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Is the following argument correct?

Proposition. Given a finite-dimensional vector space $V$ and a linear map $T:V\to V$ such that $\operatorname{rank}(T^2) = \operatorname{rank}(T)$ prove that $\operatorname{range}T\cap\operatorname{null}T = \{0\}$.

Proof. Assume that there exists at least one non-zero vector $v\in\operatorname{range}T\cap\operatorname{null}T$ and let $U$ be the linear map formed by restricting the domain of $T$ to $\operatorname{range}T$ then the rank-nullity theorem implies $$\dim\operatorname{range}T = \dim\operatorname{null}U+\dim\operatorname{range}U = \dim \operatorname{null}U+\dim\operatorname{range}T^2$$ Now since $\operatorname{range}T\cap\operatorname{null}T\subseteq\operatorname{null}U$ it follows that $v\in\operatorname{null}U$ and thus $\dim\operatorname{null}U\ge1$ consequently $\dim\operatorname{range}T^2<\dim\operatorname{range}T$.

$\blacksquare$

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  • $\begingroup$ Seems perfect to me. $\endgroup$ – Bill O'Haran Apr 26 '18 at 6:58
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Here is another take, which avoids contradiction.

$\operatorname{rank}(T^2) = \operatorname{rank}(T)$ implies $\dim \operatorname{null}(T^2) = \dim \operatorname{null}(T)$.

Since $\operatorname{null}(T) \subseteq \operatorname{null}(T^2)$, we have $\operatorname{null}(T^2) =\operatorname{null}(T)$.

Now, let $v \in \operatorname{range}T\cap\operatorname{null}T$.

Write $v = Tw$. Then $0 = Tv = T^2 w$ and so $w \in \operatorname{null}(T^2) = \operatorname{null}(T)$.

Therefore, $v=Tw=0$.

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