Let $T : V → V$ be a linear map, where $V$ is a finite-dimensional vector space. Then $T^2$ is defined to be the composite $T ◦ T$ of $T$ with itself, and similarly $T^{i+1} = T ◦ T^i$ for all $i ≥ 1$.
Suppose that $rank(T) = rank(T^2)$.
Prove that $im(T) = im(T^2)$
For $i ≥ 1$, let $U_i: im(T) → im(T)$ be defined as the restriction of $T^i$ to the subspace $im(T)$ of $V$. Show that $U_i$ is nonsingular for all $i$.
Deduce that $rank(T) = rank(T^i)$ for all $i ≥ 1$
I have solved the first part by showing that $im(T^2)$ is a subset of $im(T)$ so the basis of $im(T^2)$ can be extended to that of $im(T^2)$. But since the ranks are equal, they have equal bases.
For the second part my first approach is to proof that $U_i$ is bijective. I think I have a proof for surjectivity by induction. But I can't figure out injectivity. Also I can't figure out the third part.
Edit 1:
Proof for surjectivity. First show that $U_1$ is surjective. Suppose otherwise, then there exists $y \in V$ such that $T(T(z)) \neq T(y)$ for any $z \in V$, which contradicts the first part of the question. Now given $U_i$ (/) and $U_1$ (//) is surjective, we have $T_{i+1} = T_1(T_i(x))$ for all $x \in im(T)$ implies $U_{i+1} = U_1(U_i(x))$ by (/). But also by (/) we have that $U_i(x)$ spans the domain of $U_1(x)$ so by (//) $U_{i+1}$ is surjective. This is equivalent to invertible for linear transformations.
Any help with 3?
Edit 2:
suppose $im(T^i) = im(T)$
then $im(T^{i+1}) = im(T(T^i)) = im(T(x))$ such that $x \in im(T)$
$ ... = im(U_1) = im(T)$ so same applies to rank.