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", 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: formula_1 Tentative Proof: If the underlying Hilbert space is finite-dimensional, the spectral theorem says that "N" is of the form  formula_2 where "P" are pairwise orthogonal projections. One expects that "TN" = "NT" if and only if "TP" = "PT". Indeed it can be proved to be true by elementary arguments (e.g. it can be shown that all "P" are representable as polynomials of "N" and for this reason, if "T" commutes with "N", it has to commute with "P"...). Therefore "T" must also commute with formula_3 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 "P" to each  Borel subset of "σ"("N"). "N" can be expressed as formula_4 Differently from the finite dimensional case, it is by no means obvious that "TN = NT" implies "TP" = "P""T". Thus, it is not so obvious that "T" also commutes with any simple function of the form formula_5 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 formula_6, the most straightforward way is to assume that "T" commutes with both "N" and "N*", 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 theorem 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, "T" is bounded and "MT" = "TN". Then "M"*"T" = "TN"*. First proof (Marvin Rosenblum): By induction, the hypothesis implies that "M""T" = "TN" for all "k". Thus for any λ in formula_7, formula_8  Consider the function formula_9 This is equal to formula_10 where formula_11 because formula_12 is normal, and similarly formula_13. However we have formula_14 so U is unitary, and hence has norm 1 for all λ; the same is true for "V"(λ), so formula_15 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 formula_16 The operator "N' " is normal and, by assumption, "T' N' = N' T' ". By Fuglede's theorem, one has formula_17 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. formula_18 Take the adjoint of the above equation and we have formula_19 So formula_20 Let "S*=VR", with "V" a unitary (since "S" is invertible) and "R" the positive square root of "SS*". As "R" is a limit of polynomials on "SS*", the above implies that "R" commutes with "M". It is also invertible. Then formula_21 Corollary If "M" and "N" are normal operators, and "MN" = "NM", then "MN" is also normal. Proof: The argument invokes only Fuglede's theorem. One can directly  compute formula_22 By Fuglede, the above becomes formula_23 But "M" and "N" are normal, so formula_24 Section::::"C*"-algebras. The theorem can be rephrased as a statement about elements of C*-algebras. Theorem (Fuglede-Putnam-Rosenblum) Let "x, y" be two normal elements of a "C*"-algebra "A" and "z" such that "xz" = "zy". Then it follows that "x* z = z y*".