commutant of is
Let be a Hilbert space and its algebra of bounded operators. We denote by the identity operator of and by the set of all multiples of , that is . Let denote the commutant of , which is precisely the center of .
Theorem - We have that .
As a particular case, we see that the center of the matrix algebra consists solely of the multiples of the identity matrix, i.e. a matrix in that commutes with all other matrices is necessarily a multiple of the identity matrix.
: For each we denote by the operator given by
Let . We must have for all , hence
Choosing a non-zero and taking , we see that
Hence, must be constant for all . Denote by this constant.
We have that for all , which simply means that . Thus, .
It is clear that the multiples of the identity operator commute with all operators, hence we also have .
We conclude that .
|Title||commutant of is|
|Date of creation||2013-03-22 18:39:35|
|Last modified on||2013-03-22 18:39:35|
|Last modified by||asteroid (17536)|