4 Linear algebra

4.16 The rank-nullity theorem

4.16.1 Definition of rank and nullity

Definition 4.16.1.

Let T:VW be a linear map.

  • The nullity of T, written nullT, is dimkerT.

  • The rank of T, written rankT is dimimT.

Example 4.16.1.

Returning to the differentiation example from the end of the last lecture, D:n[x]n[x] has nullity 1 (since its kernel was one-dimensional, spanned by the constant polynomial 1) and rank n, since its image had a basis 1,x,,xn1 of size n. Notice that rank(D)+null(D)=dimn[x], this isn’t a coincidence.

4.16.2 Statement of the rank-nullity theorem

Theorem 4.16.1.

Let T:VW be a linear map. Then

rankT+nullT=dimV.

This is called the rank-nullity theorem.

Proof.

We’ll assume V and W are finite-dimensional, not that it matters. Here is an outline of how the proof is going to work.

  1. 1.

    Choose a basis 𝒦=𝐤1,,𝐤m of kerT

  2. 2.

    Extend it to a basis =𝐤1,,𝐤m,𝐯1,,𝐯n of V using Lemma 4.12.2. When we’ve done this, dimV=m+n and we need only show dimimT=n)

  3. 3.

    Show that T(𝐯1),,T(𝐯n) is a basis of imT.

The only part needing elaboration is the last part. First, I claim that T(𝐯1),,T(𝐯n) span imT. Any element of the image is equal to T(𝐯) for some 𝐯V. We have to show that any such T(𝐯) lies in the span of the T(𝐯i)s.

Since is a basis of V we may write 𝐯 as i=1mλi𝐤i+i=1nμi𝐯i for some scalars λi,μi. Then

T(𝐯) =T(i=1mλi𝐤i+i=1nμi𝐯i)
=i=1mλiT(𝐤i)+i=1nμiT(𝐯i) linearity
=i=1nμiT(𝐯i) as 𝐤ikerT
span(T(𝐯1),,T(𝐯n))

as required.

Now I claim T(𝐯1),,T(𝐯n) is linearly independent. Suppose

i=1nμiT(𝐯i)=0,

so that we need to show the μi are all 0. Using linearity,

T(i=1nμi𝐯i)=0

which means i=1nμi𝐯ikerT. As 𝒦 is a basis for kerT, we can write

i=1nμi𝐯i=i=1mλi𝐤i

for some scalars λi. But , being a basis, is linearly independent and so all the scalars are 0. In particular all the μi are 0, which completes the proof. ∎