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6.5 Basis and dimension

Understand a basis as a just-right spanning set, then use dimension to judge what can and cannot happen in a vector space.

Course contents

MATH1030: Linear algebra I

Rigorous linear algebra notes on systems, matrices, structure, and proof, with interaction used only where it clarifies the mathematics.

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All note collectionsSeries overview
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Chapter 1Systems of equations1 sections
1.11.1 Equations and solution sets
Chapter 2Matrices and elimination4 sections
2.12.1 Matrix basics2.22.2 Augmented matrices and row operations2.32.3 Gaussian elimination and RREF2.42.4 Solution-set types
Chapter 3Matrix algebra2 sections
3.13.1 Matrix multiplication and identity matrices3.23.2 Transpose and special matrices
Chapter 4Solution structure1 sections
4.14.1 Homogeneous systems and null space
Chapter 5Invertibility1 sections
5.15.1 Invertible matrices
Chapter 6Vector spaces6 sections
6.16.1 Vector spaces6.26.2 Subspaces6.36.3 Linear combinations and span6.46.4 Linear dependence and independence6.56.5 Basis and dimensionCurrent section6.66.6 Column space, row space, and rank
Chapter 7Determinants3 sections
7.17.1 Determinants and cofactor expansion7.27.2 Row operations, products, and invertibility7.37.3 Transpose, column operations, and Cramer's rule
Chapter 8Eigenvalues and diagonalization3 sections
8.18.1 Eigenvalues, eigenvectors, and eigenspaces8.28.2 Diagonalization and similarity8.38.3 Characteristic polynomials and diagonalization tests
Chapter 9Inner products and orthogonality4 sections
9.19.1 Inner products, norms, and angles9.29.2 Orthogonal sets and orthonormal bases9.39.3 Gram-Schmidt orthogonalization9.49.4 Cauchy-Schwarz and triangle inequalities

A basis is where the chapter comes together. Earlier notes taught you how to build vectors from other vectors, how to test whether a set is a subspace, and how to recognize linear dependence. This note combines those ideas into one question:

When do we have exactly enough vectors to describe a space without keeping any redundant directions?

Intuition first: enough directions, but not extra baggage

If a set of vectors spans a space, then it gives you enough directions to build every vector in that space.

If the set is also linearly independent, then none of those directions is wasted. No vector in the set can already be made from the others.

So a basis is the "just right" situation:

  • enough vectors to reach every target in the space;
  • not so many vectors that one of them is unnecessary.

That is why a basis is often the most useful coordinate system for a space. Once you choose a basis, every vector can be described in terms of those basis vectors.

Definition

Basis

A set of vectors {u1,u2,…,um}\{u_1, u_2, \ldots, u_m\} is a basis for a vector space VV if both conditions hold:

  1. u1,u2,…,umu_1, u_2, \ldots, u_m are linearly independent.
  2. Span{u1,u2,…,um}=VSpan\{u_1, u_2, \ldots, u_m\} = V.

The two conditions do different jobs.

  • The spanning condition says the set is large enough.
  • The independence condition says the set is not larger than necessary.

If you forget either condition, you can misidentify a basis.

Pause and vary the coefficients in a spanning example. Watch how changing the generators changes the set of vectors you can reach.

Read and try

Build one vector from a span

The live explorer lets you vary coefficients and watch the resulting vector move inside the span.

u

(1, 0)

v

(0, 1)

α

β

Result

αu + βv = (1, 0)

Every output vector is built from the horizontal and vertical directions.

The standard basis is the model example

The first basis most students meet is the standard basis of R3R^3:

e1=[100],e2=[010],e3=[001].e_1 = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, \quad e_2 = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}, \quad e_3 = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}.

These vectors are important because each one isolates one coordinate direction.

Worked example

Why the standard basis of R3R^3 really is a basis

To show that {e1,e2,e3}\{e_1, e_2, e_3\} is a basis for R3R^3, you must check the two basis conditions.

First, they span R3R^3. If

x=[abc],x = \begin{bmatrix} a \\ b \\ c \end{bmatrix},

then

x=ae1+be2+ce3.x = a e_1 + b e_2 + c e_3.

So every vector in R3R^3 can be built from e1,e2,e3e_1, e_2, e_3.

Second, they are linearly independent. If

α1e1+α2e2+α3e3=0,\alpha_1 e_1 + \alpha_2 e_2 + \alpha_3 e_3 = 0,

then comparing coordinates gives

α1=0,α2=0,α3=0.\alpha_1 = 0,\quad \alpha_2 = 0,\quad \alpha_3 = 0.

So the only linear relation is the trivial one.

Since both conditions hold, {e1,e2,e3}\{e_1, e_2, e_3\} is a basis for R3R^3.

