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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.

St Vincent
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    Let S: Range(T) -> Range(T) be defined by S(v) = T(v). The assumption that Ker(T) and Range (T) have trivial intersection says that Ker(S) is trivial. Now use the Rank-Nullity theorem on the transformation S. – Ned Dec 03 '15 at 00:35

2 Answers2

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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 $v \in \mathrm{range}(T^2)$.

levap
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  • Great answer! Thank you. Just want to make sure of something: Since $\ker(T)$ and $\mathrm{range}(T)$ are subspaces of $V$, their sum $\ker(T) + \mathrm{range}(T)$ is also a subspace. Then $$\dim(\ker(T) + \mathrm{range}(T))= \dim \mathrm{range}(T) + \dim \ker(T) - \dim (\mathrm{range}(T) \cap \ker(T)) = \dim \mathrm{range}(T) + \dim \ker(T) = \dim V$$ and since $\mathrm{range}(T) \cap \ker(T) = {0}$ we have $\mathrm{range}(T) \oplus \ker(T) = V$. Are these the correct details? – St Vincent Dec 03 '15 at 01:15
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    Yes, that is correct. – levap Dec 03 '15 at 01:16
  • BTW, I don't understand your proof of the first direction. $T^2(v) = 0$ implies that $Tv$ is both in the kernel and in the image and so you can deduce that $Tv = 0$ and not that $v = 0$. In fact, you did not use the condition that $v \in \mathrm{range}(T) = \mathrm{range}(T^2)$. – levap Dec 03 '15 at 01:25
  • Whoops, my first attempt was incorrect. What about this: Let $v \in \ker(T) \cap \mathrm{range}(T) = \ker(T) \cap \mathrm{range}(T^2)$ Then there is a vector $x \in V$ such that $v = T^2(x)$. But $\mathrm{range}(T) = \mathrm{range}(T^2)$, so $v = T^2(x) = T(x)$. Applying $T$ to both sides of the preceding equality yields $0 = T(v) = T^2(x)=v.$ – St Vincent Dec 03 '15 at 01:59
  • Still not good. The fact that $\mathrm{range} (T) = \mathrm{range} (T^2)$ doesn't imply $T^2(x) = T(x)$. – levap Dec 03 '15 at 02:22
  • Probably overly complicated, but I tried it again: Suppose $v \in \mathrm{range}(T) \cap \ker(T) = \mathrm{range}(T^2) \cap \ker(T)$. Then there are vectors $x_1,x_2 \in V$ such that $(\star) \hspace{1mm} v = T^2(x_1)=T(x_2)$ and $T(v)=0$. Applying $T$ to $(\star)$ yields $0 = T(v) = T^3(x_1)=T^2(x_2)$, so $T(v) = 0 = 0 + 0 = T^3(x_1) + T^2(x_2)$. Since $T^2(v) = T(T(v))=T(0) = 0 = T(v)$, we have $T(v) = T^2(v) = T(T^3(x_1)+T^2(x_2)) \Rightarrow v = T^3(x_1) + T^2(x_2) = 0 + 0 = 0$ – St Vincent Dec 03 '15 at 03:00
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Since $\text{Ker}\left(T|_{\text{range}(T)}\right)=\text{range}(T)\bigcap \text{Ker}(T),$ we have the following: \begin{align*} \text{range}\left(T^2\right)=\text{range}(T)&\Leftrightarrow T\left(\text{range}(T)\right)=\text{range}(T)\\ &\Leftrightarrow T|_{\text{range}(T)} \text{ is bijective}\\ &\Leftrightarrow \text{Ker}\left(T|_{\text{range}(T)}\right)=\{0\}\\ &\Leftrightarrow \text{range}(T)\bigcap \text{Ker}(T)=\{0\}. \end{align*}

Note, here $T|_{\text{range}(T)}$ is the restriction of $T$ on the linear subspace $\text{range}(T)$.

user760
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