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}?
Best Answer
$T:V \to V$ 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, $R(T) \cap N(T)=\{0\}$