Talk:Fuglede's theorem

Corollary: If two normal operators M and N are similar, then they are unitarily equivalent.[edit]

This proof is seriously flawed. Consider for instance any diagonal matrix M=N and any positive definite diagonal P. Let S be the positive square root of P. Then it is clear that MS=SM=SN. In the proof it is claimed that this implies that P is the identity on the range of M, but this is clearly false in general. I suspect one might fix the proof by considering the polar decomposition of S and using, say, that SS* commutes with M. Since S is invertible, the 'phase' is the sought after unitary. Isdatmaths (talk) 08:31, 9 December 2014 (UTC)[reply]

I have repaired the proof of the corollary following Isdatmaths' idea. Tinchote (talk) 16:02, 8 February 2015 (UTC)[reply]

Error in statement of Putnam's theorem?[edit]

Right now, Putnam's theorem reads as follows

   Let T, M, N be linear operators on a complex Hilbert space, and suppose that M and N are normal, M is bounded and MT = TN. Then M*T = TN*.

Going by the proof T should be required to be bounded, not M.

88.130.56.1 (talk) 20:59, 4 July 2020 (UTC)[reply]

@88.130.56.1: I agree. I just wanted to make the same comment after checking Putnam's paper (C. R. Putnam, On Normal Operators in Hilbert Space, "American Journal of Mathematics", Vol. 73, No. 2 (Apr., 1951), pp. 357--362). Putnam used N_1, N_2, and A, with the first two (here M and N) being normal and A (which here is T) bounded. As we came to the same conclusion independently, I will now make the change on the page. ESteiner (talk) 13:30, 18 August 2021 (UTC)[reply]