# commutant of $B(H)$ is $\u2102I$

Let $H$ be a Hilbert space^{} and $B(H)$ its algebra of bounded operators^{}. We denote by $I$ the identity operator of $B(H)$ and by $\u2102I$ the set of all multiples of $I$, that is $\u2102I:=\{\lambda I:\lambda \in \u2102\}$. Let $B{(H)}^{\prime}$ denote the commutant of $B(H)$, which is precisely the center of $B(H)$.

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Theorem - We have that $B{(H)}^{\prime}=\u2102I$.

As a particular case, we see that the center of the matrix algebra $Ma{t}_{n\times n}(\u2102)$ consists solely of the multiples of the identity matrix^{}, i.e. a matrix in $Ma{t}_{n\times n}(\u2102)$ that commutes with all other matrices is necessarily a multiple of the identity matrix.

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*:* For each $x,y\in H$ we denote by ${T}_{x,y}$ the operator given by

${T}_{x,y}z:=\u27e8z,x\u27e9y,z\in H$ |

Let $S\in B{(H)}^{\prime}$. We must have $S{T}_{x,y}={T}_{x,y}S$ for all $x,y\in H$, hence

$\u27e8z,x\u27e9Sy=\u27e8Sz,x\u27e9y,\forall x,y,z\in H$ | (1) |

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

$Sy={\displaystyle \frac{\u27e8Sx,x\u27e9}{\u27e8x,x\u27e9}}y,y\in H,x\in H\setminus \{0\}$ |

Hence, $\frac{\u27e8Sx,x\u27e9}{\u27e8x,x\u27e9}$ must be constant for all $x\in H\setminus \{0\}$. Denote by $\lambda \in \u2102$ this constant.

We have that $Sy=\lambda y$ for all $y\in H$, which simply means that $S=\lambda I$. Thus, $B{(H)}^{\prime}\subseteq \u2102I$.

It is clear that the multiples of the identity operator commute with all operators, hence we also have $\u2102I\subseteq B{(H)}^{\prime}$.

We conclude that $B{(H)}^{\prime}=\u2102I$. $\mathrm{\square}$

Title | commutant of $B(H)$ is $\u2102I$ |
---|---|

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 $B(H)$ |