Can this theorem be generalized to infinite dimensional vector spaces?

Question

I am reading some basic linear algebra theory, and I came across this theorem.

Let $V, V'$ be vector spaces and $\ \dim V \lt \infty$
Then for every basis $e_1, e_2, \dots, e_n$ of $V$ and
any $n$ vectors $v_1, v_2, \dots v_n$ from $V'$
there exists a unique linear map $f : V \to V'$ such that

$f(e_i) =v_i$ for every $i=1,2,\dots,n$

So it's about existence and uniqueness.

I can understand the proof just fine. But it seems to me the proof doesn't use significantly the fact that $$\ \dim V \lt \infty$$ i.e. the fact that V is finite dimensional. So I wonder if this theorem can be formulated for an infinite dimensional space $V$ somehow and would it still be true? (question 1)

Say e.g. when $V= \mathbb{R}[x]$ (the polynomials over the reals).

Also, if it can be formulated and still be true, would it matter if the basis of $V$ is countable or uncountable? (question 2)

I am mostly interested in the countable case though (question 1). I am just not sure I will understand the uncountable case i.e. question 2 (given my current knowledge).

Answer 1

Yes, it can be generalized to infinite dimensional linear spaces and the dimension doesn't have to be countable.

To be precise, the exact statement is Let $V, W$ be two vector spaces over a field $F$, and $\{e_i\}_{i\in I}$ is a (Hamel) basis of $V$, then for an arbitrary function $v: I\rightarrow W$, there is a unique linear map $T:V\rightarrow W$ such that $T(e_i) = v(i)$. Here the index set $I$ need not to be finite or countable.

The proof is also essentially the same as in the finite dimensional case. But to show a basis always exists is much harder as it requires Zorn's lemma.

Answer 2

Under appropriate set-theoretic assumptions, this is indeed true for every vector space.

Specifically, the key is the axiom of choice (most usefully in its guise as Zorn's lemma). This is usually taken as one of the standard axioms of mathematics, but is odd enough that it is in my opinion worth noting when its use is necessary. In fact, we have the following theorem due to Blass:

Over $\mathsf{ZF}$ (= set theory without the axiom of choice), the axiom of choice is equivalent to the statement "Every vector space has a basis."

This can even happen "close to home" - it is in fact consistent with $\mathsf{ZF}$ that $\mathbb{R}$ does not have a basis as a vector space over $\mathbb{Q}$. On the other hand, the existence of bases for finite-dimensional spaces is provable in $\mathsf{ZF}$ alone (taking a bit of care to phrase this so it's not outright trivial).

Summing up, the situation is the following:

Nothing special is needed to prove that if $V,W$ are $k$-vector spaces, $B$ is a basis for $V$, and $f:B\rightarrow W$ is a function, then there is exactly one linear linear map $g:V\rightarrow W$ extending $f$. However, the existence of bases in the first place does in general appeal to the axiom of choice.

Answer 3

The existence and uniqueness here follows from the very definition of basis. Namely, $(e_i)_{i\in I}$ is a basis of a vector space $V$ over field $\mathbb F$ if and only if every vector $v\in V$ can be uniquely expressed as a finite linear combination $v=\alpha_1e_{i_1}+\ldots+\alpha_ne_{i_n}$ for some $n\in\mathbb N, \alpha_1,\ldots,\alpha_n\in\mathbb F$.

Then, given the vectors $(v_i)_{i\in I}$ in another vector space $V'$ over $\mathbb F$, there is a unique linear map $L:V\to V'$ which maps each $e_i$ into $v_i$, and it is defined by:

$$L(\alpha_1e_{i_1}+\ldots+\alpha_ne_{i_n})= \alpha_1v_{i_1}+\ldots+\alpha_nv_{i_n}$$

for every $n\in\mathbb N, \alpha_1,\ldots,\alpha_n\in\mathbb F$.

To prove that, one should go through the usual steps to follow the definitions and prove that this map is well-defined (follows from unique representation in the LHS), is linear, and is unique.