Let $\mathcal{M}$ be a finite dimensional von Neumann algebra, then : $$\mathcal{M} \simeq \bigoplus_i M_{n_i}(\mathbb{C})$$
Question : Is it singly generated (as von Neumann algebra)? how ?
Allowed operations : $() \mapsto I$ , $(A) \mapsto \lambda A$ or $\mathbf{A^*}$ and $(A,B) \mapsto A+B$ or $A B$
Pick distinct complex numbers $\lambda_1,\ldots,\lambda_k$ and consider the element $$ X:=\Bigg(\,\underbrace{\begin{smallmatrix} \lambda_1&1&0&0&0\\ 0&\lambda_1&1&0&0\\ 0&0&\ddots&\ddots&0\\ 0&0&0&\lambda_1&1\\ 0&0&0&0&\lambda_1\\ \end{smallmatrix}}_{n_1}\,\Bigg) \oplus \Bigg(\,\underbrace{\begin{smallmatrix} \lambda_2&1&0&0&0\\ 0&\lambda_2&1&0&0\\ 0&0&\ddots&\ddots&0\\ 0&0&0&\lambda_2&1\\ 0&0&0&0&\lambda_2\\ \end{smallmatrix}}_{n_2}\,\Bigg) \oplus \cdots\oplus \Bigg(\,\underbrace{\begin{smallmatrix} \lambda_k&1&0&0&0\\ 0&\lambda_k&1&0&0\\ 0&0&\ddots&\ddots&0\\ 0&0&0&\lambda_k&1\\ 0&0&0&0&\lambda_k\\ \end{smallmatrix}}_{n_k}\,\Bigg) $$ I claim that for generic $\lambda_i$ that element generates your algebra.
First of all, $(X-\lambda_j)^{n_j}$ has zero for its $j$th component. Taking products of such elements, one can achieve something whose only non-zero entry is in a given summand (say the $i$th summand). The remaining term is Toeplitz and upper triangular. More precisely, it is given by $$ A:=\left(\begin{matrix} a&b&c&d&e\\ 0&a&b&c&d\\ 0&0&\ddots&\ddots&c\\ 0&0&0&a&b\\ 0&0&0&0&a\\ \end{matrix}\,\right) $$ with $a=\prod_{j\not =i}(\lambda_i-\lambda_j)^{n_j}$ and $b=$ (some horrible expression which is obviously positive if all the $(\lambda_i-\lambda_j)$ are positive, and therefore non-zero if the $\lambda_i$ are chosen generically).
Then it's a matter of playing around to see that the above matrix and its adjoint generate all of $M_{n_i}(\mathbb C)$. The first operation is to consider $N:=A^2-aA$, which is nilpotent with non-zero terms on the subdiagonal. Then you take an appropriate linear combination of powers of $N$ to generate the element $$ \left(\begin{matrix} 0&1&0&0&0\\ 0&0&1&0&0\\ 0&0&\ddots&\ddots&0\\ 0&0&0&0&1\\ 0&0&0&0&0\\ \end{matrix}\,\right). $$ Then you take the transpose of that element, and you're essentially done.
For the sake of completeness, here are more general statements:
1) Every separable type I factor is singly generated. You take the weighted shift $S=\sum_j 2^{-j}E_{j,j+1}$. Then $W^*(S)$ contains $(S^*S)^{1/2}=\sum_j2^{-j}E_{jj}$ and now by using the characteristic function of $\{2^{-j}\}$ we obtain $E_{jj}\in W^*(S)$ for all $j$. Now $E_{k,k+1}=2^kE_{kk}S\in W^*(S)$ and we can get the rest of the matrix units as in $E_{k,k+2}=E_{k,k+1}E_{k+1,k+2}$, etc. Then $W^*(S)$ contains all matrix units and is then equal to $B(H)$.
2) A countable direct sum of singly generated is singly generated. If $\mathcal M=\bigoplus_{j\in J}\mathcal M_j$, $J\subset\mathbb N$, we take generators $M_j\in\mathcal M_j$ with $\sigma(M_j)\subset[3/4,1]$ (note that $M$ is a generator if and only if $\alpha M+\beta I$ is a generator for any $\alpha,\beta\in\mathbb C$). Now $M=\bigoplus_j2^{-j}M_j$ is a generator, as we can isolate each summand via functional calculus as in 1).
So 1) and 2) address Sébastien's question.
To go further, a generalization of 1) shows that any separable infinite factor (i.e. II$_\infty$ or III) is singly generated: for type II$_\infty$, you take a dense countable subset of the positive unit ball, normalize all the elements, and form the "weighted shift" as in 1). For type III, you use that $M=M\otimes B(H)$ and you can also form the weighted shift.
3) Separable abelian von Neumann algebras are also singly generated (by a selfadjoint), by a Theorem of von Neumann himself. A nice proof can be found in II.2.8 of Davidson's "C$^*$-algebras by Example".
4) Tensor product of separable abelian and singly generated is singly generated. This is a wonderful old trick. If $\mathcal A$ is separable abelian, it has a selfadjoint generator $a$. If $\mathcal M$ is singly generated, it has a generator $b+ic$ with $b,c$ selfadjoint. Then $W^*(a)\otimes W^*(b)$ is separable and abelian, and so it has a selfadjoint generator $x$; similarly $W^*(a)\otimes W^*(c)$ has a selfadjoint generator $y$. Then $x+iy$ is a generator for $\mathcal A\otimes\mathcal M$ (since $W^*(x+iy)$ contains $a\otimes I$, and $I\otimes(b+ic)$).
5) A separable von Neumann algebra with no type II$_1$ summand is singly generated. Such an algebra is of the form $\bigoplus_j\mathcal A_j\otimes\mathcal M_j$, with each $\mathcal A_j$ abelian and separable, and each $\mathcal M_j$ a non II$_1$-factor (and so, singly generated).
So the "generator problem" for von Neumann algebras is reduced to the case of II$_1$-factors.
6) The question of whether all separable II$_1$-factors are singly generated is still open.
Take the element a=(1_{n_1},...,1_{n_k}), 1_n_i the identity matrix in M(C^{n_i}). You certainly then may generate all elements ;-).Formally: Wanting X=(X_1,...,X_k) (sum of matrices X_i, each from M(C^{n_i}), take surprisingly (X_1,...,X_k).a and you get the X. The multiplication in direct sum of algebras must be defined to make it a formal sense.
