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

# 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

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

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, $X \cong \mathbb C^n$.