# Commutant of bounded linear operators on a Hilbert space

Given a Hilbert space $H$, denote by $\mathcal{A}=\mathcal{B}(H)$ the C*-algebra of bounded linear operators on $H$. Denote further by

$$\mathcal{B}(H)' := \{A\in \mathcal{B}(H) : [A,B]=0 \;\forall B \in \mathcal{B}(H)\}$$

the commutant of $\mathcal{B}(H)$.

I think $\mathcal{B}(H)' = \{\lambda \mathbb{1}:\lambda \in \mathbb{C}\}$, but how can proof this?

Are the elements of $\mathcal{B}(H)'$ invertible in $\mathcal{B}(H)'$? (Then my claim would follow by the Gelfand-Mazur theorem...)

If $A$ is in the commutant and $x$ is any non-zero vector, then the fact that $A$ commutes with the orthogonal projection to the line spanned by $x$ implies that $Ax$ is a scalar multiple $\lambda x$ of $x$. The scalar has to be the same for all $x$ because if $x$ and $y$ are multiplied by different scalars, then $x+y$ would not be sent to a scalar multiple of itself.
I don't know if one can directly argue that all nonzero elements of $\mathcal{B}(H)'$ are invertible in $\mathcal{B}(H)'$, but it follows.
If $A \in \mathcal{B}(H)$ commutes with every bounded operator, in particular it commutes with all orthogonal projections. That means every subspace of $H$ is an invariant subspace of $A$, in particular all nonzero elements of $H$ are eigenvectors. If $A$ had more than one eigenvalue, the sum of two eigenvectors to different eigenvalues wouldn't be an eigenvector.
Combine the following: $$\mathbb{C}\mathbb{1}=\overline{\mathbb{C}\mathbb{1}}\quad\overline{\mathbb{C}\mathbb{1}}=\mathbb{C}\mathbb{1}''\quad\mathbb{C}\mathbb{1}'=\mathcal{B}(\mathcal{H})$$ The closure being the strong.