The Rank-Nullity Theorem
October 3, 2011 Leave a comment
Take a basis for , and extend it to a basis for . Now it suffices to show that form a basis for (since then we would have , , and ).
First we will check that are linearly independent. Suppose . Then, , so . This means it can be represented in terms of the . Write
But this is a linear combination of basis vectors of which sum to zero, so it must be that each of the and (we don’t really care about the , but it is true anyway). The interesting part is that for every , exactly the condition needed to say that is linearly independent.
Now we need only show that . Let . There is some vector which has . Let’s write . When we apply , we get
But since, for each , ,
Therefore, form a basis for , proving the theorem.
Notice the distinct lack of assumptions. This is true for any linear transformation over any vector spaces over any field. That’s cool.
Another thing that is cool is that this theorem generalizes wildly. This is the vector space version of what is known as Noether’s first isomorphism theorem. It’s true in the context of many other algebraic structures. We’ll probably see this result soon, at least in the context of groups.
This is the first real theorem about dimension. I wanted you to see this so that you at least understand the flavor of proof. We’ll have another similar proof when we get to Galois Theory.