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 September 20th, 2012, 07:12 AM #1 Newbie   Joined: Apr 2012 Posts: 18 Thanks: 0 Dimensions, linear transformations Prove that if A: X-->Y is a linear transformation and V is a subspace of X then dimension of AV =< dimension of V. Deduce from here that rank (AB)=
September 21st, 2012, 08:22 AM   #2
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Re: Dimensions, linear transformations

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 Originally Posted by amateurmathlover Prove that if A: X-->Y is a linear transformation and V is a subspace of X then dimension of AV =< dimension of V. Deduce from here that rank (AB)=
The rank nullity theorem that you reference says that dim V= dim(N)+ dim(I) where N is the nullity of A and I is its image. Look at the two cases, dim(N)= 0 and dim(N)> 0. All three numbers, of course, are non-negative. Tags dimensions, linear, transformations Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post matqkks Linear Algebra 1 February 7th, 2012 01:39 PM nuke Linear Algebra 3 April 14th, 2011 12:32 PM TsAmE Linear Algebra 0 October 9th, 2010 06:54 AM wontonsoup Linear Algebra 2 May 25th, 2009 04:35 PM ypatia Linear Algebra 1 March 2nd, 2009 06:28 PM

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