commutant of B(H) is I

Let H be a Hilbert spaceMathworldPlanetmath and B(H) its algebra of bounded operatorsMathworldPlanetmathPlanetmath. We denote by I the identity operator of B(H) and by I the set of all multiples of I, that is I:={λI:λ}. Let B(H) denote the commutant of B(H), which is precisely the center of B(H).

Theorem - We have that B(H)=I.

As a particular case, we see that the center of the matrix algebra Matn×n() consists solely of the multiples of the identity matrixMathworldPlanetmath, i.e. a matrix in Matn×n() that commutes with all other matrices is necessarily a multiple of the identity matrix.

: For each x,yH we denote by Tx,y the operator given by


Let SB(H). We must have STx,y=Tx,yS for all x,yH, hence

z,xSy=Sz,xy,x,y,zH (1)

Choosing a non-zero x and taking z=x, we see that


Hence, Sx,xx,x must be constant for all xH{0}. Denote by λ this constant.

We have that Sy=λy for all yH, which simply means that S=λI. Thus, B(H)I.

It is clear that the multiples of the identity operator commute with all operators, hence we also have IB(H).

We conclude that B(H)=I.

Title commutant of B(H) is I
Canonical name CommutantOfBHIsmathbbCI
Date of creation 2013-03-22 18:39:35
Last modified on 2013-03-22 18:39:35
Owner asteroid (17536)
Last modified by asteroid (17536)
Numerical id 7
Author asteroid (17536)
Entry type Theorem
Classification msc 46L10
Synonym center of B(H)