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Let $T:V\to V$ be a linear transformation from a finite dimensional vector space $V$. We are asked to show that $\textrm{rank}(T)=\textrm{rank}(T^2)\implies \textrm{N}(T)\cap\textrm{R}(T)={0}$.

My idea was as follows:

Suppose $\textrm{rank}(T)=\textrm{rank}(T^2)$. Define the transformation $T':\textrm R(T)\to V$ by $v\mapsto T(v)$. This is obviously a linear transformation. By the dimension theorem for linear transformations we have $\textrm{dim}(N(T'))+\textrm{rank}(T^2)=\textrm{dim}(N(T'))+\textrm{dim}(R(T'))=\textrm{dim}(R(T))=\textrm{rank}(T)$, so $\textrm{dim}(N(T'))=0$ which implies that $T'$ is injective.

I was hoping to be able to use this to show what I was meant to show, but I can't find how. Can someone help me with this or suggest another way of getting the required result?

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    You haven't actually defined $T'$, but I presume you mean $T'(\mathbf x) = T(\mathbf x)$ for $\mathbf x \in R(T) \subseteq V$. Perhaps this clarifies how $N(T')$ is related to $N(T) \cap R(T)$.2012-12-03
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    Are N(T), R(T) the kernel and image, resp., of T?2012-12-03
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    Yes N(T) is the kernel and R(T) the image of T2012-12-03
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    @ErickWong I think I see it now. $\textrm N(T')\subseteq \textrm N(T)$ in such a way that all elements in $\textrm N(T')$ also lie in $\textrm R(T)$ as that is the domain. $N(T')=N(T)\cap R(T)$, right? Could you tell me how I would write this down formally?2012-12-03
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    @mr.FS You got it. $T'$ is just the restriction of $T$ to $R(T)$, so $v \in N(T')$ iff $v \in R(T)$ and $T(v)=0$, which is synonymous with $v\in R(T) \cap N(T)$.2012-12-04

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