The Central Idea
The previous lesson separated two questions:
- Do these vectors span the space?
- Are these vectors linearly independent?
A basis answers both questions at once.
A basis spans the space and contains no redundant vectors.
Once a basis is chosen, every vector receives a unique list of coordinates relative to that basis.
What Is a Basis?
Let \(V\) be a vector space.
A collection
\[ B=\{\mathbf{v}_1,\mathbf{v}_2,\ldots,\mathbf{v}_n\} \]is a basis for \(V\) if:
- \(B\) spans \(V\);
- \(B\) is linearly independent.
Symbolically,
\[ \boxed{ \operatorname{span}(B)=V } \]together with linear independence.
Why Both Conditions Matter
A spanning set may contain too many vectors.
An independent set may contain too few vectors to generate the whole space.
A basis is exactly large enough:
- large enough to build every vector;
- small enough to avoid redundancy.
Example: Standard Basis of \(\mathbb{R}^2\)
The vectors
\[ \mathbf{e}_1=(1,0) \]and
\[ \mathbf{e}_2=(0,1) \]form a basis for
\[ \mathbb{R}^2. \]They span the plane because every vector
\[ (x,y) \]can be written as
\[ (x,y) = x\mathbf{e}_1+y\mathbf{e}_2. \]They are also linearly independent.
The Standard Basis
The familiar coordinate basis of
\[ \mathbb{R}^n \]is
\[ \boxed{ \mathbf{e}_1,\mathbf{e}_2,\ldots,\mathbf{e}_n }. \]Each basis vector has a single \(1\) in one coordinate and \(0\) in every other coordinate.
Standard Basis of \(\mathbb{R}^3\)
The standard basis is
\[ \mathbf{e}_1=(1,0,0), \] \[ \mathbf{e}_2=(0,1,0), \] \[ \mathbf{e}_3=(0,0,1). \]Every vector
\[ (x,y,z) \]has the representation
\[ (x,y,z) = x\mathbf{e}_1 + y\mathbf{e}_2 + z\mathbf{e}_3. \]A Basis Need Not Be the Standard Basis
Consider
\[ \mathbf{u}=(1,1) \]and
\[ \mathbf{v}=(1,-1). \]These vectors are not scalar multiples of one another, so they are linearly independent in \(\mathbb{R}^2\).
Two independent vectors in \(\mathbb{R}^2\) span the plane.
Therefore,
\[ \boxed{ \{(1,1),(1,-1)\} } \]is also a basis for \(\mathbb{R}^2\).
Vector Coordinates Depend on the Basis
The vector itself does not change when the basis changes.
Its coordinate description does.
For example, relative to the standard basis,
\[ (3,1) = 3(1,0)+1(0,1). \]So its standard coordinates are
\[ \begin{pmatrix} 3\\ 1 \end{pmatrix}. \]Coordinates Relative to Another Basis
Let
\[ B= \{(1,1),(1,-1)\}. \]To express
\[ (3,1) \]in this basis, solve
\[ a(1,1)+b(1,-1)=(3,1). \]This gives
\[ a+b=3 \]and
\[ a-b=1. \]Therefore,
\[ a=2,\qquad b=1. \]Basis Coordinates
Relative to
\[ B= \{(1,1),(1,-1)\}, \]we have
\[ (3,1) = 2(1,1)+1(1,-1). \]Therefore the coordinate vector of \((3,1)\) relative to \(B\) is
\[ \boxed{ [(3,1)]_B = \begin{pmatrix} 2\\ 1 \end{pmatrix} }. \]Why Basis Coordinates Are Unique
Suppose
\[ \mathbf{v} = a_1\mathbf{v}_1+\cdots+a_n\mathbf{v}_n \]and also
\[ \mathbf{v} = b_1\mathbf{v}_1+\cdots+b_n\mathbf{v}_n. \]Subtract the two expressions:
\[ (a_1-b_1)\mathbf{v}_1+\cdots+(a_n-b_n)\mathbf{v}_n = \mathbf{0}. \]Since the basis vectors are linearly independent,
\[ a_i-b_i=0 \]for every \(i\).
Therefore,
\[ \boxed{a_i=b_i}. \]Basis Means Unique Representation
A collection is a basis precisely when every vector in the space can be written in one and only one way as a linear combination of the basis vectors.
Spanning guarantees existence.
Independence guarantees uniqueness.
What Is Dimension?
If a vector space has a finite basis, its dimension is the number of vectors in any basis.
We write
\[ \boxed{ \dim(V)=n }. \]This means every basis of \(V\) contains exactly \(n\) vectors.
