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.
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"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.