Let $\mathbb{F}$ be a finite field.
Let $A\le \mbox{Mat}\_n(\mathbb{F})$ be a matrix algebra.


Is there a good bound on the number $k$ of random elements $a\_1,\dots,a\_k\in A$ 
that one needs to take such that, with high probability, the algebra generated by $a\_1,\dots,a\_k$ is $A$?
What if $A$ is generated by a *group* of matrices?


I am interested in a bound that applies to *all* such $A$. 


Taking $k=O(n^2)$ will provide a set that spans $A$ as a vector space. 
Would less than that suffice?


**Comment:** This is related to [this question](https://mathoverflow.net/questions/154761/finding-a-basis-for-the-linear-combinations-span-of-a-matrix-group-efficientl/206338#206338).



How many random matrices does it take to generate a matrix algebra?
At least $n^2/4$. Divide $n \times n$ matrices into four $n/2 \times n/2$ blocks, and consider the subalgebra of matrices that are zero outside the upper right block. Then because the product of any two elements is zero, your elements generate as an algebra if and only if they generate as a vector space. So you need at least $n^2/4$.


On the other hand, clearly $n^2+ c$ elements suffice for any algebra for a small constant $c$. 


Your use of big $O$ notation suggests that the constant doesn't matter. If it does, then I'd guess $n^2/4$ is optimal, as this algebra is a worst-case algebra in many respects - e.g. it is the abelian subalgebra of the largest dimension.


