TITLE: What is the minimal number of symmetric generators of the full matrix algebra?
QUESTION [8 upvotes]: Is there any lower bound known for the minimal number of generators needed to generate the full matrix algebra of real $n\times n$ matrices — when using only symmetric matrices for the generators?
Analogous question for complex matrices — when using only Hermitian matrices for the generators.
I am aware that $3$ generators suffice when using only idempotent generators. This is a result of Naum Krupnik (Minimal number of idempotent generators of matrix algebras over arbitrary field, Comm. Algebra 20 (1992), no. 11, 3251–3257). (Tandfonline link, restricted access)
I am not familiar with this type of results, so this might be well known or easy.
Thanks for any tips.

REPLY [3 votes]: Here is a more direct argument, without Burnside's Theorem.  As in M. Karimi's answer, let $A$ be diagonal with distinct entries $\lambda_1,\dotsc,\lambda_n$, and let $B$ be the matrix with all entries equal to $1$.  Note that 
$$ (A^iBA^j)_{st} = \lambda_s^i\lambda_t^j $$
Now let $M$ be an arbitrary $n\times n$ matrix.  By Lagrange interpolation, there is a unique polynomial 
$$ p(x,y) = \sum_{i,j=0}^{n-1} a_{ij}x^iy^j $$ 
with $p(\lambda_s,\lambda_t)=M_{st}$ for all $s$ and $t$.  It is now easy to check that 
$$ M = \sum_{i,j=0}^{n-1} a_{ij}A^iBA^j.  $$

REPLY [2 votes]: As an answer, the lower bound is exactly $2$. To this end take a diagonal matrix of order $n$ with all distinct and non zero entries along diagonal and the matrix of all ones of order $n$. In this way the algebra generated by these two symmetric matrices is simple(i.e. irreducible ), Now  use Burnside's theorem to deduce that it should be $M_n(C)$. See 
Laffey, Thomas J., A structure theorem for some matrix algebras, Linear Algebra Appl. 162-164, 205-215 (1992). ZBL0758.16010.