\[ \newcommand{R}{\mathbb{R}} \newcommand{defarrow}{\quad \stackrel{\text{def}}{\Longleftrightarrow} \quad} \]

Bases and dimension

Let \(X\) be a vector space.

Hamel Bases

A linearly independent set which generates \(X\) is called a (Hamel) basis for \(X\): \[ S \subset X \text{ Hamel basis for } X \defarrow \mathrm{span}(S) = X \quad\text{and}\quad S\, \text{ lin. indep.}\] Equivalently, \(S\) is a Hamel basis for \(X\) if every vector \(x \in X\) has a unique and finite representation \[ x = \sum_{\text{finite}} a_j u_j, \qquad u_j \in S.\] We shall consider only ordered Hamel bases, in which case the scalars \(a_j\), called coordinates, are well defined.

Ex.

\(\{e_1, \ldots, e_n\}\), with \[ e_j = (0,\ldots,\underbrace{1}_{\text{jth position}},0 \ldots) \] is called the standard basis for \(\R^n\).

\(\{1,x,x^2, \ldots\}\) is an ordered Hamel basis for \(P(\R)\): every real polynomial can be uniquely expressed as a finite sum, \[p(x) = \sum_{\text{finite}} a_j x^j, \qquad a_j \in \R. \]

Dimension

If \(X\) has a basis consisting of finitely many vectors, \(X\) is said to be finite-dimensional. Else, \(X\) is infinite-dimensional.

› The dimension of any finite-dimensional vector space is unique

All bases of a finite-dimensional vector space have the same number of elements. This number is called the dimension of the space.

Proof

Suppose that \(\{e_j\}_{j=1}^m\) and \(\{f_j\}_{j=1}^n\) are both bases, and that \(m > n\). Since a basis is linearly independent, the only solution of \[ \sum_{j=1}^m a_j e_j = \mathbf 0 \quad\text{ ought to be }\quad a_1, \ldots, a_m = 0. \] Since \(\{f_j\}_{j=1}^n\) is also a basis, we may represent \( e_j = \sum_{k=1}^n b_{j,k} f_{k}\) in a unique way. Then \[\sum_{j=1}^m \sum_{k=1}^n a_j b_{j,k} f_k = 0 \quad\text{ meaning that }\quad \sum_{j=1}^m a_j b_{j,k} = 0 \:\text{ for }\: k = 1, \ldots, n.\] This is a linear homogeneous system with \(n\) equations and \(m > n\) unknowns (the scalars \(a_j\)). Such a system always has a non-trivial solution (meaning that some \(a_j \neq 0\)). Hence \(\{e_j\}_{j=1}^m\) is not linearly independent, so it cannot be a basis.

Ex.

\(\R^n\) has dimension \(n\).
\(P_n(\R)\), has dimension \(n+1\). (Recall that \(P_n(\R) \cong \R^{n+1}\).)
\(\mathbb C^n\) has dimension \(n\) when considered as a complex vector space, but \(2n\) when considered a real vector space.
The \(l_p\)-, \(BC\)-, and \(L_2\)-spaces are all infinite-dimensional.

› Any finite-dimensional vector space is isomorphic to Euclidean space

Let \(X\) be a real vector space with basis \(\{e_1, \ldots, e_n\}\). Then \(X \cong \R^n\).¹⁾

Proof

By the definition of a basis, any \(x \in X\) has a unique representation \[x = \sum_{j=1}^n a_j e_j.\] Let \(T \colon X \to \R^n\) be the mapping defined by \[Tx = (a_1, \ldots, a_n).\]

\(T\) is linear: if \(x = \sum a_j e_j\) and \(y = \sum b_j e_j\), \[T(\lambda x+ \mu y) = (\lambda a_1+ \mu b_1, \ldots, \lambda a_n + \mu b_n) = \lambda (a_1, \ldots,a_n) + \mu (b_1, \ldots, b_n) = \lambda Tx + \mu Ty,\]

\(T\) is surjective: \[ \text{ for any } (a_1, \ldots, a_n) \in \R^n \quad\text{ there exists }\quad x = \sum_{j=1}^n a_j e_j; \: Tx = (a_1, \ldots, a_n)\]

\(T\) is injective: \[ Tx = Ty \quad\Longleftrightarrow\quad \forall j\colon a_j = b_j \quad\Longrightarrow\quad x = y. \] Thus \(T\) is a vector space isomorphism.

N.b. In an \(n\)-dimensional vector space, \(m > n\) vectors are linearly dependent.

¹⁾

If \(X\) is a complex vector space, \(X \cong \mathbb C^n\).