This example also explains why basis coordinates are useful. Once the basis is fixed, the coefficients a, b, and c tell you exactly how to rebuild the vector.

Why the number of basis vectors matters

The notes stress a fact that is easy to say and very important to use:

Theorem

Every basis of the same space has the same size

If B1B_1 and B2B_2 are both bases of the same vector space VV, then B1B_1 and B2B_2 contain the same number of vectors.

This statement is what makes dimension well-defined. Without it, the phrase "the number of vectors in a basis" would depend on which basis you picked, and dimension would not be a stable idea.

Dimension is a count of independent directions

Definition

Dimension

The dimension of a vector space VV, written dim(V), is the number of vectors in any basis of VV.

Dimension tells you how many genuinely independent directions the space has.

  • dim(Rm)=mdim(R^m) = m
  • dim(Mmn)=mndim(M_{mn}) = mn
  • dim(Pn)=n+1dim(P_n) = n + 1

The zero space is the exceptional case:

dim({0})=0.dim(\{0\}) = 0.

That makes sense because the zero space has no nonzero direction at all.

A basis does not have to look like the standard one

A basis is not required to use the standard coordinate vectors. Many different sets can serve as a basis for the same space, as long as they span the space and stay linearly independent.

Worked example

A basis for a plane inside R3R^3

Let

W={[xy0]:x,y∈R}.W = \left\{ \begin{bmatrix} x \\ y \\ 0 \end{bmatrix} : x, y \in R \right\}.

This is the plane z=0z = 0 inside R3R^3.

Consider

u1=[100],u2=[010].u_1 = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, \qquad u_2 = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}.

Every vector in WW has the form

[xy0]=xu1+yu2,\begin{bmatrix} x \\ y \\ 0 \end{bmatrix} = x u_1 + y u_2,

so u1u_1 and u2u_2 span WW.

They are also linearly independent, because

α1u1+α2u2=0\alpha_1 u_1 + \alpha_2 u_2 = 0

forces α1=α2=0\alpha_1 = \alpha_2 = 0.

Therefore {u1,u2}\{u_1, u_2\} is a basis for WW, and

dim(W)=2.dim(W) = 2.

So dimension does not count how many entries a vector happens to have. It counts how many independent directions the space itself contains.

How dimension becomes a shortcut

Once you know the dimension of a space, many basis questions become shorter.

If dim(V)=mdim(V) = m, then:

  • any m linearly independent vectors in VV automatically form a basis;
  • any m vectors in VV that span VV automatically form a basis;
  • fewer than m vectors cannot span VV;
  • more than m vectors cannot all remain linearly independent.

This is why dimension is powerful. It turns a long two-part basis test into a one-part check when the number of vectors already matches the dimension.

Try a few typical dependence patterns before you use that shortcut.

Read and try

Test one set for dependence

The live checker compares small vector sets and explains whether a nontrivial linear relation exists.

Verdict

Independent

The only way to solve c1e1 + c2e2 = 0 is c1 = c2 = 0, so this pair is linearly independent.

Key relation

No nontrivial linear relation appears.

A practical checklist for basis questions

When you are asked whether a set is a basis, work in this order:

  1. Identify the space you are trying to span.
  2. Count how many vectors you were given.
  3. Check linear independence, or check spanning.
  4. Use dimension to finish the argument when the count matches.

For example, in R3R^3, three independent vectors already form a basis. You do not need a second long spanning calculation after that.

Common mistake

Common mistake

Dimension is not just the number of coordinates

A subspace of R3R^3 can have dimension 1, 2, or 3. The ambient space tells you how many entries vectors have; the dimension tells you how many independent directions the subspace itself has.

Another common mistake is to check spanning and stop there. A spanning set can still fail to be a basis if one vector is redundant.

Quick checks

Quick check

Can \{(1,0), (2,0)\} be a basis for R2R^2?

Think about both basis conditions, not just the number of vectors.

Solution

Answer

Quick check

Suppose dim(V)=3dim(V) = 3 and you already found three linearly independent vectors in VV. What remains to prove?

Use the dimension shortcut from this note.

Solution

Answer

Exercises

Quick check

Do the vectors u1=(1,1,0)u_1 = (1,1,0), u2=(1,0,1)u_2 = (1,0,1), and u3=(0,1,1)u_3 = (0,1,1) form a basis for R3R^3?

Try to test independence first. If they are independent, dimension does the rest.

Solution

Guided solution

Read this first

This note depends on 6.4 Linear dependence and independence and 6.3 Linear combinations and span.

Prerequisites

6.4

6.4 Linear dependence and independence

Key terms in this unit

More notes in this series

Previous section

6.4 Linear dependence and independence

Next section

6.6 Column space, row space, and rank