $\sigma _{\mathrm {p} }(L)=\mathbb {D} _{1},\qquad \sigma (L)={\overline {\mathbb {D} _{1}}};\qquad \sigma _{\mathrm {disc} }(L)=\emptyset .$
  