Linear Transformation Let $T:V\rightarrow V$ be a linear transformation and dimension of $V$ be infinite. Then show that either the nullity or the rank of $T$ is infinite.
The problem is that I cannot use rank nullity theorem.
Thanks for any help.
 A: Let $\,\{v_1,...\}\,$ be any (obviously infinite) basis of $\,V\,$ , then:
(1) if $\,\{Tv_1,...\}\,$ contains an infinite linearly independent subset then $\,\dim Im\,\,(T)=rank\,\,T=\infty\,$ , other wise:
(2) there only exists a finite linearly independent subset in $\,\{Tv_1,...\}\,$  ,so it must be that an infinite subset of $\,\{v_1,...\}\,$ is mapped to zero by $\,T\,$ and thus $\,\dim\ker T=null\,\,T=\infty\,$
Added: Please do read the comment below by Tunococ as it proves (2) accurately.
A: You try to use a too strong version of the rank nullity theorem. A more general version (true in infinite dimension) asserts that $V$ is isomorphic fo $Ker(T)\oplus Im(T)$. In this setting saying that both $ker(T)$ and $Im(T)$ are finite dimensional would imply that $V$ would also be finite dimensional. 
A: Suppose the rank and nullity are both finite. Then there's some positive integer $n$ which exceeds both of them. Let $W$ be a subspace of $V$ of dimension $2n+1$ (I take it you can see that such a thing exists). Let $T'$ be the restriction of $T$ to $W$, $T':W\to V$. The rank of $T'$ is at most the rank of $T$, right? And the nullity of $T'$ is at most the nullity of $T$, right? But this contradicts the finite form of the rank-nullity theorem. Are you allowed/able to use that?
