[proofplan] We compute the center of $\mathcal{L}(H)$ directly. The main point is that if an operator $T\in\mathcal{L}(H)$ commutes with every rank-one operator, then every vector of $H$ is an eigenvector of $T$ with the same eigenvalue. This forces $T$ to be a scalar multiple of the identity, and scalar multiples of the identity are exactly the central elements of $\mathcal{L}(H)$. [/proofplan]

[step:Choose a unit vector and define the rank-one test operators] Since $H\ne\{0\}$, choose a vector $\xi_0\in H$ with $\xi_0\ne 0$. Define \begin{align*} \xi:=\frac{\xi_0}{\|\xi_0\|_H}. \end{align*} Then $\|\xi\|_H=1$. For each $\zeta\in H$, define the rank-one operator $R_\zeta:H\to H$ by \begin{align*} R_\zeta(\eta):=(\eta,\xi)_H\zeta \end{align*} for every $\eta\in H$. This map is linear because the [Hilbert space](/page/Hilbert%20Space) [inner product](/page/Inner%20Product) is linear in the first variable. Moreover, by the [Cauchy-Schwarz inequality](/theorems/432) applied to the vectors $\eta$ and $\xi$, \begin{align*} \|R_\zeta(\eta)\|_H=|(\eta,\xi)_H|\|\zeta\|_H\le \|\eta\|_H\|\xi\|_H\|\zeta\|_H=\|\eta\|_H\|\zeta\|_H. \end{align*} Thus $R_\zeta\in\mathcal{L}(H)$ and $\|R_\zeta\|_{\mathcal{L}(H)}\le \|\zeta\|_H$. [/step]

[step:Show that a central operator is scalar by testing against rank-one operators]Let $T\in Z(\mathcal{L}(H))$. This means that $T\in\mathcal{L}(H)$ and \begin{align*} TS=ST \end{align*} for every $S\in\mathcal{L}(H)$. In particular, for every $\zeta\in H$, the operator $T$ commutes with $R_\zeta$. Define the scalar $\lambda\in\mathbb{C}$ by \begin{align*} \lambda:=(T\xi,\xi)_H. \end{align*} For an arbitrary vector $\zeta\in H$, using $TR_\zeta=R_\zeta T$ and evaluating at $\xi$ gives \begin{align*} T\zeta=T(R_\zeta\xi)=(TR_\zeta)(\xi)=(R_\zeta T)(\xi)=R_\zeta(T\xi)=(T\xi,\xi)_H\zeta=\lambda\zeta. \end{align*} Since $\zeta\in H$ was arbitrary, $T=\lambda I_H$. Therefore \begin{align*} Z(\mathcal{L}(H))\subseteq \mathbb{C}I_H. \end{align*}[/step]

[guided]Let $T\in Z(\mathcal{L}(H))$. By definition of the center, $T$ is a [bounded linear operator](/page/Bounded%20Linear%20Operator) on $H$ and it commutes with every bounded linear operator on $H$. The point of introducing the rank-one operators $R_\zeta$ is that they isolate the vector $\zeta$ when evaluated at the chosen unit vector $\xi$. For each $\zeta\in H$, we already know that $R_\zeta\in\mathcal{L}(H)$, so centrality of $T$ gives \begin{align*} TR_\zeta=R_\zeta T. \end{align*} Define \begin{align*} \lambda:=(T\xi,\xi)_H. \end{align*} This is a complex scalar because the inner product on $H$ takes values in $\mathbb{C}$. Now fix an arbitrary vector $\zeta\in H$. Since $\|\xi\|_H=1$, we have \begin{align*} R_\zeta\xi=(\xi,\xi)_H\zeta=\zeta. \end{align*} Therefore \begin{align*} T\zeta=T(R_\zeta\xi). \end{align*} Using the commutation relation $TR_\zeta=R_\zeta T$, we obtain \begin{align*} T(R_\zeta\xi)=(TR_\zeta)(\xi)=(R_\zeta T)(\xi)=R_\zeta(T\xi). \end{align*} Finally, by the definition of $R_\zeta$, \begin{align*} R_\zeta(T\xi)=(T\xi,\xi)_H\zeta=\lambda\zeta. \end{align*} Combining these equalities gives \begin{align*} T\zeta=\lambda\zeta. \end{align*} Because $\zeta\in H$ was arbitrary, the same scalar $\lambda$ works for every vector in $H$. Hence $T=\lambda I_H$, so every central operator is a scalar multiple of the identity: \begin{align*} Z(\mathcal{L}(H))\subseteq \mathbb{C}I_H. \end{align*}[/guided]

[step:Verify that scalar operators are central and conclude factorhood] Conversely, let $\lambda\in\mathbb{C}$. For every $S\in\mathcal{L}(H)$ and every $\eta\in H$, \begin{align*} (\lambda I_H)S\eta=\lambda S\eta=S(\lambda\eta)=S(\lambda I_H\eta). \end{align*} Thus $(\lambda I_H)S=S(\lambda I_H)$ for every $S\in\mathcal{L}(H)$, so $\lambda I_H\in Z(\mathcal{L}(H))$. Hence \begin{align*} \mathbb{C}I_H\subseteq Z(\mathcal{L}(H)). \end{align*} Together with the reverse inclusion proved above, \begin{align*} Z(\mathcal{L}(H))=\mathbb{C}I_H. \end{align*} By the definition of a factor, a von Neumann algebra whose center is exactly the scalar multiples of its identity is a factor. Therefore $\mathcal{L}(H)$ is a factor. [/step]