# The Unapologetic Mathematician

## 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 $\mathbf{Vect}(\mathbb{F})$ (also $\mathbf{FinVect}(\mathbb{F})$) 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 $T:V\rightarrow W$. The first isomorphism theorem says we can factor $T$ as a surjection $E$ followed by an injection $M$. We’ll just regard the latter as the inclusion of the image of $T$ as a subspace of $W$. As for the surjection, it must be the linear map $V\rightarrow V/\mathrm{Ker}(T)$, just as in any abelian category. Then we can set up the short exact sequence

$\mathbf{0}\rightarrow\mathrm{Ker}(T)\rightarrow V\rightarrow V/\mathrm{Ker}(T)\rightarrow\mathbf{0}$

and the isomorphism theorem allows us to replace the last term

$\mathbf{0}\rightarrow\mathrm{Ker}(T)\rightarrow V\rightarrow \mathrm{Im}(T)\rightarrow\mathbf{0}$

Now since every short exact sequence splits we have an isomorphism $V\cong\mathrm{Ker}(T)\oplus\mathrm{Im}(T)$. 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 $T(v)$. 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 $T$ sends to the zero vector in $W$.

So what does the direct sum decomposition above mean? It tells us that there is a basis of $V$ which is in bijection with the disjoint union of a basis for $\mathrm{Ker}(T)$ and a basis for $\mathrm{Im}(T)$. In the finite-dimensional case we can take cardinalities and say that the dimension of $V$ is the sum of the dimensions of $\mathrm{Ker}(T)$ and $\mathrm{Im}(T)$. Or, to use our new words, the dimension of $V$ is the sum of the rank and the nullity of $T$. Thus: the rank-nullity theorem.

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Posted by John Armstrong | Algebra, Linear Algebra

## 13 Comments	»

1. [...] Let’s go back and consider a linear map . Remember that we defined its rank to be the dimension of its image. Let’s consider this a little more [...]

Pingback by | July 1, 2008 | Reply

2. [...] be the dimension of this subspace, which we called the nullity of the linear transformation . The rank-nullity theorem then tells us that we have a relationship between the number of independent solutions to the system [...]

Pingback by | July 14, 2008 | Reply

3. [...] system has no solutions at all. What’s the problem? Well, we’ve got a linear map . The rank-nullity theorem tells us that the dimension of the image (the rank) plus the dimension of the kernel (the nullity) [...]

Pingback by | July 18, 2008 | Reply

4. [...] the dimension of the cokernel add up to the dimension of the target space. But notice also that the rank-nullity theorem tells us that the dimension of the kernel and the dimension of the image add up to the dimension of [...]

Pingback by | July 22, 2008 | Reply

5. [...] the rank-nullity theorem says that , and similarly for all other linear maps. So we get — which expresses as the [...]

Pingback by | July 23, 2008 | Reply

6. [...] the time we get to the th power, where . Instead of repeating everything, let’s just use the rank-nullity theorem, which says for each power that . Now if then we [...]

Pingback by | February 18, 2009 | Reply

7. [...] is, the kernel of is trivial. Since is a transformation from the vector space to itself, the rank-nullity theorem tells us that the image of is all of . That is, must be an invertible [...]

Pingback by | July 17, 2009 | Reply

8. [...] this means that we can read off the rank of from the number of rows in , while the nullity is the number of zero columns in [...]

Pingback by | August 18, 2009 | Reply

9. [...] count the dimensions — if has dimension then each space has dimension — and use the rank-nullity theorem to see that they must be isomorphic. That is, every -multilinear functional is a linear combination [...]

Pingback by | October 22, 2009 | Reply

10. [...] this were a linear system, the rank-nullity theorem would tell us that our solution space is (generically) dimensional. Indeed, we could use [...]

Pingback by | November 19, 2009 | Reply

11. [...] we already know that their dimensions are equal, so the rank-nullity theorem tells us all we need is to find an injective linear map from one to the [...]

Pingback by | October 13, 2010 | Reply

12. [...] that this is onto, so the dimension of the image is . The dimension of the source is , and so the rank-nullity theorem tells us that the dimension of the kernel — the dimension of the space that sends back to [...]

Pingback by | November 9, 2010 | Reply

13. [...] to a constant function is automatically zero. Thus we conclude that . In fact, we can say more. The rank-nullity theorem tells us that the dimension of and the dimension of add up to the dimension of , which of course [...]

Pingback by | April 26, 2011 | Reply

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