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The following quoted text is from Evar D. Nering's Linear Algebra and Matrix Theory, 2nd Ed.

Theorem 3.5. In a finite dimensional vector space, every spanning set contains a basis.

Proof: Let $\mathcal{B}$ be a set spanning $\mathcal{V}$. If $\mathcal{V}=\{0\}$, then $\emptyset\subset\mathcal{B}$ is a basis of $\{0\}$. If $\mathcal{V}\ne\{0\}$ then $\mathcal{B}$ must contain at least one non-zero vector $\alpha_{1}$. We now search for another vector in $\mathcal{B}$ which is not dependent on $\{\alpha_{1}\}$. We call this vector $\alpha_{2}$ and search for another vector in $\mathcal{B}$ which is not dependent on the linearly independent set $\{\alpha_{1},\alpha_{2}\}$. We continue in this way as long as we can, but the process must terminate as we cannot find more than $n$ linearly independent vectors in $\mathcal{B}$. Thus suppose we have obtained the set $\mathcal{A}=\{\alpha_{1},\dots,\alpha_{n}\}$ with the property that every vector in $\mathcal{B}$ is linearly dependent on $\mathcal{A}$. Then because of Theorem 2.1 the set $\mathcal{A}$ must also span $\mathcal{V}$ and it is a basis.

The reasoning in the proof seems circular. "Seek, and ye shall find." isn't very rigorous. I certainly believe that I could find such linearly independent vectors, but that's because I believe the theorem to be true.

Is the proof satisfactory?

I might be able to come up with my own proof, but I would still like to know if Nering's proof is satisfactory to mathematicians.

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    $\begingroup$ Formally, the theorem is missing a definition of $n$ (unless some standard notation is defined somewhere else; although it is quite clear in this context). $\endgroup$
    – tylo
    Mar 2, 2017 at 17:22
  • $\begingroup$ I don't know if this is overkill, but you could formalize this kind of sequence building argument by considering all sets of linearly independent vectors from $\mathcal{B}$ ordered by inclusion and apply Zorn's Lemma. Then any maximal element must be a basis for $V$. $\endgroup$
    – subrosar
    Feb 19, 2021 at 3:02

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Yes, the proof is fine. You need not believe that you can find such linearly independent vectors. At each step, there either still exists another vector in $\mathcal B$ that is independent of what you already have (and then you can proceed), or there does not exist such a vector (and the process terminates). If you already know that a linearly independent set cannot have more than $\dim V$ elements, the process is doomed to terminate with some number $n\le \dim V$ vectors picked that way. By revisiting the termination condition, those $n$ vectors span $V$, hence must be a basis.

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  • $\begingroup$ What I find lacking is any explanation of why I should expect there to be some vector not in the current collection, but in the spanning set. I'm not even sure what "seek" means in the mathematical sense. I believe I came up with a valid proof which I find more satisfying. It's too long to post as a comment. The gist of it is that it shows that no fewer than $n$ vectors can span $\mathcal{V}$ so the difference of that set of $m<n$ vectors and $\mathcal{B}$ will be non-empty. Select one vector from that set and go again. $\endgroup$
    – Steven Thomas Hatton
    Mar 2, 2017 at 20:32
  • $\begingroup$ I'm accepting the answer because it appears to be true that some mathematicians think the proof is satisfactory. I still don't like it. It seems vague, and more like an exercise than a proof. For example, could it be expressed in the formalism of predicate logic? I hope I don't come across as ungrateful. I do value the feedback. My reluctance to fully accept it may well be my shortcoming. $\endgroup$
    – Steven Thomas Hatton
    Mar 2, 2017 at 21:46

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