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Basis &
Dimension

Discover how linearly independent spanning sets unlock the structure of vector spaces, and learn to determine the dimension of any subspace.

Linear Algebra

What Is a Vector Space?

A vector space \( V \) over a field \( F \) is a collection of objects (vectors) that can be added together and scaled by elements of \( F \), satisfying closure, associativity, commutativity, and distributivity.

\( \mathbb{R}^2 \)
All ordered pairs \( (x, y) \) with real entries
\( \mathbb{R}^3 \)
All ordered triples \( (x, y, z) \) with real entries
\( P_n \)
All polynomials of degree \( \leq n \)

The Span of a Set

Given vectors \( \mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_k \) in a vector space \( V \), their span is the set of all possible linear combinations:

\( \mathbf{v}_1 \)
\( \mathbf{v}_2 \)
\( c_1\mathbf{v}_1 + c_2\mathbf{v}_2 \)
 O
\( \text{span}\{\mathbf{v}_1, \mathbf{v}_2\} = \) entire plane

If two vectors in \( \mathbb{R}^2 \) point in different directions, their span covers the entire plane.

Linear Independence

Vectors \( \mathbf{v}_1, \dots, \mathbf{v}_k \) are linearly independent if the only solution to

Independent
No vector is a linear combination of the others: no redundancy
Dependent
At least one vector can be written in terms of the rest: redundant

Definition of a Basis

A basis for a vector space \( V \) is a set of vectors \( \mathcal{B} = \{\mathbf{v}_1, \dots, \mathbf{v}_n\} \) that satisfies two conditions simultaneously:

1
Spanning: \( \text{span}(\mathcal{B}) = V \). Every vector in \( V \) can be written as a linear combination of basis vectors.
2
Linear Independence: The vectors in \( \mathcal{B} \) are linearly independent: no vector is redundant.

The Standard Basis

Every \( \mathbb{R}^n \) has a natural standard basis made of unit vectors along each axis:

\( \mathbb{R}^2 \)
\( \mathbf{e}_1 \)
\( \mathbf{e}_2 \)
 (1,0) (0,1)
\( \mathbb{R}^3 \)
\( \mathbf{e}_1 \)
\( \mathbf{e}_2 \)
\( \mathbf{e}_3 \)

The Dimension of a Vector Space

\( \dim(\mathbb{R}^n) = n \)
Standard basis has \( n \) vectors
\( \dim(P_n) = n+1 \)
Basis: \( \{1, x, x^2, \dots, x^n\} \)
\( \dim(\{\mathbf{0}\}) = 0 \)
The zero space has only the empty basis

Dimension tells you the "degrees of freedom" in a vector space: the number of independent directions you can move.

The Basis Theorem

If \( \dim(V) = n \), then any set of vectors in \( V \) with exactly \( n \) elements is a basis if it satisfies either condition:

1
The \( n \) vectors are linearly independent.
2
The \( n \) vectors span \( V \).

Spanning vs Independent vs Basis

Property Spanning Set Indep. Set Basis
Spans \( V \)? ✓ Not necessarily ✓
Linearly independent? Not necessarily ✓ ✓
Number of vectors \( \geq \dim(V) \) \( \leq \dim(V) \) \( = \dim(V) \)
Unique representation? No Not applicable ✓

How to Find a Basis

1
Arrange your vectors as columns of a matrix \( A \).
2
Row reduce \( A \) to its reduced row echelon form (RREF).
3
Identify pivot columns: The original vectors corresponding to the pivot columns form a basis for the span of the vectors.
4
Count pivots to find the dimension.

Example 1: Basis of \( \mathbb{R}^2 \)

Is \( S = \{(1, 2),\; (3, 4)\} \) a basis for \( \mathbb{R}^2 \)?

Example 2: Dependent Vectors in \( \mathbb{R}^3 \)

Find a basis for the subspace of \( \mathbb{R}^3 \) spanned by \( S = \{(1,0,1),\; (2,1,0),\; (3,1,1)\} \).

Example 3: Null Space Basis

Find a basis for \( \text{Null}(A) \) where \( A = \begin{pmatrix} 1 & 2 & 1 \\ 2 & 4 & 2 \end{pmatrix} \).

The Rank-Nullity Theorem

\( \text{rank}(A) \)
\( \text{nullity}(A) \)
 \( n \) (number of columns) + =

This connects the column space dimension (rank) to the null space dimension (nullity). Together they account for every column of \( A \).

Common Mistakes

✕ Confusing Span with Basis
A spanning set can have extra vectors. A basis must be independent too.
✕ Using RREF Vectors as Basis
When finding a basis for the column space, the RREF only tells you which original column vectors to pick. Don't use the vectors from the RREF itself.
✕ Forgetting the Zero Space
\( \{\mathbf{0}\} \) has dimension \( 0 \), not \( 1 \). The zero vector is never part of a basis.

Summary Table

Vector Space Standard Basis Dimension
\( \mathbb{R}^n \) \( \{\mathbf{e}_1, \dots, \mathbf{e}_n\} \) \( n \)
\( P_n \) (polynomials) \( \{1, x, x^2, \dots, x^n\} \) \( n + 1 \)
\( M_{m \times n} \) (matrices) \( \{E_{ij}\} \) (matrix units) \( mn \)
\( \{\mathbf{0}\} \) \( \varnothing \) (empty set) \( 0 \)
\( \text{Col}(A) \) Pivot columns of \( A \) \( \text{rank}(A) \)
\( \text{Null}(A) \) From free variables \( \text{nullity}(A) \)

Question 1 of 5

What is \( \dim(\mathbb{R}^4) \)?

Question 2 of 5

Which of the following is a basis for \( \mathbb{R}^2 \)?

Question 3 of 5

A \( 3 \times 5 \) matrix \( A \) has \( \text{rank}(A) = 2 \). What is the nullity of \( A \)?

Question 4 of 5

What is \( \dim(P_3) \), the space of polynomials of degree \( \leq 3 \)?

Question 5 of 5

If \( V \) has dimension \( n \) and \( S \) is a set of \( n \) linearly independent vectors in \( V \), what can you conclude?