proof question: Let V be a finite-dimensional vector space, and let $T:V \rightarrow V$ be linear. 
Let V be a finite-dimensional vector space, and let $T:V \rightarrow V$ be linear.

a). if $\mathrm{rank} (T)= \mathrm{rank} (T^2)$, prove $R(T) \bigcap N(T)$={0}. 
$N(T)$ is defined by $T(0_V)=0_W$ for any finite dimensional $T: V\rightarrow W$. 
Since $V$ is finite-dimensional vector space, from dimensional theorem,
$$\dim(V)=\dim(N(T)+\dim(R(T))-\dim(N(T) \cap R(T))$$
How can I go further ahead and deduce that $R(T) \bigcap N(T)$={0}?
 A: $T:V \to V$ is a linear map. Now when you're considering $T^2=T \circ T$ , note that $\tilde{T} =T|_{R(T)} :R(T) \to R(T)$ is basically an isomorphism by Rank-nullity theorem since, $Rank(T)=Rank(T^2)$ . 
Let, $v \in R(T) \cap N(T)$ , then $T(v)=0$ and $\exists w \in V$ such that $Tw=v$ . Thus $T^2w=Tv=0 \implies \tilde{T}(Tw)=0\implies \tilde{T}v=0$ . As we showed earlier that $\tilde{T}$ is an isomorphism on $R(T)$ it follows that $\tilde{T}v=0 \implies v=0$
Hence, $R(T) \cap N(T)=\{0\}$
A: We know from rank-nullity theorem that $\text{rank}(T)+\text{nulity}(T)=\text{dim}(V)$ ($N(T)$ is the null-space of $T$ and $R(T)$ is the image of $T$, $V$ is domain and co-domain, both). Assume $\text{rank}(T) =r$, $\text{nulity}(T)=n$ and $\text{dim}(V)=v$.
Let's $\{ x_1,x_2,\cdots ,x_r,x_{r+1}\cdots,x_v \}$ be a complete basis for $V$, thus any relation $\sum_{i=1}^{m}c_ix_i=0$ indicates $\forall i: c_i=0$ (where $1\le m\le v$);This basis has the property that $\{x_{r+1},\cdots ,x_v\}$ forms a basis for null space of $T$ (in other words we expanded the basis of null space of $T$ to get a complete basis for $V$). Thus $\forall i, r+1 \le i \le v: T(x_i)=0$. The image of $T$ , i.e. $y_i=T(x_i) , i=1,\cdots , r$ can be written in this basis as $\forall i: y_i = \sum_{j=1}^{v}d_{ij}x_j$.
Finally considering $z\in R(T)\cap N(T)$, we can write $z = \sum_{k=1}^{r}f_ky_k(\in R(T))=\sum_{l=r+1}^{v}e_lx_l(\in N(T))$. Applying $T$ on $z$ gives $T(z)=\sum_{l=r+1}^{v}e_lT(x_l)=0$, but we know that $z=\sum_{k=1}^{r}f_ky_k= \sum_{k=1}^{r}f_k\left(\sum_{j=1}^{v}d_{kj}x_j\right)=\sum_{j=1}^{v}g_jx_j$. Applying $T$ on $z$ now gives
$$T(z)=\sum_{j=1}^{v}g_jT(x_j)=\sum_{j=1}^{r}g_jT(x_j)=T\left(\sum_{j=1}^{r}g_jx_j\right)=0$$
Considering the fact that $T\left(\sum_{j=1}^{r}g_jx_j\right)=0$ and $\forall i, i=r+1,\cdots ,v:T(T(x_i))=T(0)=0$ means $\text{nullity}(T^2)>\text{nullity}(T)$ and thus $\text{rank}(T^2)<\text{rank}(T)$ which is a contradiction unless $\forall i, 1\le i \le r: g_i=0$ implying $z\in N(T)$ and $z\notin R(T)$, but that also contradicts with $z\in R(T)\cap N(T)$. Therefore our only option is $z=0$.
