Let $T$ be a linear operator on a finite dimensional space $V$. Prove that if $rank(T^n) = rank(T^{n+1})$ for some positive integer $n$, then $rank(T^n) = rank(T^m)$ for all positive integer $m \geq n$.
I can show the range of $T^n$ is equal to the range of $T^{n+1}$ i.e, $R(T^n) = R(T^{n+1})$. Then I'm stuck.
What should I do?
Thank you.
By induction: the inductive step assuming that $Rank(T^m)=Rank(T^{m+1})$ for $m\ge n$. It's easy to prove that
$$Im(T^{m+2})\subset Im(T^{m+1})=Im(T^m)$$ Now let $y\in Im(T^{m+1})$ and $x$ such that $y=T^{m+1}(x)=T(T^mx) $ but $T^m x\in Im(T^m)=Im(T^{m+1})$ so there's $z$ such that $$T^m x=T^{m+1}z$$ hence
$$y=T(T^{m+1}z)=T^{m+2}(z)\in Im(T^{m+2})$$ so by double inclusion we get the result.
I don't understand what you can show, except the assumption...
Hint: let $x$ be in the image of $T^{n+1}$ Can you show that it is in the image of $T^{n+2}$ by using that elements in the image of $T^n$ are in the image of $T^{n+1}$?
