Fuglede's theorem
Encyclopedia
In mathematics
, Fuglede's theorem is a result in operator theory
, named after Bent Fuglede
.
. If TN = NT, then TN* = N*T, where N* denotes the adjoint
of N.
Normality of N is necessary, as is seen by taking T=N. When T is self-adjoint, the claim is trivial regardless of whether N is normal:
Tentative Proof: If the underlying Hilbert space is finite-dimensional, the spectral theorem
says that N is of the form
where Pi are pairwise orthogonal projections. One aspects that
TN = NT if and only if TPi = PiT.
Indeed it can be proved to be true by elementary arguments.
Therefore T must also commute with
In general, when the Hilbert space is not finite-dimensional, the normal operator N gives rise to a projection-valued measure
P on its spectrum, σ(N), which assigns a projection PΩ to each Borel subset of σ(N). N can be expressed as
Differently from the finite dimensional case, it is by no means obvious that TN = NT implies TPΩ = PΩT. Thus, it is not so obvious that T also commutes with any simple function of the form
Indeed, following the construction of the spectral decomposition for a bounded, normal, not self-adjoint, operator T, one sees that to verify that T
commutes with
, the most straightforward way is to assume that T commutes with both N and 
, giving rise to a vicious circle!
That is the relevance of Fuglede's theorem: The latter hypothesis is not really necessary.
when assuming N=M.
Theorem (Calvin Richard Putnam) Let T, M, N be linear operators on a complex Hilbert space
, and suppose that M and N are normal
and MT = TN.
Then M*T = TN*.
First proof (Marvin Rosenblum):
By induction, the hypothesis implies that MkT = TNk for all k.
Thus for any λ in
,
Consider the function
This is equal to
,
where
and 
. However we have
so U is unitary, and hence has norm 1 for all λ; the same is true for V(λ), so
So F is a bounded analytic vector-valued function, and is thus constant, and equal to F(0) = T. Considering the first-order terms in the expansion for small λ, we must have M*T = TN*.
The original paper of Fuglede appeared in 1950; it was extended to the form given above by Putnam in 1951. The short proof given above was first published by Rosenblum in 1958; it is very elegant, but is less general than the original proof which also considered the case of unbounded operators. Another simple proof of Putnam's theorem is as follows:
Second proof: Consider the matrices
The operator N' is normal and, by assumption, T' N' = N' T' . By Fuglede's theorem, one has
Comparing entries then gives the desired result.
From Putnam's generalization, one can deduce the following:
Corollary If two normal operators M and N are similar, then they are unitarily equivalent.
Proof: Suppose MS = SN where S is a bounded invertible operator. Putnam's result implies M*S = SN*, i.e.
Take the adjoint of the above equation and we have
So
Therefore, on Ran(M), SS* is the identity operator. SS* can be extended to Ran(M)⊥ = Ker(M). Therefore, by normality of M, SS* = I, the identity operator. Similarly, S*S = I. This shows that S is unitary.
Corollary If M and N are normal operators, and MN = NM, then MN is also normal.
Proof: The argument invokes only Fuglede's theoerm. One can directly compute
By Fuglede, the above becomes
But M and N are normal, so
Mathematics
Mathematics is the study of quantity, space, structure, and change. Mathematicians seek out patterns and formulate new conjectures. Mathematicians resolve the truth or falsity of conjectures by mathematical proofs, which are arguments sufficient to convince other mathematicians of their validity...
, Fuglede's theorem is a result in operator theory
Operator theory
In mathematics, operator theory is the branch of functional analysis that focuses on bounded linear operators, but which includes closed operators and nonlinear operators.Operator theory also includes the study of algebras of operators....
, named after Bent Fuglede
Bent Fuglede
Bent Fuglede is a Danish mathematician and, since 1992, retired professor from University of Copenhagen.He is known for his contributions to mathematical analysis, in particular functional analysis, where he has proved Fuglede's theorem and stated Fuglede's conjecture.Fuglede graduated from Skt....
.
The result
Theorem (Fuglede) Let T and N be bounded operators on a complex Hilbert space with N being normalNormal operator
In mathematics, especially functional analysis, a normal operator on a complex Hilbert space H is a continuous linear operatorN:H\to Hthat commutes with its hermitian adjoint N*: N\,N^*=N^*N....
