Basis and Dimension

A basis is a linearly independent spanning set -- the most efficient way to describe a vector space. The dimension counts the number of vectors in any basis.

Definition

Definition1.8Basis

A subset $\beta = \{v_1, v_2, \ldots, v_n\}$ of a vector space $V$ is a basis for $V$ if:

$\beta$ is linearly independent, and
$\text{span}(\beta) = V$ .

Equivalently, $\beta$ is a basis if and only if every vector $v \in V$ can be written uniquely as

$v = a_1 v_1 + a_2 v_2 + \cdots + a_n v_n$

for some scalars $a_1, \ldots, a_n \in F$ . The scalars $a_i$ are called the coordinates of $v$ with respect to $\beta$ .

RemarkUniqueness of coordinates

The uniqueness of the representation $v = \sum a_i v_i$ is equivalent to linear independence. If there were two representations $v = \sum a_i v_i = \sum b_i v_i$ , then $\sum (a_i - b_i) v_i = 0$ , and independence forces $a_i = b_i$ for all $i$ .

Examples of bases

ExampleStandard basis for F^n

The standard basis for $F^n$ is $\{e_1, e_2, \ldots, e_n\}$ where $e_i$ has $1$ in position $i$ and $0$ elsewhere. Every $(a_1, \ldots, a_n) = a_1 e_1 + \cdots + a_n e_n$ uniquely.

ExampleNon-standard basis for R^2

$\{(1, 1), (1, -1)\}$ is a basis for $\mathbb{R}^2$ . To express $(a, b)$ :

$(a, b) = \frac{a+b}{2}(1, 1) + \frac{a-b}{2}(1, -1).$

The change-of-coordinates formula gives unique coefficients.

ExampleStandard basis for P_n(F)

$\{1, x, x^2, \ldots, x^n\}$ is the standard basis for $P_n(F)$ . Every polynomial of degree at most $n$ is $a_0 + a_1 x + \cdots + a_n x^n$ uniquely. So $\dim P_n(F) = n + 1$ .

ExampleStandard basis for M_{m x n}(F)

The matrices $E_{ij}$ (with $1$ in position $(i,j)$ and $0$ elsewhere) form the standard basis for $M_{m \times n}(F)$ . There are $mn$ such matrices, so $\dim M_{m \times n}(F) = mn$ .

ExampleBasis for symmetric 2 x 2 matrices

$\text{Sym}_2(\mathbb{R})$ has basis

$\left\{\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix},\; \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix},\; \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}\right\},$

so $\dim \text{Sym}_2(\mathbb{R}) = 3$ . In general, $\dim \text{Sym}_n(F) = n(n+1)/2$ .

ExampleBasis for trace-zero matrices

The subspace $W = \{A \in M_{2 \times 2}(F) \mid \text{tr}(A) = 0\}$ has basis

$\left\{\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix},\; \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix},\; \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}\right\},$

so $\dim W = 3 = 4 - 1 = \dim M_{2 \times 2}(F) - 1$ .

ExampleBasis for a solution space

The solution space of $x + 2y - z = 0$ in $\mathbb{R}^3$ has parametrization $(x, y, z) = s(-2, 1, 0) + t(1, 0, 1)$ and thus basis $\{(-2, 1, 0), (1, 0, 1)\}$ with dimension 2.

ExampleBasis for C over R

$\{1, i\}$ is a basis for $\mathbb{C}$ as a vector space over $\mathbb{R}$ , so $\dim_{\mathbb{R}} \mathbb{C} = 2$ . But $\{1\}$ is a basis for $\mathbb{C}$ over $\mathbb{C}$ , so $\dim_{\mathbb{C}} \mathbb{C} = 1$ .

ExampleLagrange interpolation basis

For distinct points $c_0, c_1, \ldots, c_n \in F$ (with $|F| > n$ ), the Lagrange polynomials

$L_i(x) = \prod_{j \neq i} \frac{x - c_j}{c_i - c_j}$

form a basis for $P_n(F)$ . They satisfy $L_i(c_j) = \delta_{ij}$ , so the coordinates of $p(x)$ with respect to this basis are simply $p(c_0), p(c_1), \ldots, p(c_n)$ .

ExampleBernstein basis

The Bernstein polynomials $B_{i,n}(x) = \binom{n}{i} x^i (1-x)^{n-i}$ for $i = 0, 1, \ldots, n$ form another basis for $P_n(\mathbb{R})$ . They are used extensively in computer-aided geometric design (Bezier curves).

ExampleInfinite-dimensional spaces

$P(F)$ has basis $\{1, x, x^2, x^3, \ldots\}$ (infinite but countable). $C[0,1]$ is infinite-dimensional: for any $n$ , the functions $1, x, x^2, \ldots, x^n$ are linearly independent, so $\dim C[0,1] \geq n$ for all $n$ .

ExampleCoordinate vectors

With respect to the basis $\beta = \{(1, 1), (1, -1)\}$ of $\mathbb{R}^2$ , the vector $v = (5, 3)$ has coordinate vector $[v]_\beta = (4, 1)$ because $v = 4(1, 1) + 1(1, -1)$ .

Dimension

Definition1.9Dimension

A vector space $V$ is finite-dimensional if it has a finite spanning set. The dimension of a finite-dimensional vector space $V$ , denoted $\dim V$ (or $\dim_F V$ ), is the number of vectors in any basis for $V$ .

A vector space that is not finite-dimensional is infinite-dimensional.

Theorem1.6Every basis has the same size

If $V$ is a finite-dimensional vector space, then any two bases for $V$ have the same number of elements. Thus the dimension is well-defined.

This theorem is a consequence of the Replacement Theorem.

Theorem1.7Basis existence and extension

Let $V$ be a finite-dimensional vector space.

Every spanning set contains a basis (by removing redundant vectors).
Every linearly independent set can be extended to a basis (by adding vectors).
If $W \subseteq V$ is a subspace, then $\dim W \leq \dim V$ , with equality iff $W = V$ .

Dimension tables

Vector space	Standard basis	Dimension
$F^n$	$e_1, \ldots, e_n$	$n$
$P_n(F)$	$1, x, \ldots, x^n$	$n + 1$
$M_{m \times n}(F)$	$E_{ij}$	$mn$
$\text{Sym}_n(F)$	$E_{ii},\; E_{ij}+E_{ji}$	$n(n+1)/2$
$\text{Skew}_n(F)$	$E_{ij}-E_{ji}$ ( $i < j$ )	$n(n-1)/2$
$U_n(F)$ (upper triangular)	$E_{ij}$ ( $i \leq j$ )	$n(n+1)/2$
$D_n(F)$ (diagonal)	$E_{ii}$	$n$

RemarkLooking ahead

The concept of basis gives rise to coordinate systems, which connect abstract vector spaces to the concrete space $F^n$ . This connection is formalized through the notion of isomorphism: every $n$ -dimensional vector space over $F$ is isomorphic to $F^n$ .