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.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.