Dimension Counts Independent Directions
Geometrically:
- a line through the origin has dimension \(1\);
- a plane through the origin has dimension \(2\);
- ordinary three-dimensional space has dimension \(3\).
Algebraically, dimension counts how many independent basis vectors are required to generate the space.
Dimension of \(\mathbb{R}^n\)
The standard basis of
\[ \mathbb{R}^n \]contains \(n\) vectors.
Therefore,
\[ \boxed{ \dim(\mathbb{R}^n)=n }. \]The Zero Vector Space
The vector space containing only the zero vector is
\[ \{\mathbf{0}\}. \]Its basis is the empty set:
\[ \varnothing. \]Therefore,
\[ \boxed{ \dim\{\mathbf{0}\}=0 }. \]Why the Empty Set Can Be a Basis
The span of the empty set is defined to be the zero subspace:
\[ \operatorname{span}(\varnothing) = \{\mathbf{0}\}. \]The empty set is also vacuously linearly independent.
Thus it serves as the unique basis of the zero vector space.
Example: Dimension of a Line Through the Origin
Consider
\[ L= \{(t,2t):t\in\mathbb{R}\}. \]We can write
\[ L= \operatorname{span}\{(1,2)\}. \]Since
\[ (1,2)\ne(0,0), \]the one-vector set is independent.
Therefore,
\[ \boxed{\dim(L)=1}. \]Example: Dimension of a Plane in \(\mathbb{R}^3\)
Consider
\[ W= \{(x,y,z)\in\mathbb{R}^3:z=x+y\}. \]Write
\[ (x,y,z) = (x,y,x+y). \]Then
\[ (x,y,x+y) = x(1,0,1)+y(0,1,1). \]Therefore
\[ W= \operatorname{span} \{(1,0,1),(0,1,1)\}. \]The two spanning vectors are independent, so
\[ \boxed{\dim(W)=2}. \]Every Basis of a Finite-Dimensional Space Has the Same Size
This fact is fundamental.
A vector space cannot have one basis with \(3\) vectors and another basis with \(5\) vectors.
If one basis contains \(n\) vectors, every basis contains \(n\) vectors.
This common number is the dimension.
Too Many Vectors Force Dependence
If
\[ \dim(V)=n, \]then every collection of more than \(n\) vectors in \(V\) is linearly dependent.
Therefore, in
\[ \mathbb{R}^3, \]every collection of four or more vectors is dependent.
Too Few Vectors Cannot Span
If
\[ \dim(V)=n, \]then fewer than \(n\) vectors cannot span \(V\).
For example, one vector cannot span all of
\[ \mathbb{R}^2. \]Exactly \(n\) Vectors in an \(n\)-Dimensional Space
Suppose
\[ \dim(V)=n \]and we have exactly \(n\) vectors.
Then either one of the following conditions is enough to prove the collection is a basis:
- the vectors are linearly independent;
- the vectors span \(V\).
In an \(n\)-dimensional space, \(n\) independent vectors automatically span, and \(n\) spanning vectors are automatically independent.
Example: Two Independent Vectors in \(\mathbb{R}^2\)
Since
\[ \dim(\mathbb{R}^2)=2, \]any two linearly independent vectors in \(\mathbb{R}^2\) automatically form a basis.
We do not need to prove spanning separately.
Example: Three Spanning Vectors in \(\mathbb{R}^3\)
Since
\[ \dim(\mathbb{R}^3)=3, \]any three vectors that span \(\mathbb{R}^3\) automatically form a basis.
Their independence follows from the dimension count.
Basis from a Spanning Set
If a spanning set contains redundant vectors, those vectors can be removed until an independent spanning set remains.
The remaining set is a basis.
This process is sometimes called reducing a spanning set to a basis.
Example: Remove a Redundant Vector
Consider
\[ (1,0),\qquad (0,1),\qquad (1,1). \]These vectors span \(\mathbb{R}^2\), but
\[ (1,1) = (1,0)+(0,1). \]Therefore the third vector is redundant.
Removing it leaves the basis
\[ \boxed{ \{(1,0),(0,1)\} }. \]Extending an Independent Set to a Basis
If a linearly independent collection does not yet span the whole space, vectors can be added from outside the current span.
Continue adding independent directions until the collection spans the entire space.
The result is a basis.
Example: Extend One Vector in \(\mathbb{R}^2\)
Begin with
\[ \mathbf{v}=(1,1). \]This single nonzero vector is independent but does not span \(\mathbb{R}^2\).
Add a vector not in its span, such as
\[ \mathbf{w}=(1,0). \]Then
\[ \{(1,1),(1,0)\} \]is a basis for \(\mathbb{R}^2\).
