$A$ and $B$ are nilpotent matrices in $\mathbb M_n(F)$ that commute. Let $P$ be an invertible matrix such that $PAP^{-1}$ is strictly upper triangular. Does it also imply that $PBP^{-1}$ is also strictly upper triangular?
I was trying to prove that $A+B$ is nilpotent and can't think of a proof or a contradiction
No. Since $A$ and $B$ commute (and they have full spectra over $F$), they can be simultaneously triangualarised by some similarity transform. However, that doesn't mean any similarity transform that triangularises $A$ will also triangularise $B$. For instance, consider $$ X=\pmatrix{1&-1\\ 1&-1},A=\pmatrix{0&X\\ 0&0},B=\pmatrix{X&X\\ X&X} \text{ and } P=I_4. $$