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 May 19th, 2009, 06:19 AM #1 Newbie   Joined: May 2009 Posts: 20 Thanks: 0 linear algebra- vector spaces i really need help with this question: Let V be a finite-dimensional vector space. suppose that T : V-->V is a linear operator. Show that T is injective if and only if T is surjective. and are there any results for sets and functions?
 May 20th, 2009, 06:53 AM #2 Senior Member   Joined: May 2008 From: York, UK Posts: 1,300 Thanks: 0 Re: linear algebra- vector spaces Use the rank-nullity theorem - i.e. the one that says $\dim(\ker(T))+\dim(\operatorname{im}(T))=\dim(V).$ You can find $\dim(\ker(T))$ by noting that if $T$ is injective, then $T(x)-T(y)=0\Leftrightarrow x-y=0,$ and using the properties of a linear map.
 May 20th, 2009, 08:23 AM #3 Newbie   Joined: May 2009 Posts: 20 Thanks: 0 Re: linear algebra- vector spaces sorri could u possibly expand you reply im not getting you cheers
 May 20th, 2009, 08:43 AM #4 Senior Member   Joined: May 2008 From: York, UK Posts: 1,300 Thanks: 0 Re: linear algebra- vector spaces The rank-nullity theorem states that for a linear map $T:V\to W,$ the dimension of the kernel of $T$ and the dimension of the image of $T$ sum to the dimension of $V.$ The kernel of $T$ is defined to be the space of vectors $x\in V$ such that $Tx=0.$ The image of $T$ is the space of vectors $y\in W$ such that there exists an $x\in V$ such that $Tx=y.$ Clearly, a linear operator $T:V\to W$ is surjective iff $\operatorname{im}(T)=W.$ You need to show that for $T:V\to V,$ we have $\operatorname{im}(T)=V$ iff $T$ is injective, using the rank-nullity theorem. To find the kernel of an injective map, note that if $T$ is injective, then $T(x)=T(y)\Leftrightarrow x=y$ for all $x,y\in V,$ or $T(x)-T(y)=0\Leftrightarrow x-y=0.$ If you let $v=x-y,$ you therefore have that $T$ is injective iff $T(y+v)-T(y)=0\Leftrightarrow v=0$ for all $v,y\in V.$ Then use the properties of a linear operator to show that $\ker(T)=\{0\}.$ Once this is established, you get the desired result by a simple application of the theorem.
 May 26th, 2009, 12:48 AM #5 Newbie   Joined: May 2009 Posts: 20 Thanks: 0 Re: linear algebra- vector spaces thankyou so much for your help it was much appreciated

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