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

Question

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.

Answer 1

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

Answer 2

Every square matrix satisfies its characteristic equation. Since $0$ is the only eigen value of $B$ its characteristic polynomial is $\lambda ^{n}$ so $B^{n}=0$.

Answer 3

If $A^n=0$ and $B=P^{-1}AP$, then $B^n=P^{-1}A^nP=0$.

Answer 4

Let $A^n=0$. Since $A$ and $B$ are similar , there is an invertible matrix $T$ such that $B=T^{-1}AT$. Since $B^k=T^{-1}A^kT$ for all k, we get $B^n=0$.