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The full question:

$T^2$ means the composition, i.e. $T(T(x))$

Let $V$ be a vector space and let $T: V \rightarrow V$ be linear. Prove that $T^2$ is the zero transformation if and only if $range(T) \subseteq \ker(T)$.

So far I know that for ($\implies$). Assume $T^2$ is the zero transformation. Then $ker(T^2)=\{0 \in V: T^2(v)=0\}$. Note that this is $\forall v \in V$. We also know that the $range(T)=\{0\in V: \exists v \in V: T^2(v)=0\}$.

Not sure where to go from here. Also for the other way, I don't know where to go.

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What you've written for $\ker(T^2)$ and $\mathrm{range}(T)$ is a bit strange, and for the range, it's not correct. The definitions are $\ker(T^2)=\{v\in V:T^2(v)=0\}$ and $\mathrm{range}(T)=\{T(v):v\in V\}$. – Keenan Kidwell Sep 27 '12 at 2:03
up vote 3 down vote accepted

Suppose that $\operatorname{ran}(T)\subseteq\ker(T)$. Then for any $x\in V$, $$T^2(x)=T\big(T(x)\big)\in T[\operatorname{ran}(T)]\subseteq T[\ker(T)]=\{0\}\;,$$ so $T^2(x)=0$.

Conversely, suppose that $T^2$ is the zero transformation, and let $x\in\operatorname{ran}(T)$. Then $x=T(y)$ for some $y\in V$, and $T(x)=T^2(y)=0$, so ... ?

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This is very straightforward.

Suppose $T^2$ is the zero map. Then for any $v \in V$, $T^2(v) = T(T(v)) = 0$, which implies that $T(v) \in \ker T$ for any $v$. Hence $\{T(v): v \in V\} = T(V) \subset \ker T$. Now suppose, $T(V) \subset \ker T$. For any $v \in V$, $T^2(v) = T(T(v)) = 0$, so $T^2$ is the zero map.

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Hint: Assume by negation that $Range(T)\not\subset ker(T)$ and take $v\in Range(T)$ s.t $v\not\in ker(T)$.

What can you say about $T(v)$ ? can you find a vector $u$ s.t $T^{2}(u)\neq0$ ?

The other direction should be straightforward using the definitions (of composition, $ker(T)$ etc')

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