I can see that the standard base is not of nilpotent element (the elements on the diagonal are not), but can't prove that proposition, or be sure it is even true.
"No" because of the following hints I'm giving you here.
The trace of a nilpotent matrix is zero. What do you know about the traces of $A+B$ and $\lambda A$ for matrices $A,B$ and scalar $\lambda$?
This should make it clear that you won't be able to generate just any matrix with a linear combination of nilpotent matrices.