Let $V$ be a finite-dimensional vector space over a field $F$ and let $T$ be an operator on $V$. Prove $\text{range}(T^2) = \text{range}(T)$ if and only if $\text{range}(T) \cap \ker(T) = \{0 \}$.
Proof. Assume $v \in \text{range}(T) \cap \ker(T) = \text{range}(T^2) \cap \ker(T)$. Then there are vectors $v_1,v_2 \in V$ such that $(\star) \hspace{1mm} v = T^2(v_1)=T(v_2)$ and $T(v)=0$ Thus $T(v) = T^3(v_1) = T^2(v_2) = 0$, implying that $$T(v) = T^3(v_1)+T^2(v_2)= T(T^2(v_1)+T(v_2)) \Longleftrightarrow v = T^2(v_1)+T(v_2)= v + v = 2v$$ $$\hspace{5.6cm} \Longleftrightarrow v = 2v \Rightarrow v = 0.$$
Conversely, assume that the intersection of $\text{range}(T)$ and $\ker(T)$ is trivial and let $v \in \text{range}(T^2)$. Then there exists $u \in V$ such that $v = T^2(u)=T(T(u))$. Since $T$ is an operator, $T(u) \in V$, implying that there exists a vector $w \in V$ such that $w = T(u)$. Hence $T^2(u)=T(w) = v$, implying that $v \in \text{range}(T)$, and so $\text{range}(T^2) \subset \text{range} (T)$.
This is where I get stuck. I'm trying to show that $\dim \text{range} (T^2) = \dim \text{range}(T)$, so the above inclusion implies equality, but I am having trouble. I know we probably need to apply the rank-nullity theorem combined with our assumption that the intersection of the range and kernel is trivial, but I am not sure how.
Best Answer
Let $v \in \mathrm{range}(T)$ so $Tu = v$ for some $u \in V$. Since $\mathrm{range}(T) \cap \ker(T) = \{ 0 \}$, by the rank-nullity theorem we have $\mathrm{range}(T) \oplus \ker(T) = V$. Decompose $u = Tw + x$ where $w \in V$ and $x \in \ker(T)$. Then $v = Tu = T(Tw + x) = T^2(w)$ showing that $v \in \mathrm{range}(T^2)$.