Abstract Nonsense

Crushing one theorem at a time

Every Vector Space Has a Basis


Point of post: In this post we will give a careful proof that every vector space does, in fact, have a basis. This is mainly so that when I reference this fact in further posts, I can link back to this one, for the interested reader.

Theorem: Let \mathscr{V} be a vector space (or more generally and R-module over a division ring)  and L\subseteq\mathscr{V} any linearly independent subset of \mathscr{V}. Then, there exists a basis for \mathscr{V} containing L.

Proof: Let

\text{ }

\mathscr{P}=\left\{S\subseteq\mathscr{V}:L\subseteq S\text{ and }S\text{ is linearly independent}\right\}

\text{ }

Then, \left(\mathscr{P},\subseteq\right) is a partially ordered set. So, let \mathcal{C} be a chain in \left(\mathscr{P},\subseteq\right). Clearly, if \displaystyle \bigcup_{C\in\mathcal{C}}C\in\mathcal{P} we have that \displaystyle \bigcup_{C\in\mathcal{C}}C is an upper bound for \mathcal{C}. So, to see that \displaystyle \bigcup_{C\in\mathcal{C}}C is, in fact, in \mathscr{P} it suffices to prove that it is linearly independent. So, let \displaystyle x_1,\cdots,x_n\in\bigcup_{C\in\mathcal{C}}C. Then, there exists C_1,\cdots,C_n\in\mathcal{C} such that x_j\in C_j,\text{ }j=1,\cdots,n. But, since \mathcal{C} is a chain we may conclude that C_1\cup\cdots\cup C_n=C_{s} for some s\in\{1,\cdots,n\}. But, this means that x_1,\cdots,x_n\in C_s and thus \{x_1,\cdots,x_n\} is a subset of a linearly independent set (namely C_s) and so it itself is linearly independent. It follows that any finite subset of \displaystyle \bigcup_{C\in\mathcal{C}}C is linearly independent, and thus by definition \displaystyle \bigcup_{C\in\mathcal{C}}C is, itself, linearly independent.

\text{ }

From this we see that every chain in \left(\mathscr{P},\subseteq\right) has an upper bound, and so by Zorn’s Lemma, we are able to conclude that \left(\mathscr{P},\subseteq\right) has some maximal element, call it \mathfrak{M}. Now, since \mathfrak{M} is linearly independent (since it’ in \mathscr{P}) to prove it’s a basis it suffices to prove that \text{span }\mathfrak{M}=\mathscr{V}. To do this, suppose to the contrary that there is some v\in\mathscr{V}-\text{span }\mathfrak{M}. We claim that \mathfrak{M}\cup\{v\} is in \mathscr{P}. Since L\subseteq\mathfrak{M} it suffices to prove that \mathfrak{M}\cup\{v\} is linearly independent. To see this let \{x_1,\cdots,x_n\} be any finite subset of \mathfrak{M}\cup\{v\}. If v\notin\{x_1,\cdots,x_n\} then linear independence of \{x_1,\cdots,x_n\} follows immediately. If not, then v=x_j for some j\in\{1,\cdots,n\}, assume without loss of generality that v=x_1. Then, if

\text{ }

\alpha_1 v+\alpha_2x_2+\cdots+\alpha_nx_n=\bold{0}

\text{ }

we must have that \alpha_1=0 otherwise

\text{ }

\displaystyle v=-\alpha^{-1}\alpha_2x_1+\cdots+-\alpha_1^{-1}\alpha_nx_n

\text{ }

contradicting that v\notin\text{span }\mathscr{M}. Thus, \alpha_1=0 and so the above equation may be rewritten as

\text{ }

\alpha_2x_2+\cdots+\alpha_nx_n=\bold{0}

\text{ }

but since \{x_2,\cdots,x_n\}\subseteq\mathfrak{M} is linearly independent it follows that \alpha_1=\cdots=\alpha_n=0. Thus, since the finite subset was arbitrary it follows that \mathfrak{M}\cup\{v\} is linearly independent, and so \mathfrak{M}\cup\{v\}\in\mathscr{P}. But, this contradicts the maximality of \mathfrak{M} since \mathfrak{M}\cup\{v\}\not\subseteq\mathfrak{M}. Thus, the assumption that v\in\mathscr{V}-\text{span }\mathfrak{M} must have been incorrect, in other words \mathscr{V}-\text{span }\mathfrak{M}=\varnothing. But, this is only true if \mathscr{V}\subseteq\text{span }\mathfrak{M}, but since \text{span }\mathfrak{M}\subseteq\mathscr{V} we may conclude that \mathscr{V}=\text{span }\mathfrak{M}. Thus, \mathfrak{M} is a basis. \blacksquare

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October 19, 2010 - Posted by | Algebra, Linear Algebra | , , , , , , ,

5 Comments »

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