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 February 6th, 2012, 11:42 AM #1 Member   Joined: Apr 2011 Posts: 36 Thanks: 0 Linear Transformations in Linear algebra What is the most tangible way to introduce linear transformations? Most books tend to start with a very abstract view which is off putting for some students. Is there a short and simple proof of the Nullity - Rank Theorem which claims that if T: U->V is a linear transformation then rank(T)+Nullity(T)=n where n is the n dimension vector space U.
February 7th, 2012, 01:39 PM   #2
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Re: Linear Transformations in Linear algebra

$\mbox{I think in most books they start with the definition of a linear transformation.}$

Quote:
 Originally Posted by matqkks Is there a short and simple proof of the Nullity - Rank Theorem which claims that if T: U \to V is a linear map then rank(T)+Nullity(T)=n where n is the n dimension vector space U.
There're different ways to prove this. One way:
Let $T: U \to V$ be a linear map. We have to prove:
$dim(\ker(T))+\dim(Im(t))=dim(U)$

You should know the quotient space $U/\ker(T)$ and $Im(T)$ are isomorfic vector spaces and therefore they have the same dimension. So we have:
$dim(U/\ker(T))=dim(Im(T))$
Applying the dimension formula: $dim(U/\ker(T))=dim(U)-dim(\ker(T))$ we obtain:
$dim(U)-dim(\ker(T))=dim(Im(t))$
$\Rightarrow dim(U)=dim(Im(t))+dim(\ker(T))$

This dimension formula is the most general form. Assume you have a $(m,n)-$ matrix. $A$. We can associate a linear map defined as:
$f_A: K^n \to K^m: x \mapsto f_A(x)= A\cdot x$
Try to figure out how this implies into rank(T)+nulity(T)=n.

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