# Finding a basis of an infinite dimensional vector space with a given vector

If $$K$$ is a field and $$V=K^n$$ a finite dimensional $$K$$-vector space with basis $$A=\{e_1,\dots,e_n\}$$, then, given a vector $$v\in V$$ we can find a new basis $$E$$ of $$V$$ such that $$v\in E$$; this is done as follows:

First we form the set $$\{v\}\cup A$$ and observe that it has $$n+1$$ elements, therefore it spans all of $$V$$. Now we start the following process: We set $$E_0=\{v\}$$. If $$e_1\not\in\text{span}(E_0)$$, then we set $$E_1=\{v\}\cup\{e_1\}$$. Otherwise, $$E_1=E_0$$. Then, if $$e_2\not\in\text{span}(E_1)$$, we set $$E_2=E_1\cup\{e_2\}$$; otherwise, $$E_2=E_1$$. We go on until we eliminate every element of $$A$$; the set $$E_n$$ is our basis $$E$$; obviously it is linearly independent and note that it is impossible to "rule out" two elements of $$A$$, since if $$e,e'$$ were ruled out, that would mean that there are not-all-zero scalars $$\lambda_i,\mu_i$$ such that $$e=v+\sum\lambda_ie_i$$, $$e'=v+\sum\mu_ie_i$$, then $$e=e'+\sum(\lambda_i-\mu_i)e_i$$, which is impossible, therefore $$E$$ has $$n$$ elements.

My question is, can we do the same for infinite dimensional vector spaces? That is, given a vector $$v$$ on an infinite dimensional $$K$$-vector space $$V$$, can we construct (feels like it is too much to ask)/ can we prove the existence of a basis $$E$$ such that $$v\in E$$?

P.S: I live in a world where the axiom of choice is true, so $$V$$ does have a basis.

• BTW. A property $P(x)$, which any set $x$ may or may not have, is called a property of finite character when $P(x)\iff \forall$ finite $y\subset x\,(P(y))$ holds for every $x$. For example, being a linearly independent subset of $V$ is a property of finite character. The Teichmuller-Tukey Lemma: If $P$ is of finite character and $P(\{x\})$ for some $x\in X$ then there is a $\subset$-maximal $Y\subset X$ such that $x\in Y$ and $P(Y).$ In the axiom system ZF, this "lemma" is equivalent to the axiom of choice. – DanielWainfleet Feb 28 at 10:41

Yes (as long as $$v \neq 0$$). You prove that a basis exists via Zorn's Lemma, proving that a maximal element (which must exist by Zorn's Lemma) in the collection of linearly independent subsets of $$V$$ (partially ordered by inclusion) is a basis. This proof shows that any linearly independent set can be extended to a basis -- just take a maximal element that extends the linearly independent set. But the proof necessarily uses the Axiom of Choice so it won't be constructive.
Let $$A$$ be a basis of $$V$$ (using AC) and $$v \in V$$. Then $$v$$ is a finite linear combination of elements of $$A$$. wlog assume that combination is $$v = a + k_2a_2 + \cdots + k_na_n .$$
Then $$B = A/\{a\} \cup \{v\}$$ is a basis: it spans, and any finite subset is linearly independent.