Attempt: Firstly, we note that the $n \times n$ matrices of the form $c_{ij}=1$ for $ (i,j)=(r,r)$, and zero everywhere else are idempotent matrices. [For eg, the diagonal matrix $\text{diag}(0,1,0,...,0)$]. Denote these type of matrices as $C_{rr,rr}$.
Secondly, we also take note of the fact that a matrix with $c_{rr}=c_{rs}=1$, $c_{ij}=0$ everywhere else is also an idempotent matrix. Let us denote this matrix as $C_{rr,rs}$.
Now, consider an arbitrary matrix $A \in M_n(F)$. We can write $A=(a_{ij})_{n\times n}=\displaystyle\sum_{i,j} P_{ij}$, where $P_{lm}=(p_{ij})_{n \times n}$ with $p_{ij}=a_{lm}$ for $(i,j)=(l,m)$ and $ p_{ij}=0$ otherwise.
We can write $P_{lm}=a_{lm}C_{ll,lm}-a_{lm}C_{ll,ll}$ for $l \neq m$ and $P_{ll}=a_{ll}C_{ll,ll}$
Thus, we can write $\displaystyle\ A=\sum_{i\neq j}(a_{ij}C_{ii,ij}-a_{ij}C_{ii,ii})+\sum_ia_{ii}C_{ii,ii}$, where $a_{ij}\in F$ and $C_{ii,ij} \in \text {Set of idempotent matrices}$ for all $1\leq i, j \leq n$.
Which implies that a subset of the set of idempotent matrices $(=K$, say$)$ generate $M_n(F)$ which further implies that $K$ generates $M_n(F)$.
Is this correct?
Kindly VERIFY.