Let $V$ be a finite-dimensional vector space and let $T$ be a linear operator on $V$ . Suppose that $\mathrm{rank}(T^2)=\mathrm{rank}(T)$. Prove that the range and null space of $T$ are disjoint, i.e., have only the zero vector in common.
Approach(1): Since $T:V\to V$ is linear map, we have $T^2=T\circ T:V\to V$ is linear map, by theorem 6 section 3.2. $V$ is finite dimensional vector space over field $F$. By rank nullity theorem, $\mathrm{dim}(V)$ $= \mathrm{dim}(R_T)+ \mathrm{dim}(N_T)$ $= \mathrm{dim}(R_{T^2})+ \mathrm{dim}(N_{T^2})$. Since $\mathrm{dim}(R_T)= \mathrm{dim}(R_{T^2})$, we have $\mathrm{dim}(N_T)= \mathrm{dim}(N_{T^2})$. We claim $N_T\subseteq N_{T^2}$. Let $x\in N_T$. Then $T(x)=0_V$. So $T^2(x)$ $=T(T(x))$ $=T(0_V)$ $=0_V$. Thus $x\in N_{T^2}$. Hence $N_T\subseteq N_{T^2}$. So $N_T\leq N_{T^2}$. By contrapositive of theorem 5 corollary 1 section 2.3, $N_T=N_{T^2}$. If $T(T(\alpha))=0_V$, then $\alpha \in N_{T^2}$. So $\alpha \in N_T$. Thus $T(\alpha)=0_V$. By exercise 13 section 3.1, $R_T\cap N_T=\{0_V\}$. Is my proof correct?
Approach(2): Let $x\in N_T\cap R_T$. Then $T(x)=0_V$ and $\exists u\in V$ such that $T(u)=x$. So $T^2(u)$ $=T(T(u))$ $=T(x)$ $=0_V$. So $u\in N_{T^2}$. Since $N_{T^2}=N_T$, we have $u\in N_T$. So $T(u)=x=0_V$. Thus $N_T\cap R_T\subseteq \{0_V\}$. Hence $N_T\cap R_T=\{0_V\}$. Is my proof correct?