Every Vector Space has a basis

July 23, 2023

Every Vector Space has a basis

Filed under: advanced mathematics, linear albegra, set theory, vector spaces — Tags: Zorn's Lemma, Axiom of Choice, Basis — oldgote @ 1:27 am

Note: I will probably make a video for this. AND this post (and video to come) is for beginners. The pace will be too slow for the seasoned math student.

Let $V$ be a vector space with some scalars $\alpha \in F$ . I’ll assume you know what a spanning set is and what it means to be linearly independent.

Recall: a basis is a linearly independent spanning set. We’d like to prove that every vector space HAS a basis.

Those new to linear algebra might wonder why this even requires proof. So, I’ll say a few words about some vector spaces that have an infinite number of basis vectors in their basis.

Polynomial space: here the vectors are polynomials in $x$ , say, $3 + 2x -x^2 +\pi x^3$ . The coefficients are real numbers and the set of scalars $F$ will be the real numbers. (note: the set of scalars are sometimes called a “field”). So one basis (there are many others) is $1, x , x^2, ....x^k....$ and since there is no limit to the degree of the polynomial, the basis must have an infinite number of vectors.

Note: vector addition is FINITE. So, though you may have learned in calculus class that ${ 1 \over 1-x} =1+x+x^2 + x^3.....+x^k + x^{k+1} ....$ ( $x \in (-1, 1)$ ) this definition depends on infinite series, which in turn requires a limiting process, which then requires a notion of “being close” (the delta- epsilon stuff) In some vector spaces there IS a way to do that, but you need the notion of “inner product” to define size and closeness (remember the dot product from calculus?) and then you can introduce ideas like convergence of an infinite sum. For example, check out Hilbert Spaces . But such operations can be thought of as an extension of vector space addition; they are NOT the addition operation itself.

2. The vector space of all real variable functions that are continuous $f$ on $[0,1]$ . The scalars will be the real numbers. Now this is a weird vector space. Remember that what is included are things like polynomials, rational functions without roots in $[0,1]$ , exponential functions, trig functions whose domain includes the unit interval, functions like $ln(2+x)$ and even piecewise defined functions whose graphs meet up at all points. And that is only the beginning.

Any basis of this beast will be impossible to list exactly, and, in fact, no basis will be able to be put into one to one correspondence with the positive integers (we say the basis is “uncountably infinite“)

But this vector space indeed has a basis, as we shall see.

So, how do we prove our assertion that every vector space has a basis?

Let $V$ be our non-empty vector space. There is some vector in it, say $\vec{v_1}$ . Let $V_1$ denote the span of this vector. Now if the span is not all of $V$ we can find $\vec{v_2}$ not in the span. Let the span of $\{\vec{v_1}, \vec{v_2} \}$ be denoted by $V_2$ . If $V_2 = V$ we have our basis, we are done. Otherwise, we continue on.

We continue indefinitely. And here is where some set theory comes in: our index set might well become infinite. But that is ok; by looking at the span of vectors $\cup_{\gamma} \vec{v_{\gamma}} =V_{\gamma}$ we obtain a chain of nested subsets $V_1 \subset V_2 \subset .... V_k \subset......V_{\gamma} ....$ and this chain has an upper bound, namely $V$ , the given vector space itself.

Now we have to use some set theory. Zorn’s Lemma (which is equivalent to the axiom of choice) implies that any ordered chain of subsets that has an upper bound has a maximal element; that is, a set that contains ALL of the other sets in the order. So, in this chain (the one that we just constructed), call that set $V^*$ .

Now we claim that the set of vectors that span $v^*$ are linearly independent and span all of $V$ .

Proof of claim: remembering that vector addition has only a positive number of summands, any FINITE sum of vectors, say

$\vec{v_{n1}} + ... \vec{v_{nk}}$

must lie in some $V_{\gamma}$ and, by construction, are linearly dependent (order these vectors and remember how we got them: we added vectors by adding what was NOT in the span of the previous vectors.

Now to show that they span: suppose $\vec{x}$ is NOT in the span. Then let $W = V^{*} \cup \{\vec{x} \}$ . This is then in the chain and contains $V^*$ which violates the fact that $V^{*}$ is maximal.

So, we now have our basis.

Leave a Comment »

No comments yet.

RSS feed for comments on this post. TrackBack URI

College Math Teaching

July 23, 2023