# If $U=Im(T^k)$ then $T|_U$ is invertible

Let $$V$$ be a vector space, $$dim(V)=n<\infty$$, and $$T:V\to V$$ linear map. (A) Prove there exists $$k\in \mathbb{N}$$ such that $$Im(T^k)=Im(T^{k+1})$$.

(B) Let $$U=Im(T^k)$$. Prove that (i) $$U$$ is an invariant subspace. (ii) $$T|_U:U\to U$$ is invertible.

I will be happy to receive help in $$(B)ii$$. Here are my ideas for $$A,B(i)$$:

(A) $$\forall k: Im(T^{k+1})\subseteq Im(T^k)\subseteq V$$. $$V$$ is finite-dimensional, so we cannot have an infinite chain of subspaces strictly contained in one another.

(B) i) $$u\in Im(T^k)\Rightarrow \exists v\in V$$ such that $$T^k v=u$$. This implies $$Tv = T^{k}(Tu) \in Im(T^k)$$ so $$U$$ is invariant.

I thought of proving (B)(ii) by showing that $$0$$ is not an eigenvalue of $$T|_U$$, but did not really manage to do it.

Thank you!

From rank nullity theorem: $$dim V=dim(Im(T^k))+dim(Ker(T^k))$$ $$dim V=dim(Im(T^{k+1}))+dim(Ker(T^{k+1}))$$
As $$Im(T^k)=Im(T^{k+1})$$ (from (A)), we have $$dim(Ker(T^{k+1}))=dim(Ker(T^k))$$ But $$Ker(T^k)\subseteq Ker(T^{k+1})$$, therefore, $$Ker(T^{k+1})=Ker(T^k)$$
Therefore, if $$T|_Uv=0$$, then $$T(T^{k}w)=0$$ for some $$w\in V$$ and $$v=T^{k}w$$, i.e. $$w\in Ker(T^{k+1})$$, so $$w\in Ker(T^k)$$, hence $$v=T^{k}w=0$$, we have proved that $$Ker(T|_U)=\{0\}$$, hence $$T|_U$$ is invertible (as $$U$$ is finite dimensional).