Spectral theorem

From Wikinfo

In mathematics, the spectral theorem is an important decomposition theorem applying to normal operators in linear algebra and functional analysis. The stated decomposition is called the spectral decomposition. Since the class of normal operators includes a number of special kinds of operators such as symmetric operators or unitary operators or special kinds of matrices for example, a symmetric matrix, the spectral theorem can be applied in a wide range of situations. In broad terms, what the spectral theorem does is to identify cases where linear operators can be modelled by a multiplication operator; which as is simple as one can hope to find. See also spectral theory for a historical perspective.

Contents 1 Finite-dimensional case 2 The spectral theorem for compact symmetric operators 3 Functional analysis 4 See also 5 Reference 5.1 References

Finite-dimensional case

We begin by considering a symmetric operator A on a finite dimensional inner product space V; the symmetry condition means <A x, y> = <x, A y> for all x,y elements of V. Recall that an eigenvector of a linear operator A is a vector x such that A x = r x for some scalar r. r is the corresponding eigenvalue.

Theorem. There is an orthonormal basis of V consisting of eigenvectors of A. Each eigenvalue is real.

This result is of such importance in many parts of mathematics, that we provide a sketch of a proof. First the property that all the eigenvalues are real. Indeed if λ is an eignevalue of A, for some non-zero vector x

<math> \overline{\lambda} \langle x, x \rangle= \langle A x, x \rangle = \langle x, A x \rangle = \lambda \langle x, x \rangle .</math>

It follows λ equals its own conjugate and is therefore real.

To prove the existence of an eigenvector basis, we use induction on the dimension of V. In fact it suffices to show A has at least one non-zero eigenvector e. For then we can consider the space K of vectors v orthogonal to e. This is finite dimensional, and A has the property that it maps every vector w in K into K:

<math> \langle A w, e \rangle = \langle w, A e \rangle = \lambda \langle w, e \rangle = 0. </math>

Moreover, A considered as a linear operator on K is also symmetric so by the induction hypothesis this completes the proof.

It remains however to show A has at least one eigenvector. The easiest way to do that is to consider the case in which the field of scalars is complete. Then the polynomial function p(x) = det(A − x I) has a complex zero r. This means the linear operator A − r I is not invertible and hence maps a non-zero vector e to 0. This vector e is a non-zero eigenvector of A. This completes the proof.

The spectral theorem is also true for symmetric operators on finite dimensional real inner product spaces.

The spectral decomposition of an operator A which has a orthornormal basis of eigenvectors, is obtained by grouping together all vectors corresponding to the same eigenvalue. Thus

<math> V_\lambda = \{v \in V: A v = \lambda v\}.</math>

Note these spaces are invariantly defined.

As an immediate consequence of the spectral theorem for symmetric operators we get the spectral decomposition theorem: V is the orthogonal direct sum of the spaces V_λ where the index ranges over eigenvalues. Another equivalent formulation is letting P_λ be the orthogonal projection onto V_λ

<math> P_\lambda P_\mu=0 \quad \mbox{if } \lambda \neq \mu </math>

and if λ₁,..., λ_m are the eigenvalues of A,

<math>A =\lambda_1 P_{\lambda_1} +\cdots+\lambda_m P_{\lambda_m}.</math>

If A is a normal operator on a finite dimensional inner product space, A also has a spectral decomposition and the decomposition theorem holds for A. The eigenvalues will be complex numbers in general. The proof is somewhat more complicated and is discussed in the Axler reference below.

These results translate immediately into results about matrices: For any normal matrix, there exists a unitary matrix U such that

<math>A=U \Sigma U^* \;</math>

where Σ is the diagonal matrix where the entries are the eigenvalues of A. Furthermore, any matrix which diagonalizes in this way must be normal.

The column vectors of U are the eigenvectors of A and they are orthogonal.

It could be viewed as a special case of Schur decomposition.

If A is a real symmetric matrix, it follows by the real version of the spectral theorem for symmetric operators that there is an orthogonal matrix such that U A U* is diagonal and all the eigenvalues of A are real.

The spectral theorem for compact symmetric operators

In Hilbert spaces in general, the statement of the spectral theorem for cmpact symmetric operators is virtually the same as in the finite-dimensional case.

Theorem. Suppose A is a compact symmetric operator on a Hilbert space. There is an orthonormal basis of V consisting of eigenvectors of A. Each eigenvalue is real.

Again the key point is to prove the existence of at least one nonzero eigenvector. To prove this, we cannot rely on determinants to show existence of eigenvalues, but instead we use a maximization argument analogous to proving the min-max theorem for eigenvalues.

Note that the above spectral theorem holds for real or complex Hilbert spaces.

Functional analysis

The next generalization we consider is that of bounded self-adjoint operators A on a Hilbert space V. Such operators may have no eigenvalues: for instance let A be the operator multiplication by t on L²[0,1], that is

<math> [A \phi](t) = t \phi(t). \;</math>

Theorem. Let A be a bounded self-adjoint operator on a Hilbert space H. Then there is a measure space (X, M, μ) and a real-valued measurable function f on X and a unitary operator U:H → L²_μ(X) such that

<math> U^* T U = A \;</math>

where T is the multiplication operator:

<math> [T \phi](x) = f(x) \phi(x). \;</math>

This is the beginning of the vast research area of functional analysis called operator theory.

A normal operator on a Hilbert space may have no eigenvalues; for example, the bilateral shift on the Hilbert space l²(Z) has no eigenvalues. There is also a spectral theorem for normal operators on Hilbert spaces, though, in which the sum in the finite-dimensional spectral theorem is replaced by an integral of the coordinate function over the spectrum against a projection-valued measure.

When the normal operator in question is compact, this spectral theorem reduces to the finite-dimensional spectral theorem above, except that the operator is expressed as a linear combination of possibly infinitely many projections.

See also

• Matrix decomposition
• Jordan decomposition, an "algebraic" analogue to spectral decomposition.
• Singular value decomposition, a generalisation of spectral theorem to arbitrary matrices.

Reference

• Sheldon Axler, Linear Algebra Done Right, Springer Verlag, 1997

References

• Adapted from the Wikipedia article, "Spectral_theorem" http://en.wikipedia.org/wiki/Spectral_theorem, used under the GNU Free Documentation License