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Let $V$ be a vector space , $T:V \to V$ be a linear operator , then is it true that

$\ker (T) \cap R(T) \cong R(T)/R(T^2) $ ( where $R(T)$ denotes the range of $T$ ) ?

I know that the statement is true when $V$ is finite dimensional as I can show that if $V$ is finite

dimensional , then $\dim (\ker (T) \cap R(T))=rank (T) - rank (T^2)=\dim R(T)/R(T^2)$ ,

but I don't know what happens in general . Please help . Thanks in advance

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It doesn't appear to hold in the infinite-dimensional setting. Let $H=\ell^2(\mathbb{N})$ (where $\mathbb{N}=\{1,2,3,\ldots\}$ and we denote members of this set by functions $x:\mathbb{N}\to\mathbb{C}$), and define the left-shift operator $S_L\in\mathcal{B}(H)$ by $S_L(x)(n)=x(n+1)$ for all $n\in\mathbb{N}$. Then \begin{align*} R(S_L)&=R(S_L^2)=\ell^2(\mathbb{N}), \\ \ker(S_L)&= \{x\in\ell^2(\mathbb{N}):x(n)=0\ \forall n\geq2\} = \ker(S_L)\cap R(S_L). \end{align*} Thus $R(S_L)/R(S_L^2)\cong \{0\}$, while $\ker(S_L)\cap R(S_L)\cong\mathbb{C}$.

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    $\begingroup$ you appear to be right ... though I just realized your example doesn't violate $(\ker(T) \cap R(T) ) \times R(T^2) \cong R(T)$ which holds in finite dimensional case due to the same formula I mentioned . Do you have any thoughts on this later isomorphism claim ? $\endgroup$ – user228168 Jul 22 '16 at 5:43
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    $\begingroup$ That I will have to think about. Good questions! $\endgroup$ – Aweygan Jul 22 '16 at 5:47
  • $\begingroup$ thanks , your example motivated me to ask if there is any possible way out ( I am not very optimistic though :p) $\endgroup$ – user228168 Jul 22 '16 at 5:49
  • $\begingroup$ I have more faith about this next claim, but it may take a while as I am getting very tired (its nearly 1:00a.m. local). $\endgroup$ – Aweygan Jul 22 '16 at 5:59
  • $\begingroup$ Okay no problem :) I will ask a different question with linking this question ... $\endgroup$ – user228168 Jul 22 '16 at 6:02

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