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.

  • 2
    $\begingroup$ It is true, how do you prove the rank-nullity theorem for finite dimensional v.s.? It will give you some insight. $\endgroup$
    – sansae
    Jan 27, 2022 at 1:00
  • $\begingroup$ I see, thank you! $\endgroup$
    – Isochron
    Jan 27, 2022 at 1:05

1 Answer 1


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 $\dim V < \infty$.

  • $\begingroup$ What is a Hamel basis? $\endgroup$
    – user26857
    Jan 27, 2022 at 6:41
  • $\begingroup$ @user26857 the usual basis; there are other notions of basis for infinite dimensional topological vector spaces. $\endgroup$ Jan 27, 2022 at 12:47
  • $\begingroup$ I said usual, not canonical. The concept of Hamel basis is the one taught in elementary linear algebra, that is, a set of l.i. vectors that span the space through finite linear combinations. All bases you meet on a course of linear algebra are Hamel bases. Hence, usual. Alternative notions such as Schauder basis and Total Orthonomal sets arise in connection to infinite dimensional vector spaces (the realm of functional analysis), where the usual notion of Hamel basis is not that useful. $\endgroup$ Jan 27, 2022 at 16:26
  • $\begingroup$ 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) $\endgroup$ Jan 27, 2022 at 16:39
  • $\begingroup$ 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. $\endgroup$ Jan 27, 2022 at 16:42

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