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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!

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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).

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  • $\begingroup$ Thank you @AGH! Really elegant $\endgroup$ Commented Jun 20, 2021 at 14:19

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