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 "P_i" are pairwise orthogonal projections. "TN = NT" if and only if "TP_i = P_iT". Therefore "T" must also commute with

:

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

: A limiting argument then shows that "T" commutes with

:

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 $mathbb\{C\}$,:.

Consider the function:$F(lambda)\; =\; e^\{lambda\; M^*\}\; T\; e^\{-lambda\; N^*\}$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:$|F(lambda)|\; le\; |T|\; forall\; lambda$

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

:

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

:$T\text{'}\; (N\text{'})^*\; =\; (N\text{'})^*T\text{'}.\; ,$

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

:$=\; M^*\; N^*\; MN\; =\; (MN)^*\; MN.\; ,$