Basis and Dimension

Overview

A basis provides a way to uniquely represent every vector in a vector space. The dimension of a space is the number of vectors in any basis.

Linear Independence

Vectors $\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_k$ are linearly independent if:

$c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + \cdots + c_k\mathbf{v}_k = \mathbf{0} \implies c_1 = c_2 = \cdots = c_k = 0$

Otherwise, they are linearly dependent.

Testing Linear Independence

Set up the equation $c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + \cdots + c_k\mathbf{v}_k = \mathbf{0}$ and solve:

• If only trivial solution (all $c_i = 0$): linearly independent
• If non-trivial solution exists: linearly dependent

Span

The span of vectors $\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_k$ is:

$\text{span}\{\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_k\} = \{c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + \cdots + c_k\mathbf{v}_k : c_i \in F\}$

This is the set of all linear combinations of the vectors.

Basis

A basis for a vector space $V$ is a set $B = \{\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_n\}$ that is:

1. Linearly independent
2. Spans $V$

Properties of a Basis

• Every vector in $V$ can be written uniquely as a linear combination of basis vectors
• Removing any vector makes $B$ no longer span $V$
• Adding any vector makes $B$ linearly dependent

Standard Bases

$\mathbb{R}^2$ Standard Basis

$\mathbf{e}_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \quad \mathbf{e}_2 = \begin{bmatrix} 0 \\ 1 \end{bmatrix}$

$\mathbb{R}^3$ Standard Basis

$\mathbf{e}_1 = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, \quad \mathbf{e}_2 = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}, \quad \mathbf{e}_3 = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$

$\mathbb{R}^n$ Standard Basis

$\mathbf{e}_i$ has 1 in position $i$ and 0 elsewhere.

Dimension

The dimension of a vector space $V$, denoted $\dim(V)$, is the number of vectors in any basis for $V$.

Space	Dimension
$\mathbb{R}^n$	$n$
$P_n$ (polynomials degree $\leq n$)	$n + 1$
$M_{m \times n}$ ($m \times n$ matrices)	$m \times n$
$\{\mathbf{0}\}$ (zero space)	0

Key Theorems

Uniqueness of Dimension

All bases for a vector space $V$ have the same number of vectors.

Spanning Set Theorem

If $S$ spans $V$ and $\dim(V) = n$, then:

• $S$ has at least $n$ vectors
• Any $n$ linearly independent vectors in $S$ form a basis

Independence Theorem

If $S$ is linearly independent in $V$ and $\dim(V) = n$, then:

• $S$ has at most $n$ vectors
• $S$ can be extended to a basis

Finding a Basis

For Column Space

1. Write matrix with columns as the vectors
2. Row reduce to echelon form
3. The original columns corresponding to pivot columns form a basis

For Null Space

1. Solve $A\mathbf{x} = \mathbf{0}$
2. Write solution in parametric form
3. The vectors multiplied by free variables form a basis

Examples

Example 1: Verify Basis for $\mathbb{R}^2$

$B = \left\{\begin{bmatrix} 1 \\ 1 \end{bmatrix}, \begin{bmatrix} 1 \\ -1 \end{bmatrix}\right\}$

1. Linear Independence: $c_1\begin{bmatrix} 1 \\ 1 \end{bmatrix} + c_2\begin{bmatrix} 1 \\ -1 \end{bmatrix} = \mathbf{0}$

   • $c_1 + c_2 = 0$
   • $c_1 - c_2 = 0$
   • Only solution: $c_1 = c_2 = 0$ ✓

2. Spans $\mathbb{R}^2$: Any $\begin{bmatrix} a \\ b \end{bmatrix} = c_1\begin{bmatrix} 1 \\ 1 \end{bmatrix} + c_2\begin{bmatrix} 1 \\ -1 \end{bmatrix}$

   • $c_1 = (a+b)/2$
   • $c_2 = (a-b)/2$ ✓

$B$ is a basis, $\dim(\mathbb{R}^2) = 2$.

Example 2: Dimension of Subspace

$W = \{(x, y, z) : x + y + z = 0\}$

Basis: $\{(-1, 1, 0), (-1, 0, 1)\}$

$\dim(W) = 2$