Epistemic status: Mostly the result of a late night procrastination session. Proceed with caution. Might be too verbose or not verbose enough.

I have been thinking about linear algebra quite a lot lately. There is a common proof strategy, which usually goes something like:

  1. You have a subspace of a vector space
  2. Get a basis for that subspace
  3. Use the exchange lemma to extend that basis to a basis of some larger subspace
  4. Try to figure out if your new basis, or maybe the basis of the complementary subspace, has some desirable properties

However, after reading a particularly cool proof (3.106) in Linear Algebra Done Right, I noticed a fascinating concept: many linear algebra theorems, which are commonly proven using the startegy described above, can often be proven by picking a bunch of appropriate linear maps between the relevant vector spaces and then using a bunch of “elementary” theorems, usually proven using the above method.

This usually takes a bit of effort to figure out, but the results are quite satisfying. Let me illustrate:

Proposition (Dimension of a sum): Let VV and UU be finitely generated subspaces of a common vector space. Then we have

dim(V×U)=dim(V+U)+dim(VU). \dim(V \times U) = \dim (V + U) + \dim (V \cap U).

Proof. Consider a pair of linear maps: f:V×UV+U(v,u)v+u,g:VUV×Uv(v,v). \begin{aligned} f : V \times U &\to V + U \\ (v, u) &\mapsto v + u, \\ g : V \cap U &\to V \times U \\ v &\mapsto (v, -v). \end{aligned}

It is clear that ff is surjective, and it is also clear that gg is injective. We just need to prove that Rang=Kerf\operatorname{Ran}{g} = \operatorname{Ker} f .

Take g(v)Rangg(v) \in \operatorname{Ran}{g} . Then we compute f(g(v))=f(v,v)=vv=0, f(g(v)) = f(v, -v) = v - v = 0, therefore RangKerf\operatorname{Ran} g \subseteq \operatorname{Ker}{f} .

Now we consider (v,u)Kerf(v, u) \in \operatorname{Ker}{f} . This means that f(v,u)=v+u=0,thusv=u. f(v, u) = v + u = 0, \quad \text{thus}\quad v = -u. Additionally, UU is a subspace, hence it contains the additive inverse for each of its elements, meaning v=uVUv = -u \in V \cap U . This means we can use gg to get g(v)=(v,v)=(v,u), g(v) = (v, -v) = (v, u), and therefore KerfRang\operatorname{Ker}{f} \subseteq \operatorname{Ran}{g} .

We finish by using the Rank-Nullity theorem twice, first on ff and then on gg dim(V×U)=dimRanf+dimKerf=dim(V+U)+dimRang=dim(V+U)+dim(VU). \begin{aligned} \dim (V \times U) &= \dim \operatorname{Ran}{f} + \dim\operatorname{Ker}{f} \\ &= \dim(V + U) + \dim{\operatorname{Ran}{g}} \\ &= \dim(V + U) + \dim(V \cap U). \end{aligned}

The idea is ultimately simple: we know that every element in V+UV + U can be expressed as a sum of a vector from VV and a vector from UU . By investigating the kernel of ff , we are essentially asking how many ways this can be done.

I am not really sure if constructing proofs like this one is really “practical”. However, I find this approach somewhat more aesthetic and enlightening than picking a bunch of basis and then getting lost in a sea of coordinates.

Anyway, this blog post is pretty different from what I used to write in the past. I have finished my Electrical Engineering bachelor’s degree and started working towards getting a bachelor’s degree in Mathematics, so I ultimately hope to post more in this vein in the future. I do hope to write some technical posts as well, I have not given up engineering completely and still continue to work as a freelance embedded systems engineer.