# Why do we only focus on finite-dimensional vector space?

I have self-learned (theoretical) linear algebra and read various books about this topic, such as Linear Algebra Done Right (Sheldon Axler) and Linear Algebra (Stephen Friedberg et al.). Also, I think I am familiar with set theory and the axiom of choice since I have completed chapters 3 and 8 of Terence Tao’s book Analysis I.

My question: Why these books focus on finite-dimensional vector space but not arbitrary ones? What is the main difference between finite-dimensional vector spaces and infinite-dimensional ones?

• Infinite dimensional vector spaces are studied, we usually call it Functional Analysis rather than Linear Algebra Sep 29 at 14:17
• On of the main practical differences is the definition/use of the notion of "basis" Sep 29 at 14:19
• Of course people study infinite dimensional vector spaces. Generally you want to assume some more structure (like an inner product) but not always.
– lulu
Sep 29 at 14:19
• It is better to start with learning something simpler, before going to generalities and more abstract methods. You need to learn to walk before being able to run.
– KBS
Sep 29 at 14:25
• "We" don't. But beginning learning concentrates on finite dimensional spaces because it has a lot of computational tools that make it easier to discuss, particularly, matrices and a basis. (Yes, technically, infinite dimensional spaces have bases, but they tend to exist only theoretically, not as something we can use for computation, and more commonly we have a kind of dense basis, which requires us to go into the area of no red vector vector spaces.) Sep 29 at 14:27

The finite-dimensional case is much simpler, one can say much more about it, and it already suffices for many applications (systems of equations, first-order linear homogeneous constant-coefficient ODEs). It's a great place to start for practical, theoretical, and pedagogical reasons.

In the infinite-dimensional case a few things still work (bases (although this requires the axiom of choice), rank-nullity), but:

1. A linear map $$T : V \to V$$ can be injective without being surjective, and vice versa;
2. Linear maps $$T : V \to V$$ can have no eigenvalues; there is no analogue of Jordan normal form;
3. It is no longer possible to define the trace or determinant in general; there is no analogue of the characteristic polynomial;
4. Row rank is no longer equal to column rank;
5. A linear map can be left-invertible without being right-invertible and vice versa; that is, $$TS = I$$ no longer implies $$ST = I$$;
6. The double dual map $$V \to (V^{\ast})^{\ast}$$ is never an isomorphism (assuming the axiom of choice)

and probably other terrible stuff I'm forgetting.

Also, if you don't believe the axiom of choice, vector spaces can have no bases or trivial dual. Here is an illustrative example: consider the uncountable-dimensional vector space $$\mathbb{R}^{\mathbb{N}}/\mathbb{R}^{(\mathbb{N})}$$ given by the quotient of the space of functions $$\mathbb{N} \to \mathbb{R}$$ by the subspace of functions with finite support (meaning they are zero except at finitely many points). Can you explicitly write down a basis of this vector space? (Already it's not trivial and a nice exercise to write down uncountably many linearly independent vectors.) Can you explicitly write down a nonzero linear functional on it?

In order to deal with infinite-dimensional linear algebra, which historically was almost entirely motivated by the desire to solve differential equations such as the heat equation or wave equation, we need to bring in tools from analysis; this involves talking about topological vector spaces, normed vector spaces, Banach spaces, Hilbert spaces, etc. depending on the desired application. This is much more machinery than is needed in the finite-dimensional case and is covered in a course on functional analysis, not an introduction to linear algebra.