Left shift operators Let $T$ be the left shift operator on $B(l^{2}(\mathbb{N}))$. How to see that von Neumann algebra generated by $T$ is $B(l^{2}(\mathbb{N}))$?
 A: Let $A$ be a bounded operator. If I understand correctly, then it is enough to construct a sequence of operators $A_n$, which are polynomial in $T$ and $T^*$, such that
$$
\langle (A-A_n)e_i, e_j\rangle \to 0 \quad \forall i,j.
$$
Given $i,j$ let me construct $A_{ij}$ such that
$$
\langle (A_{ij}-A)e_i,e_j\rangle = 0, \ A_{ij}e_k = 0\ \forall k\ne i.
$$
Since $T^*$ is equal to right-shift, we have
$$
(I-T^*T)e_1 = e_1, \ 
(I-T^*T)e_k = 0 \ \forall k\ne 1.
$$
Hence the operator $(T^*)^{j-1}(I-T^*T)T^{i-1}$ maps $e_i$ to $e_j$, and $e_k$ to zero for all $k\ne 0$. Then
$$
A_{ij}:= \langle Ae_i,e_j\rangle\dot (T^*)^{j-1}(I-T^*T)T^{i-1}
$$
Now define
$$
A_n:=\sum_{i,j: i+j\le n} A_{ij}.
$$
A: The von Neumann algebra generated by $T$ is the double commutant of the $*$-algebra generated by $T$, let's call that $S$. Now, I claim that the commutant of $S$ is $\mathbb{C}1$. For any operator $A$ on $l^{2}(\mathbb{N})$ we can compute its matrix coeffecients $A_{ij} = \langle e_{i}, A e_{j} \rangle$. Given an operator $A$ in the commutant of $S$, i.e. $[T^{n},A] = 0$. We get
\begin{align*}
 0 = \langle e_{0}, [T^{n},A]e_{j} \rangle &= \langle (T^{*})^{n}e_{0}, Ae_{j} \rangle - \langle e_{i}, A T^{n}e_{j} \rangle\\
&= \langle e_{n}, Ae_{j} \rangle - \langle e_{0}, A e_{j-n} \rangle
\end{align*}
Similarly, we compute the matrix coefficients of $[(T^{*})^{n},A]$,
\begin{equation*}
0 = \langle e_{i}, [(T^{*})^{n},A]e_{0} \rangle = \langle e_{i-n},Ae_{0} \rangle - \langle e_{i}, A e_{n} \rangle.
\end{equation*}
Now if $n > j$, then $e_{j-n} = 0$. Hence, these equations imply, respectively, that for $n > j$
\begin{equation*}
 0 = \langle e_{n}, A e_{j} \rangle,
\end{equation*}
and for $n > i$
\begin{equation}
 0 = \langle e_{i}, A e_{n} \rangle.
\end{equation}
In other words, all off-diagonal entries of $A$ are equal to zero. We complete the proof by showing that all diagonal entries need to agree as well:
\begin{equation*}
 0 = \langle e_{j-1}, [T,A] e_{j} \rangle = \langle e_{j}, A e_{j} \rangle - \langle e_{j - 1}, A e_{j-1} \rangle, \quad j > 1.
\end{equation*}
