If matrix $A$ in $\mathbb{R}^{n\times n}$ is nilpotent, what I know:
- all the eigenvalues of $A$ are $0$
- the determinant of $A$ is $0$ (because the minimal polynomial has to be $0$)
- the rank of $A$ has to be less than $n$
All of that applies also to the matrix $B$ because they are similar.
I have read in some answers to similar questions that I need to compare the Jordan normal form of $A$ and $B$.
I haven't learned anything about the Jordan normal form yet, so if anyone is arguing with it some little details about it would be great.
Now I don't know how to work with that and how to show that $B$ also must be nilpotent.
Best Answer
$A$ nilpotent means $A^k=0$ for some $k$; hence $$\eqalign{ \hbox{$B$ similar to $A$}\quad &\Rightarrow\quad B=PAP^{-1}\qquad\hbox{for some $P$}\cr &\Rightarrow\quad B^k=PA^kP^{-1}=0\ .\cr}$$