Namespaces
Variants
Actions

Spectral decomposition of a linear operator

From Encyclopedia of Mathematics
Revision as of 08:22, 6 June 2020 by Ulf Rehmann (talk | contribs) (tex encoded by computer)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
Jump to: navigation, search


The representation of a linear operator in the form of an integral with respect to the spectral measure (the spectral resolution). For any self-adjoint operator $ T $ on a Hilbert space $ H $ there is a spectral resolution $ P ( \cdot ) $ such that

$$ T = \int\limits _ {- \infty } ^ { {+ } \infty } t d P ( t) . $$

This means that

$$ D _ {T} = \left \{ {x \in H } : { \int\limits _ {- \infty } ^ { {+ } \infty } t ^ {2} d ( P ( t) x , x ) < \infty } \right \} , $$

$$ ( T x , y ) = \int\limits _ {- \infty } ^ { {+ } \infty } t d ( P ( t) x , y ) $$

for all $ x \in D _ {T} $, $ y \in H $. The spectral resolution of a self-adjoint operator $ T $ can be calculated from its resolvent $ R ( \lambda , T ) $( at points of continuity) by the formulas

$$ P ( b ) - P ( a) = $$

$$ = \ \lim\limits _ {\delta \rightarrow 0 } \lim\limits _ {\epsilon \rightarrow 0 } \frac{1}{2 \pi i } \int\limits _ {a + \delta } ^ { {b } - \delta } R ( \lambda - i \epsilon , T ) - R ( \lambda + i \epsilon , T ) d \lambda . $$

It follows from the spectral decomposition theorem for self-adjoint operators that operators can be realized by multiplication operators, and that there is a functional calculus on Borel functions.

Using the spectral decomposition of a self-adjoint operator and the theory of extensions to a larger space (see [2]), one can obtain an integral representation of a symmetric operator by a generalized spectral resolution. Integral representations of isometric operators (cf. Isometric operator) are constructed in a similar way. The analogy between spectral decompositions of self-adjoint and unitary operators on the one hand, and integral representations of symmetric and isometric operators on the other, is far from complete (due to the lack of uniqueness of the generalized spectral resolutions, the lack of strong convergence of the integrals, the comparative narrowness of the functional calculus, etc.).

For any bounded normal operator $ T $ on a Hilbert space $ H $ there is a self-adjoint spectral measure $ E ( \cdot ) $, countably-additive in the strong operator topology on the $ \sigma $- algebra of Borel subsets of the complex plane, such that

$$ T = \int\limits _ { \mathbf C } z E ( d z ) , $$

where $ \supp E ( \cdot ) = \sigma ( T) $, the spectrum of $ T $( cf. Spectrum of an operator), $ T E ( \alpha )= E ( \alpha ) T $, and $ \sigma ( T \mid _ {E ( \alpha ) H } ) \subset \overline \alpha \; $. It is convenient to reformulate this theorem as follows: Every bounded normal operator is unitarily equivalent to the operator of multiplication by some essentially-bounded function in the space $ L _ {2} ( S , \Sigma , \mu ) $, where $ \mu $ can be chosen to be a finite measure if the space is separable.

It follows from the spectral decomposition theorem that there is a functional calculus for normal operators, that is, there is a homomorphism $ f \mapsto f ( T) $ from the algebra of essentially-bounded Borel functions on $ \sigma ( T) $ into the algebra of bounded operators that satisfies the condition $ \mathop{\rm id} ( T) = T $ and that maps every bounded pointwise-convergent sequence of functions into a strongly-converging sequence of operators. The image of this homomorphism (that is, the set of all functions of the operator $ T $) coincides with the set of all operators commuting with every operator that commutes with $ T $. Since the existence of a functional calculus in turn implies the spectral decomposition theorem, this result can be regarded as a form of the spectral theorem. The spectral decomposition theorem can also be generalized to unbounded normal operators (see [2]).

In the case of spectral decomposition of a unitary operator, which is a special case of a normal operator, the spectral measure is given on the unit circle. The spectral decomposition of a unitary operator $ U $ is sometimes written in the form

$$ U = \int\limits _ { 0 } ^ { {2 } \pi } e ^ {i \theta } d E ( \theta ) , $$

where $ E ( \cdot ) $ is a spectral resolution concentrated on the interval $ [ 0 , 2 \pi ] $. Thus, a spectral decomposition enables one to represent a unitary operator in the form $ \mathop{\rm exp} ( i A ) $, where $ A $ is a self-adjoint operator. This result is generalized by Stone's theorem: Every strongly-continuous one-parameter group of unitary operators can be written in the form

$$ U ( t) = \mathop{\rm exp} i t A , $$

where $ A $ is a self-adjoint (possibly unbounded) operator.

References

[1] N. Dunford, J.T. Schwartz, "Linear operators. Spectral theory" , 2 , Interscience (1963)
[2] N.I. Akhiezer, I.M. Glazman, "Theory of linear operators in Hilbert space" , 1–2 , Pitman (1981) (Translated from Russian)
How to Cite This Entry:
Spectral decomposition of a linear operator. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Spectral_decomposition_of_a_linear_operator&oldid=19240
This article was adapted from an original article by V.S. Shul'man (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article