Is the Matrix Nilpotent? 
Let $A, B$ be nilpotent matrices over $\Bbb C$ where $AB=BA$. Are the following matrices nilpotent $$A-2B $$
$$A^3B$$

I don't really  know where to start here. I saw a solution with the binomial expansion, however it seemed hard to understand.
Is there a different way?
Thanks,
Alan

I try a solution without the binomial theorem. I suppose that we are working in $\mathbb{C}$.
Let $p\geq 1$ such that $A^p=0$. Let $\lambda $ an eigenvalue of $A-2B$, and $x\not = 0$ an eigenvector for the eigenvalue $\lambda$, ie $Ax-2Bx=\lambda x$. We have $A^p x=0$; hence there exists $h\geq 1$ such that $A^hx=0$ and $A^{h-1}x\not =0$. Multiplying by $A^{h-1}$, (and using that $AB=BA$) we get $-2BA^{h-1}x=\lambda A^{h-1}x$. This show that $y=A^{h-1}x$ is an eigenvector for $B$, with eignevalue $-\lambda/2$. As $B$ is nilpotent, we get $\lambda=0$. We have show that all eigenvalues of $A-2B$ are $0$, hence $A-2B$ is nilpotent.   
