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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

Quote:
 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.

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