[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*}
[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]
[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]
