Fuglede's theorem

Fuglede's theorem

In mathematics, Fuglede's theorem is a result in operator theory, named after Bent Fuglede.

The result

Theorem (Fuglede) Let "T" and "N" be bounded operators on a complex Hilbert space with "N" being normal. If "TN = NT", then "TN*" = "N*T".

Colloquially, the theorem claims that commutativity between operators is transitive under the given assumptions. The claim does not hold in general if "N" is not normal. A simple counterexample is provided by letting "N" be the unilateral shift and "T = N". Also, when "T" is self-adjoint, the claim is trivial regardless of whether "N" is normal:

:TN^* = (NT)^* = (TN)^* = N^*T.,

Proof: If the underlying Hilbert space is finite-dimensional, the spectral theorem says that "N" is of the form

:N = sum_i lambda_i P_i ,

where "Pi" are pairwise orthogonal projections. "TN = NT" if and only if "TPi = PiT". Therefore "T" must also commute with

:N^* = sum_i {ar lambda_i} P_i.

In general, the normal operator "N" gives rise to a projection-valued measure "P" on its spectrum, "σ"("N"), which assigns a projection "P"Ω to each Borel subset of "σ"("N"). "N" can be expressed as

:N = int_{sigma(N)} lambda d P(lambda). ,

As in the finite dimensinal case, "TN = NT" implies "TP"Ω = "P"Ω"T". Thus "T" commutes with any simple function of the form

: ho = sum_i {ar lambda} P_{Omega_i}. A limiting argument then shows that "T" commutes with

:N^* = int_{sigma(N)} {ar lambda} d P(lambda). ,

Putnam's generalization

The following contains Fuglede's result as a special case.

Theorem (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 (Rosenblum ):By induction, the hypothesis implies that "M""k""T" = "TN""k" for all k. Thus for any λ in mathbb{C},:e^{arlambda M}T = T e^{arlambda N}.

Consider the function:F(lambda) = e^{lambda M^*} T e^{-lambda N^*}This is equal to :e^{lambda M^*} left [e^{-arlambda M}T e^{arlambda N} ight] e^{-lambda N^*} = U(lambda) T V(lambda)^{-1},where U(lambda) = e^{lambda M^* - arlambda M} and V(lambda) = e^{lambda N^* - arlambda N}. However we have:U(lambda)^* = e^{arlambda M - lambda M^*} = U(lambda)^{-1}so U is unitary, and hence has norm 1 for all λ; the same is true for "V"(λ), so:|F(lambda)| le |T| forall lambda

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

:T' = egin{bmatrix}0 & 0\ T & 0end{bmatrix}

quad mbox{and} quad

N' = egin{bmatrix}N & 0 \ 0 & Mend{bmatrix}.

The operator "N' " is normal and, by assumption, "T' N' = N' T' ". By Fuglede's theorem, one has

:T' (N')^* = (N')^*T'. ,

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.

:S^{-1} M^* S = N^*. ,

Take the adjoint of the above equation and we have

:S^* M (S^{-1})^* = N. ,

So

:S^* M (S^{-1})^* = S^{-1} M S quad Rightarrow quad SS^* M (SS^*)^{-1} = M.

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

:(MN) (MN)^* = MN (NM)^* = MN M^* N^*. ,

By Fuglede, the above becomes

:= M M^* N N^* = M^* M N^*N. ,

But "M" and "N" are normal, so

:= M^* N^* MN = (MN)^* MN. ,


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