Kernel, range and nullity + rank = dim(domain) theorem

Forum for the GRE subject test in mathematics.
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Kernel, range and nullity + rank = dim(domain) theorem

Postby fullofquestions » Fri Oct 26, 2007 11:11 am

I believe I understood this at some point but it is when I re-read the text that I get a bit hung up. Seeking reinforcement/clarification...

Let T: R^n -> R^m by a linear transformation.

Kernel T: set of all vectors in R^n that get mapped to the zero vector.
Nullity T: dimension (Kernel T), i.e. the # of vectors in Kernel T

Range T: set of all images of T. In symbolic form: R(T) = {T(x): x E R^n}
So we have that x E R^n but the images of T/ Range of T is a subspace of R^m.
Rank T = dimension (Range T), i.e. the # of images/elements produced by T.

Rank plus nullity theorem
nullity(T) + rank(T) = n = dimension (domain T)

Since we are talking about vector spaces we have to remember that "every vector space must contain the zero vector"

From the definitions I've been reading I regard the range of T simply as all the images/elements, whether or not they are the zero vector. Therefore, it seems that in some cases, when the nullity(T) is greater than zero, we are double counting those vectors that map to the zero vector image in both the nullity and rank number. Does this make sense to anyone?

Perhaps my definition of images and maps are incorrect. For f: A -> B, I see a map as the translation of an element in A to an element in B. Therefore the linear transformation has x number of maps where x is the total number of elements in A. Likewise, an image is the number of elements in B.

Lastly, there is a dense paragraph in the book that I cannot say I understand unless if I re-read it a couple of times. Perhaps an alternate explanation might be better. Here it goes:

"If T: R^n -> R^n is a linear operator, then the nullity of T is zero precisely when the rank of T is n. It can be shown that T is one-to-one when the nullity is zero, and that T is onto when the rank of T is n. Therefore, the linear operator T is one-to-one if and only if it's onto and this happens when T has a full rank. So, if T(x) = Ax , then T is one-to-one and onto when the rank of A is n, that is, when det A != 0"

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