If matrix $A$ is similar to $B$ and $A$ is nilpotent, does that imply that $B$ is also nilpotent

matricesnilpotence

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 $$\eqalign{ \hbox{$B$ similar to $A$}\quad &\Rightarrow\quad B=PAP^{-1}\qquad\hbox{for some $P$}\cr &\Rightarrow\quad B^k=PA^kP^{-1}=0\ .\cr}$$

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