# The rank-nullity theorem

Posted on January 28, 2021
Tags: math, linear-algebra, abstract-algebra

This post is about the what the rank-nullity theorem, a rather basic and fundamental result in linear algebra, really means.

First, I will state a result about vector spaces from which one could prove the rank-nullity theorem:

Theorem 1: every short exact sequence of vector spaces splits.

This looks nothing like the rank-nullity theorem! But first we can look at how this implies rank-nullity. A short exact sequence is just a sequence $$0 \rightarrow U \xrightarrow{f} V \xrightarrow{g} W \rightarrow 0$$ of vector spaces (or $R$-modules, which are exactly the same as vector spaces except that they are defined over a base ring $R$ that is not necessarily a field), where $f$ is injective, $g$ is surjective, and $\mathrm{im} \ f = \ker g$ (hence the name “exact”).

This sequence is said to be split when $V \cong U \oplus W$. A short exact sequence is interesting, because it tells us how $V$ is built from $U$ and $W$: if the exact sequence is split, then we have the simplest building method (namely, by direct sum), which make things much easier. Another way to think about this is by looking at the splitting lemma:

Lemma (splitting lemma): for a short exact sequence $$0 \rightarrow L \xrightarrow{f} M \xrightarrow{g} N \rightarrow 0$$ of $R$-modules, where $R$ is any ring, the following are equivalent:

• $f$ is left-invertible, i.e. there is a morphism $h : B \rightarrow A$ such that $h \circ f = \mathrm{id}_A$;
• $g$ is right-invertible;
• the sequence is split, i.e. $M \cong L \oplus N$.

I won’t repeat the proof here, because it could be found in any graduate algebra text. On a side note, this facts actually holds over any abelian category, not just $\mathbf{Mod} R$.

The fact that every short exact sequence of vector spaces splits is rooted in a much simpler fact: any vector space has a basis. We can use this fact to prove Theorem 1:

Proof of Theorem 1: let $\{w_i\}$ be a basis of $W$. Since $g$ is necessarily surjective,. we can find a set $\{v_i\} \subseteq V$ such that $f(v_i) = w_i$. Then we define a function $h : W \rightarrow V$ by setting $h(w_i) = v_i$. This is well-defined because a linear map is determined by its action on basis elements. $h$ is a right inverse to $g$, so by the splitting lemma this short exact sequence is split.

Proof of rank-nullity: at this point, the rank-nullity theorem is almost trivial. Simply consider the short exact sequence: $0 \rightarrow \ker f \hookrightarrow V \twoheadrightarrow \mathrm{im} \ f \rightarrow 0$ and observe that, since the sequence is split, $V \cong \ker f \oplus \mathrm{im} \ f$. Taking dimension on both sides yields the rank-nullity theorem.

In fact, we have proven a statement that is more general than the rank-nullity theorem! When $V$ is infinite-dimensional, the rank-nullity theorem fails, but $V \cong \ker f \oplus \mathrm{im} \ f$ still holds.

This, again, tells us one important fact about vector spaces: that linear maps contain all the information we need to study a space. This is, again, due to the fact that every vector space has a basis, i.e. a generating set that is linearly independent! This is what makes vector spaces so much simpler than general modules.

On the other hand, the very familiar first isomorphism theorem also implies rank-nullity:

Theorem 2 (the first isomorphism theorem): let $M, N$ be modules over a base ring, and $f : M \rightarrow N$ a linear map. Then $\mathrm{im} \ f \cong M/\ker f$.

This is slightly weaker than Theorem 1, but enough to prove rank-nullity:

Alternate proof: simply take the dimension on both sides in the isomorphism given by the first isomorphism theorem.

This reiterates the essence of rank-nullity: the rank-nullity theorem is just the first isomorphism theorem in disguise, telling us that a linear map divides the information about a vector space into two complementary parts!