# 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.

June 27, 2008 Posted by | Algebra, Linear Algebra | 13 Comments