Proving that range(T) $\cap$ ker(T)={0} for a linear transformation T: V $\rightarrow$ V

Question

My question is related to this other question:V is a n-dimensional vector space and $T : V \rightarrow V$ is a LT s.t $rank(T) = rank(T^2)$. Prove that $range(T)\cap ker(T) = \{\mathbb{0}\}$. I solved it in a different way and i would like to know if my reasoning is correct.

My demonstration:

Since V is both the domain and codomain of T, T is an isomorphism. This means that T is inyective and surjective at the same time. Given that T is inyective we have that ker(T)={0}. Since range(T) is a vector space, it must contain the vector 0. Here we conclude that range(T) $\cap$ ker(T)={0}.

Answer 1

consider rank-nullity theorem for linear maps $T:V \rightarrow V$ and $T^2 : V \rightarrow V$: $$ rank(T) + nul(T) = dim(V) \\ rank(T^2) + nul(T^2) = dim(V) $$ since $rank(T)=rank(T^2)$ ,then $dim(ker(T))=dim(ker(T^2))$. we know that $ker(T) \subset ker(T^2)$ because : $$ x \in ker(T) \rightarrow Tx=0 \rightarrow T^2 x= T0=0 \rightarrow x \in ker(T^2) $$ and kernel of a linear transformation is a linear subspace of $V$; therefore $$ker(T)=ker(T^2). \tag{*}$$ now let's prove the statement, we want to show if $x \in range(T) \cap ker(T)$ then $x=0$. so assume $x \in range(T) \cap ker(T)$. since $x\in range(T)$ then there exit a $y \in V$ s.t $Ty=x$. and since $x \in ker(T)$ then $Tx=T^2y=0$. so $y\in ker(T^2)=ker(T)$(*). by last result we get $x=Ty=0$. $\square$

Answer 2

The key is to first prove that if $x\in R(T)$ then $x = Tx$. Then find the value of $x\in R(T)\cap N(T)$.

Since $T=T^2$ it follows that if $x\in R(T)$ then $x = T\xi = T^2\xi = Tx$. Thus if $x\in N(T)\cap R(T)$ then $0 = Tx = x \implies N(T)\cap R(T)=\{0\}$.