commutant of $B(H)$ is $ℂI$

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

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

$T_{x,y}z:=⟨z,x⟩y,z\in H$

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

$⟨z,x⟩Sy=⟨Sz,x⟩y,\forall x,y,z\in H$(1)

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

$Sy={\displaystyle \frac{⟨Sx,x⟩}{⟨x,x⟩}}y,y\in H,x\in H\setminus \{0\}$

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

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

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

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