Spectral Theorem Article Index for
Spectral 		Website Links For
Spectral 		 

Information About

Spectral Theorem
  CATEGORIES ABOUT SPECTRAL THEOREM
   
spectral theory
   
mathematical theorems
   
linear algebra
   
matrix theory
   
singular value decomposition
   
APPAREL
BABY
BEAUTY
BOOKS
CAR TOYS
CELL PHONES
DVD'S
ELECTRONICS
GOURMET FOOD
GROCERIES
HEALTH & PERSONAL
HOME & GARDEN
JEWELRY
MUSIC
MUSIC INSTRUMENTS
OFFICE PRODUCTS
SOFTWARE
SPORTING GOODS
TOOLS & HARDWARE
TOYS
VIDEO GAMES
SHOPPING HOME
MORE SHOPPING...



Examples of operators to which the spectral theorem applies are Self-adjoint Operator s or more generally Normal Operator s on Hilbert Space s.

The spectral theorem also provides a Canonical decomposition, called the spectral decomposition, or '''eigendecomposition''', of the underlying vector space on which it acts.

In this article we consider mainly the simplest kind of spectral theorem, that for a Self-adjoint operator on a Hilbert space. However, as noted above, the spectral theorem also holds for normal operators on a Hilbert space.


FINITE-DIMENSIONAL CASE



Hermitian matrices


We begin by considering a Hermitian Matrix ''A'' on a finite-dimensional Real or Complex Inner Product Space ''V'' with the standard Hermitian Inner Product ; in Dirac's Bra-ket Notation , the Hermitian condition means

: \langle A x \mid y angle = \langle x \mid A y angle

for all ''x'', ''y'' elements of ''V''.

  • = ''A'', where ''A''--- is the Conjugate Transpose of ''A''. If ''A'' is a real matrix, this is also equivalent to ''A''T = ''A''


Recall that an Eigenvector of a linear operator ''A'' is a (non-zero) vector ''x'' such that ''Ax'' = λ''x'' for some scalar λ. The value λ 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 for the case wherein the underlying field of scalars is the Complex Number s. First we show that all the eigenvalues are real. Suppose that λ is an eigenvalue of ''A'' with corresponding eigenvector ''x''. Thus

: \overline{\lambda} \langle x \mid x angle= \langle A x \mid x angle = \langle x \mid A x angle = \lambda \langle x \mid x angle .

Since ''x'' is non-zero, it follows that 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 because it is a subspace of a finite dimensional space, and ''A'' has the property that it maps every vector ''w'' in ''K'' into ''K''. This is shown as follows: If ''w'' ∈ ''K'', then using the symmetry property of ''A'',

: \langle A w \mid e angle = \langle w \mid A e angle = \langle w \mid \lambda e angle = \lambda \langle w \mid e angle = 0.

Moreover, ''A'' considered as a linear operator on ''K'' is also symmetric, so by the induction hypothesis there is a basis for ''V'' consisting of eigenvectors of ''A''.

It remains, however, to show that ''A'' has at least one eigenvector. Since the ground field is Algebraically Closed , the polynomial function (called the Characteristic Polynomial of ''A'')

:p(\lambda) = \det(\lambda I - A)

has a ''complex'' root ''r''. This implies the linear operator ''A'' − ''rI'' is not invertible and hence maps a non-zero vector ''e'' to 0. This vector ''e'' is a non-zero eigenvector of ''A''. This implies that ''r'' is an eigenvalue, so is actually a real number. This completes the proof.

Notice the second part of the proof works for ''any'' square matrices. Clearly any square matrix has at least one eigenvector. Therefore crucial to the argument is the following consequence of the Hermiticity of ''A'': If ''A'' is Hermitian and ''e'' is an eigenvector of ''A'', then not only is the linear span of ''e'' an invariant subspace of ''A'', but so is its orthogonal complement.

The argument is also valid for symmetric operators on finite-dimensional real inner product spaces, but the existence of an eigenvector is harder to establish. A real symmetric matrix has real eigenvalues, therefore eigenvectors with real entries.

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

: V_\lambda = \{\,v \in V: A v = \lambda v\,\}.

Note that these spaces are invariantly defined, in that the definition does not depend on any choice of specific eigenvectors.

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, letting ''P''λ be the Orthogonal Projection onto ''V''λ ( P_\lambda P_\mu=0 \quad \mbox{if } \lambda
eq \mu ) and λ1,..., λ''m'' the eigenvalues of ''A'', is

:A =\lambda_1 P_{\lambda_1} +\cdots+\lambda_m P_{\lambda_m}.

The spectral decomposition is a special case of the Schur Decomposition . It is also a special case of the Singular Value 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 ''U'' such that ''UAUT'' is diagonal and all the eigenvalues of ''A'' are Real .


Normal matrices


  • ''A'' = ''A A''---. One can show that ''A'' is normal if and only if it is unitarily diagonalizable: By the Schur decomposition, we have ''A'' = ''U T U''---, where ''U'' is unitary and ''T'' upper-triangular.

  • = ''T''--- ''T''. Therefore ''T'' must be diagonal. The converse is also obvious.


In other words, ''A'' is normal if and only if there exists a Unitary Matrix ''U'' such that

  • \;


where Λ is the Diagonal Matrix the entries of which are the Eigenvalue s of ''A''. The column vectors of ''U'' are the eigenvectors of ''A'' and they are orthonormal. Unlike the Hermitian case, the entries of Λ need not be real.


THE SPECTRAL THEOREM FOR COMPACT SELF-ADJOINT OPERATORS

See Also: Compact operator on Hilbert space


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

Theorem. Suppose ''A'' is a compact self-adjoint operator on a Hilbert space ''V''. There is an Orthonormal Basis of ''V'' consisting of eigenvectors of ''A''. Each eigenvalue is real.

As for Hermitian matrices, 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 one can use a maximization argument analogous to the variational characterization of eigenvalues. The above spectral theorem holds for real or complex Hilbert spaces.

If the compactness assumption is removed, it is not true that a self adjoint operator has eigenvectors.


GENERALIZATION TO NON-SYMMETRIC MATRICES


For a non-symmetric but square (N imes N dimensional) matrix \mathbf{A}, the ''right eigenvectors''
\mathbf{r}_{k} are defined by

:
\mathbf{A} \cdot \mathbf{r}_{k} = \lambda_{k} \mathbf{r}_{k}


whereas the ''left eigenvectors'' \mathbf{l}_{k} are defined by

:
\mathbf{l}_{k} \cdot \mathbf{A} = \lambda_{k} \mathbf{l}_{k}


or in the special case of a Real matrix, equivalently,

:
\mathbf{A}^{T} \cdot \mathbf{l}_{k} = \lambda_{k} \mathbf{l}_{k}


where \mathbf{A}^{T} represents the Transpose
of \mathbf{A}.

In other words, the right eigenvectors are the eigenvectors of ''A'' in the usual sense, while the left eigenvectors are the eigenvectors of A^T (the dual operator, thought of as an operator on the same space).

In these equations, the Eigenvalues
\lambda_{k} are the same, being the roots of the same Characteristic Polynomial

:
  \det \left \mathbf{A}^{T} - \lambda I Ight 0