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 February 11th, 2011, 06:12 AM #1 Member   Joined: Feb 2011 Posts: 58 Thanks: 0 linear map T, finite dimensional vector space proof if we let T: V$\mapsto$V be a linear map, where V is a finite dimensional vector space, and suppose Rank($T$)=Rank($T^2$) prove that Im($T$)=Im($T^2$)
 February 11th, 2011, 09:19 AM #2 Senior Member   Joined: May 2008 From: York, UK Posts: 1,300 Thanks: 0 Re: linear map T, finite dimensional vector space proof $\text{Rank}(T)=\dim(\text{Im}(T)).$ We know that for $v\in V$ we have $T^2v=T(Tv)\in\text{Im}(T),$ so $\text{Im}(T^2)\subseteq\text{Im}(T).$ But the image of a linear operator is a vector space in its own right, and if vector spaces $V,W$ have the properties $V\subseteq W$ and $\dim(V)=\dim(W)$ it follows that $V=W.$

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