# commutant of $B(H)$ is $\mathbb{C}I$

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 $\mathbb{C}I$ the set of all multiples of $I$, that is $\mathbb{C}I:=\{\lambda I:\lambda\in\mathbb{C}\}$. 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 $B(H)^{\prime}=\mathbb{C}I$.

As a particular case, we see that the center of the matrix algebra $Mat_{n\times n}(\mathbb{C})$ consists solely of the multiples of the identity matrix, i.e. a matrix in $Mat_{n\times n}(\mathbb{C})$ 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

 $\displaystyle T_{x,y}z:=\langle z,x\rangle y\,,\qquad\qquad z\in H$

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

 $\displaystyle\langle z,x\rangle Sy=\langle Sz,x\rangle y\,,\qquad\qquad\forall x% ,y,z\in H$ (1)

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

 $\displaystyle Sy=\frac{\langle Sx,x\rangle}{\langle x,x\rangle}y\,,\qquad% \qquad y\in H,x\in H\setminus\{0\}$

Hence, $\displaystyle\frac{\langle Sx,x\rangle}{\langle x,x\rangle}$ must be constant for all $x\in H\setminus\{0\}$. Denote by $\lambda\in\mathbb{C}$ this constant.

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

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

We conclude that $B(H)^{\prime}=\mathbb{C}I$. $\square$

Title commutant of $B(H)$ is $\mathbb{C}I$ CommutantOfBHIsmathbbCI 2013-03-22 18:39:35 2013-03-22 18:39:35 asteroid (17536) asteroid (17536) 7 asteroid (17536) Theorem msc 46L10 center of $B(H)$