. If TN = NT, then TN* = N*T, where N* denotes the adjoint
Hermitian adjoint
In mathematics, specifically in functional analysis, each linear operator on a Hilbert space has a corresponding adjoint operator.Adjoints of operators generalize conjugate transposes of square matrices to infinite-dimensional situations...
of N.
Normality of N is necessary, as is seen by taking T=N. When T is self-adjoint, the claim is trivial regardless of whether N is normal:
Tentative Proof: If the underlying Hilbert space is finite-dimensional, the spectral theorem
Spectral theorem
In mathematics, particularly linear algebra and functional analysis, the spectral theorem is any of a number of results about linear operators or about matrices. In broad terms the spectral theorem provides conditions under which an operator or a matrix can be diagonalized...
says that N is of the form
where Pi are pairwise orthogonal projections. One aspects that
TN = NT if and only if TPi = PiT.
Indeed it can be proved to be true by elementary arguments.
Therefore T must also commute with
In general, when the Hilbert space is not finite-dimensional, the normal operator N gives rise to a projection-valued measure
Projection-valued measure
In mathematics, particularly functional analysis a projection-valued measure is a function defined on certain subsets of a fixed set and whose values are self-adjoint projections on a Hilbert space...
P on its spectrum, σ(N), which assigns a projection PΩ to each Borel subset of σ(N). N can be expressed as
Differently from the finite dimensional case, it is by no means obvious that TN = NT implies TPΩ = PΩT. Thus, it is not so obvious that T also commutes with any simple function of the form
Indeed, following the construction of the spectral decomposition for a bounded, normal, not self-adjoint, operator T, one sees that to verify that T
commutes with
, the most straightforward way is to assume that T commutes with both N and , giving rise to a vicious circle!That is the relevance of Fuglede's theorem: The latter hypothesis is not really necessary.
Putnam's generalization
The following contains Fuglede's result as a special case. The proof by Rosenblum pictured below is just that presented by Fuglede for his theoremwhen assuming N=M.
Theorem (Calvin Richard Putnam) Let T, M, N be linear operators on a complex Hilbert space
Hilbert space
The mathematical concept of a Hilbert space, named after David Hilbert, generalizes the notion of Euclidean space. It extends the methods of vector algebra and calculus from the two-dimensional Euclidean plane and three-dimensional space to spaces with any finite or infinite number of dimensions...
, and suppose that M and N are normal
Normal operator
In mathematics, especially functional analysis, a normal operator on a complex Hilbert space H is a continuous linear operatorN:H\to Hthat commutes with its hermitian adjoint N*: N\,N^*=N^*N....
and MT = TN.
Then M*T = TN*.
First proof (Marvin Rosenblum):
By induction, the hypothesis implies that MkT = TNk for all k.
Thus for any λ in
,Consider the function
This is equal to
,where
and . However we haveso U is unitary, and hence has norm 1 for all λ; the same is true for V(λ), so
So F is a bounded analytic vector-valued function, and is thus constant, and equal to F(0) = T. Considering the first-order terms in the expansion for small λ, we must have M*T = TN*.
The original paper of Fuglede appeared in 1950; it was extended to the form given above by Putnam in 1951. The short proof given above was first published by Rosenblum in 1958; it is very elegant, but is less general than the original proof which also considered the case of unbounded operators. Another simple proof of Putnam's theorem is as follows:
Second proof: Consider the matrices
The operator N' is normal and, by assumption, T' N' = N' T' . By Fuglede's theorem, one has
Comparing entries then gives the desired result.
From Putnam's generalization, one can deduce the following:
Corollary If two normal operators M and N are similar, then they are unitarily equivalent.
Proof: Suppose MS = SN where S is a bounded invertible operator. Putnam's result implies M*S = SN*, i.e.
Take the adjoint of the above equation and we have
So
Therefore, on Ran(M), SS* is the identity operator. SS* can be extended to Ran(M)⊥ = Ker(M). Therefore, by normality of M, SS* = I, the identity operator. Similarly, S*S = I. This shows that S is unitary.
Corollary If M and N are normal operators, and MN = NM, then MN is also normal.
Proof: The argument invokes only Fuglede's theoerm. One can directly compute
By Fuglede, the above becomes
But M and N are normal, so