# Rank Nullity "Converse"?

The rank nullity theorem requires a linear map $$T:V \longrightarrow W$$ between a finite dimensional domain VS and some VS W.

The conclusion of the theorem is that $$\text{Dim}(V) = \text{Dim}(\text{Ker}(T))+\text{Dim}(\text{Ran}(T))$$.

My question is, if I have a linear map $$T: V \longrightarrow W$$ between any two vector spaces and $$\text{Dim}(\text{Ker}(T))+\text{Dim}(\text{Ran}(T)) < \infty$$, then is $$\text{Dim}(V) < \infty$$ with $$\text{Dim}(V) = \text{Dim}(\text{Ker}(T))+\text{Dim}(\text{Ran}(T))$$?

Apologies if this is blatantly true or false.

Thank you.

• It is true, how do you prove the rank-nullity theorem for finite dimensional v.s.? It will give you some insight. Jan 27, 2022 at 1:00
• I see, thank you! Jan 27, 2022 at 1:05

Suppose $$\dim{V} = \infty$$, then we can write $$V = \ker{T} \oplus U$$, where $$U$$ is an infinite dimensional subspace of $$V$$. Let $$\{u_i\}_{i \in I}$$ be a Hamel basis for $$U$$, then $$\{T(u_i)\}$$ would be a Hamel basis for $$\operatorname{Ran}{T}$$. This means $$\operatorname{Ran}{T}$$ would need to be infinite dimensional, which it isn't. Hence, we have a contradiction, and $$\dim V < \infty$$.
• You only start hearing the name Hamel basis during a course where such alternative notions are needed, so that you can differentiate one from the other. Let me give you a motivating example: pick the set of real sequences such that $\sum_{i=1}^\infty x_i^2$ is finite. Convince yourself that this is indeed a vector space. Now try to find a basis for this space. The natural choice is $e_1 = (1,0,...), e_2 = (0,1,0,...), ...$, but the span of this set is only the sequences with finitely many non-zero terms; it doesn't span the whole space! If you allow for infinite linear combinations (series) Jan 27, 2022 at 16:39
• Then the $(e_n)$ indeed form a basis, but in a slightly unusual way. Hence, it gets called a Schauder basis, instead of Hamel basis. Note that, for this to work, you need some notion of convergence; here is where the norm kicks in. Schauder bases are only available in normed vector spaces. Worst still, not all normed vector spaces admit such a basis (in sharp contrast, every vector space has a Hamel basis, regardless of dimension). The ones that admit are nicer than the ones that don't. Jan 27, 2022 at 16:42