Let be a linear map.
The nullity of , written , is .
The rank of , written is .
Returning to the differentiation example from the end of the last lecture, has nullity 1 (since its kernel was one-dimensional, spanned by the constant polynomial 1) and rank , since its image had a basis of size . Notice that , this isn’t a coincidence.
Let be a linear map. Then
This is called the rank-nullity theorem.
We’ll assume and are finite-dimensional, not that it matters. Here is an outline of how the proof is going to work.
Choose a basis of
Extend it to a basis of using Lemma 4.12.2. When we’ve done this, and we need only show )
Show that is a basis of .
The only part needing elaboration is the last part. First, I claim that span . Any element of the image is equal to for some . We have to show that any such lies in the span of the s.
Since is a basis of we may write as for some scalars . Then
linearity | ||||
as required.
Now I claim is linearly independent. Suppose
so that we need to show the are all 0. Using linearity,
which means . As is a basis for , we can write
for some scalars . But , being a basis, is linearly independent and so all the scalars are 0. In particular all the are 0, which completes the proof. ∎