A von Neumann algebra is the norm closed linear span of its projections. If we restrict to the case of matrix algebras which are themselves von Neumann algebras why is this the case? I think it should be easy to see but in the 2x2 case one has that the orthogonal projections are:
$\begin{bmatrix}1&0\\0&0\end{bmatrix}$$\begin{bmatrix}0&0\\0&1\end{bmatrix}$$\begin{bmatrix}1&0\\0&1\end{bmatrix}$
but all these matrices commute with eachother, how can the closed linear span of them be $M_{2\times2}(\mathbb{C})?$
You are missing $$ \begin{bmatrix} t&e^{i\theta}\sqrt{t-t^2}\\ e^{-i\theta}\sqrt{t-t^2}&1-t\end{bmatrix},\ \ t\in(0,1),\ \ \ \theta\in[0,2\pi]. $$ Any 4 of them, with different $t$, will be linearly independent, and as $\dim M_2(\mathbb C)=4$, they will span the algebra.