Composite linear map Rank and Image

Question

I have been pondering on this question, I did part $(a)$ wherein you had to prove that $\operatorname{Im}(T)= \operatorname{Im}(T^{2})$ , but I am struggling to get the concept of part $(b)$, any help would be aprreciated

$$ \text{Let } T: V \rightarrow V \text{ be a linear map, where } V \text{ is a finite-dimensional vector space. Then } T^2 \\ \text{ is defined to be the composite } TT \text{ of } T \text{ with itself, and similarly } T^{i+1} = T^{i} \\ \text{ for all } i \geq 1. \text{ Suppose that } \operatorname{Rank}(T) = \operatorname{Rank}(T^2). \text{ (b) For } i \geq 1, \text{ let } U_{i}: \operatorname{Im}(T) \rightarrow \operatorname{Im}(T) \\ \text{ be defined as the restriction of } T^{i} \text{ to the subspace } \\ \operatorname{Im}(T) \text{ of } V \text{ . Show that } U^{i} \text{ is nonsingular for all } i. \text{ Then deduce that } \operatorname{Rank}(T) = \operatorname{Rank}(T^{i}) \\ \text{ for all } i \geq 1. $$

Answer 1

If $U^i$ is singular, then there is some vector $v \in Im(T)$ so that $U^iv=T^iv=0$. But then $T(T^{i-1}v)=0$. So if $i \geq 2$, then $T^{i-1} v \in \ker U$ and otherwise, $v \in \ker U$. Thus, $U$ is singular. Now think about what you showed in part a).

Answer 2

Note that $Im(U_1) = U_1(Im(T)) = T(Im(T)) = T^2(V) = Im(T^2) \leq Im(T)$.

Since $Rank(T) = Rank(T^2)$, it's actually the case that $Im(T^2) = Im(T)$ and we therefore have $Im(U_1) = Im(T)$. Hence $U_1$ is a surjective map $Im(T) \rightarrow Im(T)$ and therefore nonsingular by rank-nullity. $U_i = U_1^i$ and is therefore a composition of nonsingular maps, hence nonsingular itself.

Can you get the final deduction now? Have you got a way of linking $Rank(T^i)$ with the rank of some $U_j$?

PS: Did your supervisor not explain this?... :)

Answer 3

For $i=1$ we have $$\operatorname{Im}(U_1)=T(\operatorname{Im}(T))=\operatorname{Im}(T^2)=\operatorname{Im}(T)$$ and the last equality is true since we have (simple to prove) $$\operatorname{Im}(T^2)\subset \operatorname{Im}(T)$$ and by hypothesis $$\operatorname{rank}(T^2)=\operatorname{rank}(T)$$ hence we have proved that $ \operatorname{Im}(U_1)=\operatorname{Im}(T)$ and by the rank-nullity theorem we see that $U_1$ is non-singular.

Can you generalize this result for $i\ge2$?