Normal eigenvalue

In mathematics, specifically in spectral theory, an eigenvalue of a closed linear operator is called normal if the space admits a decomposition into a direct sum of a finite-dimensional generalized eigenspace and an invariant subspace where $$A-\lambda I$$ has a bounded inverse. The set of normal eigenvalues coincides with the discrete spectrum.

Root lineal
Let $$\mathfrak{B}$$ be a Banach space. The root lineal $$\mathfrak{L}_\lambda(A)$$ of a linear operator $$A:\,\mathfrak{B}\to\mathfrak{B}$$ with domain $$\mathfrak{D}(A)$$ corresponding to the eigenvalue $$\lambda\in\sigma_p(A)$$ is defined as


 * $$\mathfrak{L}_\lambda(A)=\bigcup_{k\in\N}\{x\in\mathfrak{D}(A):\,(A-\lambda I_{\mathfrak{B}})^j x\in\mathfrak{D}(A)\,\forall j\in\N,\,j\le k;\, (A-\lambda I_{\mathfrak{B}})^k x=0\}\subset\mathfrak{B}, $$

where $$I_{\mathfrak{B}}$$ is the identity operator in $$\mathfrak{B}$$. This set is a linear manifold but not necessarily a vector space, since it is not necessarily closed in $$\mathfrak{B}$$. If this set is closed (for example, when it is finite-dimensional), it is called the generalized eigenspace of $$A$$ corresponding to the eigenvalue $$\lambda$$.

Definition of a normal eigenvalue
An eigenvalue $$\lambda\in\sigma_p(A)$$ 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 called normal (in the original terminology, $$\lambda$$ corresponds to a normally splitting finite-dimensional root subspace), if the following two conditions are satisfied:
 * 1) The algebraic multiplicity of  $$\lambda$$ is finite: $$\nu=\dim\mathfrak{L}_\lambda(A)<\infty$$, where $$\mathfrak{L}_\lambda(A)$$ is the root lineal of $$A$$ corresponding to the eigenvalue $$\lambda$$;
 * 2) The space $$\mathfrak{B}$$ could be decomposed into a direct sum $$\mathfrak{B}=\mathfrak{L}_\lambda(A)\oplus \mathfrak{N}_\lambda$$, where $$\mathfrak{N}_\lambda$$ is an invariant subspace of $$A$$ in which $$A-\lambda I_{\mathfrak{B}}$$ has a bounded inverse.

That is, the restriction $$A_2$$ of $$A$$ onto $$\mathfrak{N}_\lambda$$ is an operator with domain $$\mathfrak{D}(A_2)=\mathfrak{N}_\lambda\cap\mathfrak{D}(A)$$ and with the range $$\mathfrak{R}(A_2-\lambda I)\subset\mathfrak{N}_\lambda$$ which has a bounded inverse.

Equivalent characterizations of normal eigenvalues
Let $$A:\,\mathfrak{B}\to\mathfrak{B}$$ be a closed linear densely defined operator in the Banach space $$\mathfrak{B}$$. The following statements are equivalent (Theorem III.88):
 * 1) $$\lambda\in\sigma(A)$$ is a normal eigenvalue;
 * 2) $$\lambda\in\sigma(A)$$ is an isolated point in $$\sigma(A)$$ and $$A-\lambda I_{\mathfrak{B}}$$ is semi-Fredholm;
 * 3) $$\lambda\in\sigma(A)$$ is an isolated point in $$\sigma(A)$$ and $$A-\lambda I_{\mathfrak{B}}$$ is Fredholm;
 * 4) $$\lambda\in\sigma(A)$$ is an isolated point in $$\sigma(A)$$ and $$A-\lambda I_{\mathfrak{B}}$$ is Fredholm of index zero;
 * 5) $$\lambda\in\sigma(A)$$ is an isolated point in $$\sigma(A)$$ and the rank of the corresponding Riesz projector $$P_\lambda$$ is finite;
 * 6) $$\lambda\in\sigma(A)$$ is an isolated point in $$\sigma(A)$$, its algebraic multiplicity $$\nu=\dim\mathfrak{L}_\lambda(A)$$ is finite, and the range of $$A-\lambda I_{\mathfrak{B}}$$ is closed.

If $$\lambda$$ is a normal eigenvalue, then the root lineal $$\mathfrak{L}_\lambda(A)$$ coincides with the range of the Riesz projector, $$\mathfrak{R}(P_\lambda)$$.

Relation to the discrete spectrum
The above equivalence shows that the set of normal eigenvalues coincides with the discrete spectrum, defined as the set of isolated points of the spectrum with finite rank of the corresponding Riesz projector.

Decomposition of the spectrum of nonselfadjoint operators
The spectrum of a closed operator $$A:\,\mathfrak{B}\to\mathfrak{B}$$ in the Banach space $$\mathfrak{B}$$ can be decomposed into the union of two disjoint sets, the set of normal eigenvalues and the fifth type of the essential spectrum:

\sigma(A)=\{\text{normal eigenvalues of}\ A\}\cup\sigma_{\mathrm{ess},5}(A). $$