The Rank-Nullity Theorem
Today we start considering how a given linear transformation acts on a vector space. And we start with the “rank-nullity” theorem. This sounds really fancy, but it’s actually almost trivial.
We already said that 
(also 
) is a abelian category. Now in any abelian category we have the first isomorphism theorem.
So let’s take this and consider a linear transformation 
. The first isomorphism theorem says we can factor 
as a surjection 
followed by an injection 
. We’ll just regard the latter as the inclusion of the image of as a subspace of 
. As for the surjection, it must be the linear map 
, just as in any abelian category. Then we can set up the short exact sequence
and the isomorphism theorem allows us to replace the last term
Now since every short exact sequence splits we have an isomorphism 
. This is the content of the rank-nullity theorem.
So where do “rank” and “nullity” come in? Well, these are just jargon terms. The “rank” of a linear transformation is the dimension of its image — not the target vector space, mind you, but the subspace of vectors of the form 
. That is, it’s the size of the largest set of linearly independent vectors in the image of the transformation. The “nullity” is the dimension of the kernel — the largest number of linearly independent vectors that sends to the zero vector in .
So what does the direct sum decomposition above mean? It tells us that there is a basis of 
which is in bijection with the disjoint union of a basis for 
and a basis for 
. In the finite-dimensional case we can take cardinalities and say that the dimension of is the sum of the dimensions of and . Or, to use our new words, the dimension of is the sum of the rank and the nullity of . Thus: the rank-nullity theorem.
June 27, 2008 -
Posted by John Armstrong |
Algebra, Linear Algebra
The Unapologetic Mathematician 
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