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In Halmos' Finite-Dimensional Vector Spaces, section I.8 has a proof of the Steinitz exchange lemma, which says that if $V$ is a vector space, $S$ is a finite independent subset of $V$, and $T$ is a finite generating subset of $V$, then the cardinality of $T$ is not less than the cardinality of $S$.

Having proved this, Halmos goes on to assert that "The number of elements in any basis of a finite-dimensional vector space $V$ is the same as in any other basis." But this is not quite right, because the lemma only proves that any two finite bases have the same number of elements. I don't see where it has been ruled out that there might still somehow be infinite bases lurking in a vector space with a finite basis.

There is a theorem in section I.7 which says that any independent set can be extended to a basis. Halmos assumes that the independent set is finite, but I think the proof still goes through if we assume it possibly infinite, so that if there are any infinite independent subsets, we can extend it to an infinite basis. So it seems we need to rule out the possibility that a vector space with a finite basis might have an infinite independent set.

Have I made a mistake anywhere in the above? And if not, can we repair the gap in this theorem easily?

I took a quick look at Roman's Advanced Linear Algebra and it seems to have the same defect of assuming that in a vector space with a finite spanning set, all bases must be finite. I'm not able to find where this is proved. Roman has a proof that works for arbitrary vector spaces, but it uses cardinal arithmetic, which I am not familiar with. Is there any easier way?

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3 Answers 3

up vote 8 down vote accepted

There is no gap. Suppose a $V$ is a vector space with a finite (with say $n$ vectors) basis and an infinite basis. Take $n+1$ vectors from the infinite basis and you have a set of $n+1$ linearly independent vectors. By the lemma you claim you agree to the proof it follows that $n$ (the size of a spanning set of vectors (namely the finite basis)) is not smaller than $n+1$ (the size of the linearly independent set we extracted from the infinite basis). This is impossible, so no finite dimensional vector space can have a subset of infinitely many independent vectors, let alone an infinite basis.

This should resolve your question, but it should just be mentioned that one can prove that given any vector space $V$, a spanning set $S$ and a linearly independent set $I$, there exists an injection $I\to S$ (using the axiom of choice (typically in the form of Zorn's lemma)). This proves that all bases in a vectors spaces have the same cardinality (finite or not). Another typical application of Zorn's lemma proves any vector space has a basis, and thus the two results together show that the notion of dimension is meaningful for all vector spaces: Every vector space has a well-defined dimension, the common cardinality (finite or infinite) of its bases.

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Dear Ittay, " bijection $I \to S$" should possibly read "injection". Cheers, –  Matt E Apr 10 at 23:18
thanks (corrected). –  Ittay Weiss Apr 11 at 0:54
Thanks, that's a good argument. I neglected to consider finite subsets of the infinite basis. –  nham Apr 11 at 15:04

Perhaps this will clarify Halmos's proof of his Theorem 1.8.1. Halmos does not seem to address your specific concern directly, rather, he gives an argument which is easily extensible. Note that Halmos never uses the finiteness of $Y$ in his proof of the theorem.

Let $V$ be a finite dimensional vector space. The first is the set $X=\{x_1,x_2,...,x_n\}$ which spans $V$, i.e. every $v\in V$ is a linear combination of elements in $X$. The other set is $Y=\{y_1,y_2,...\}$, which is a linearly independent set. Halmos's goal is to show that $| Y|$ is finite. So, consider the ordered set: $$ S=\{y_1,x_1,x_2,...,x_m\} $$ Clearly, $S$ is not linearly independent, as $X$ is a spanning set, so we drop the first $x_i$ we meet that is part of a linear relation, to create a new set: $$ S'=\{y_1,x_1,x_2,...,x_{i-1},x_{i+1},...,x_m\} $$ Note that $S'$ is still a spanning set. We now attach $y_2$ to the beginning of $S'$ to find: $$ S''=\{y_2,y_1,x_1,x_2,...,x_{i-1},x_{i+1},...,x_m\} $$ As $S'$ spans $V$, we can similarly drop another $x_j$ from the list. Note that repeating this process we will never be required by Thm 1.6.1 to drop a $y_j$ because by hypothesis the elements of $Y$ are linearly independent.

As $|X|=n$ is finite, we will eventually drop all of the $x$'s from the list, and be left with a list of the first $n$ entries of $Y$ that spans $V$. If we try to add another $y$ at this point, then we will have a linear relation among the elements of $Y$, which will be a contradiction, thus we see that $|Y|\le |X|$, and thus $|Y|$ is finite.

Now, note that if we're considering two bases, then we can swap the roles of $X$ and $Y$ in this argument, and so we see that $|X|=|Y|$ is finite.

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The question already has good answers, but perhaps it will be useful to offer another with a slightly different argument.

Suppose that $V$ has a finite spanning set, and that $S$ is a spanning set. Then I claim that $S$ contains a finite subset which also spans. Indeed, let $v_1,\ldots,v_n$ be the given finite spanning set. Each $v_i$ is a finite linear combination of some of the vectors in $S$ (by assumption), and there are only finitely many $v_i$, so altogether we can find a finite subset of $S$ whose span contains each $v_i$, and hence whose span is equal to all of $V$.

If $S$ is infinite, then if $S_0$ is a finite spanning subset, any $s \in S_0$ is a linear combination of the elements in $S_0$, and so we obtain a non-trivial linear dependence relationship in the set $\{s\} \cup S_0$, and hence in $S$.

Thus if a vector space $V$ admits a finite spanning set, it cannot admit an infinite set of linearly independent vectors. This argument uses nothing but the definition of spanning, and so shows that in a developing the theory of finite dimensional vector spaces, it is reasonable to restrict attention to finite sets of spanning and/or linearly independent vectors.

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