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Let $A\subseteq M_n(\mathbb{R})$ a finite set, where $M_n(\mathbb{R})$ is the set of all $n\times n$ matrices. Let $S(A)$ the generated algebra by $A$.

What is the minimun number of elements of $A$ in order that $S(A)=M_n(\mathbb{R})$?

In other words: what is the minimum $k$ such that there are $A_1,...,A_k$ elements of $M_n(\mathbb{R})$ with the property that every element of $M_n(\mathbb{R})$ is a polynomial on $A_1,...,A_n$?

I know that there is a lot of articles about the exact number of elements of $A$, but i believe this is a simpler question.

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I can do it with $2$ matrices. One matrix $X$ with all entries $0$ except for one (say the top left) where we have a $1$, and one order-$n$ cyclic permutation matrix $Y$.

Any matrix with a single entry $1$ and the rest $0$ may be written as a product $Y^aXY^b$ for natural numbers $a,b$. Any matrix in $M_n(\Bbb R)$ may be written as a linear combination of those.

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