Vector space as direct sum of kernel and image Let $T: V\to V$ be a linear map on an infinite-dimensional vector space $V$ over a field $F$. Suppose that $T(V)$ is finite dimensional, and $T^2(V)=T(V)$.

Show that $V=\ker T\oplus T(V)$.

(Note that this is not a duplicate of Show that the direct sum of a kernel of a projection and its image create the originating vector space., it seems to be more difficult since we only have $T^2(V)=T(V)$ rather than $T^2=T$.)



What I tried: I can prove $V=\ker T+T(V)$, what is difficult seems to be showing the direct sum, i.e. the intersection is 0.

Let $v\in V$. Since $T(v)\in T(V)=T^2(V)$, $T(v)=T^2(x)$ for some $x\in V$. Then $T(v-T(x))=0$, so that $v-T(x)=y\in\ker T$. Thus $v=y+T(x)\in\ker T+T(V)$. Hence $V\subseteq\ker T+T(V)$. $\ker T+T(V)\subseteq V$ is clear since $V$ is a vector space, so $V=\ker T+T(V)$.

I also have a previous result Linear Transformation from Infinite dimensional to Finite dimensional Space which may or may not be useful.

Thanks for any help!
The restriction of $T$ to $T(V)$ is an automorphism $T(V)\rightarrow T(V)$, in particular its kernel is $0$.
Proof. Since $v = \pi (v) + (v - \pi(v))$ and π(v – π(v)) = 0, V = im V + ker V. To show that the sum is direct, let v = π(w) ∈ ker π. Then 0 = π(v) = π^2(w) = π(w) = v, so im π∩ker π = 0.

Ran across this in a book I am reading.