Problem: How many $n \times n$ pairs $(A, B)$ of (0 , 1) matrices are there which generate the whole algebra $M_{n \times n}(\mathbb{\mathbb{C}})$? I don't know if this question is open or hopelessly hard. I would like to know related results and such. To give some background each such pair will lead to an irreducible representation for a quiver with one vertex and two loops or in more common language irreducible $\mathbb{C}<x, y>$ modules of dimension $n$
I could find one as follows:
From what we know it is enough that $A$ and $B$ generate All the elementary matrices $e_{i, j}$. The idea is to choose $A$ such that $A$ by its own is able to create shifts of rows (on left multiplication by $A$) and $B$ does analgously for columns!. This way we can use one $e_{i ,j}$ to generate the rest! What is the matrix that shifts the $1st$ row to the $2nd$ row?
\begin{bmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}
But we need one matrix that does all the row shifts, i.e., that is shifts the 1st row to 2nd, 2nd to 3rd and so on! so we take ( 4 by 4 case):
\begin{bmatrix} 0 & 0 & 0 & 0\\ 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ \end{bmatrix}
We let this be our $A$.
So, when we multiply $A.C$ then the rows of $C$ get shifted. So, $A.e_{1, 1} = e_{2, 1}$ so on! Wonderful! We can do analgously for columns by choosing a $B$ that does similar thing to columns! Easy to see that $B$ will just be the transpose of $A$.
So, from $e_{1, 1}$ we can get all $e_{i, j}$.
Now $A^{n-1} =e_{n ,1} $ and $B^{n-1} = e_{1, n}$.
and $e_{1, n}e_{n, 1} = e_{1, 1}$
I can find more by conjugating with permutation matrices. But not sure how to count since i dont know the stabilizers and the main question is are there others?