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Examples of operators to which the spectral theorem applies are self-adjoint operators or more generally normal operators on Hilbert spaces.
The spectral theorem also provides a canonical decomposition, called the spectral decomposition 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, for a Hilbert space, the spectral theorem also holds for normal operators.
We begin by considering a symmetric operator A on a finite dimensional inner product space V; the symmetry condition means
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. The value 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 eigenvalue of A, for the corresponding eigenvector x
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:
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 implies 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 an orthonormal basis of eigenvectors, is obtained by grouping together all vectors corresponding to the same eigenvalue. Thus
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 Vλ where the index ranges over eigenvalues. Another equivalent formulation is letting Pλ be the orthogonal projectionIn geometry, an orthogonal projection of a k dimensional object onto a d dimensional hyperplane d < k is obtained by intersections of k − d dimensional hyperplanes drawn through the points of an object orthogonally to the d hyperplane. In particular onto Vλ
and if λ1,..., λm are the eigenvalues of A,
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 matrixA complex square matrix A is a normal matrix iff : where A is the conjugate transpose of A (if A is a real matrix, this is the same as the transpose of A . Examples All unitary matrices, hermitian matrices and positive definite matrices are normal. If A i, there exists a unitary matrixIn mathematics, a unitary matrix is a n by n complex matrix U satisfying the condition U ''U UU I where I is the identity matrix and U is the conjugate transpose (also called the Hermitian adjoint) of U''. Note this condition says that a matrix U is unita U such that
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.
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 such that U A U* is diagonal and all the eigenvalues of A are real.