# $V$ be a vector space , $T:V \to V$ be a linear operator , then is $\ker (T) \cap R(T) \cong R(T)/R(T^2)$?

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

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 $\ker(S_L)\cap R(S_L)\cong\mathbb{C}$.
• 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 ? – user228168 Jul 22 '16 at 5:43