Basis of Polynomial Spaces
In
\[ P_n(\mathbb{R}), \]the standard polynomial basis is
\[ \boxed{ \{1,x,x^2,\ldots,x^n\} }. \]Every polynomial of degree at most \(n\) can be written uniquely as
\[ a_0+a_1x+\cdots+a_nx^n. \]Dimension of \(P_n(\mathbb{R})\)
The standard polynomial basis contains
\[ n+1 \]vectors.
Therefore,
\[ \boxed{ \dim P_n(\mathbb{R})=n+1 }. \]Why the Dimension Is \(n+1\), Not \(n\)
The polynomial space includes the constant term.
The powers are
\[ x^0,\ x^1,\ x^2,\ldots,x^n. \]Since
\[ x^0=1, \]there are
\[ n+1 \]basis vectors.
Basis of Matrix Spaces
The vector space
\[ M_{m\times n}(\mathbb{R}) \]has a standard basis consisting of matrices with exactly one entry equal to \(1\) and every other entry equal to \(0\).
There is one such basis matrix for each matrix position.
Dimension of Matrix Spaces
An \(m\times n\) matrix has
\[ mn \]entries.
Therefore,
\[ \boxed{ \dim M_{m\times n}(\mathbb{R})=mn }. \]Example: \(2\times2\) Matrices
The standard basis of
\[ M_{2\times2}(\mathbb{R}) \]contains the four matrices
\[ E_{11},\ E_{12},\ E_{21},\ E_{22}. \]Therefore,
\[ \boxed{ \dim M_{2\times2}(\mathbb{R})=4 }. \]Function Spaces and Bases
Some function spaces are finite-dimensional.
For example, the functions of the form
\[ a+b\cos x+c\sin x \]are spanned by
\[ \{1,\cos x,\sin x\}. \]If these three functions are linearly independent, they form a basis of that function space.
Infinite-Dimensional Spaces
Not every vector space has a finite basis.
The space of all real polynomials
\[ \mathbb{R}[x] \]cannot be spanned by any finite collection because polynomials may have arbitrarily high degree.
Such a space is infinite-dimensional.
Dimension of a Subspace
If
\[ W\le V \]is a subspace and \(V\) is finite-dimensional, then
\[ \boxed{ \dim(W)\le\dim(V) }. \]A subspace cannot contain more independent directions than the space that contains it.
Equal Dimension Can Force Equality
Suppose
\[ W\le V \]and both spaces are finite-dimensional.
If
\[ \dim(W)=\dim(V), \]then
\[ \boxed{W=V}. \]A proper subspace must have strictly smaller dimension.
Coordinates Turn Abstract Vectors into Tuples
Let
\[ B= \{\mathbf{v}_1,\ldots,\mathbf{v}_n\} \]be a basis.
Every vector
\[ \mathbf{v}\in V \]has a unique expression
\[ \mathbf{v} = a_1\mathbf{v}_1+\cdots+a_n\mathbf{v}_n. \]Its coordinate vector is
\[ \boxed{ [\mathbf{v}]_B = \begin{pmatrix} a_1\\ \vdots\\ a_n \end{pmatrix} }. \]Coordinate Maps Preserve Linear Structure
Relative to a fixed basis,
\[ \mathbf{v} \longmapsto [\mathbf{v}]_B \]converts abstract vectors into coordinate columns.
Vector addition and scalar multiplication become ordinary coordinate operations.
This idea foreshadows linear transformations and isomorphisms.
Basis Change Preview
A vector space may have many different bases.
Changing the basis changes the coordinates of vectors but does not change the vectors themselves.
Basis-change matrices provide a systematic way to convert one coordinate description into another.
This becomes important in linear transformations and matrix representations.
Basis and Pivot Columns
In matrix problems, row reduction can help extract a basis from a spanning set.
Pivot columns identify columns that contribute independent directions.
The corresponding original columns often form a basis for the column space.
Rank as a Dimension Preview
The rank of a matrix is the dimension of its column space.
Equivalently, it is the number of pivot columns.
Therefore rank measures the number of independent column directions carried by the matrix.
Nullity as a Dimension Preview
The solution set of
\[ A\mathbf{x}=\mathbf{0} \]is called the null space of \(A\).
Its dimension is called the nullity.
Rank and nullity will later connect dimension to linear transformations.
Common Error: Thinking a Basis Is Unique
A vector space usually has many different bases.
For \(\mathbb{R}^2\),
\[ \{(1,0),(0,1)\} \]is a basis, but so is
\[ \{(1,1),(1,-1)\}. \]What is fixed is the number of basis vectors, not the basis itself.
Common Error: Confusing Basis with Span
A span is a set of vectors generated by linear combinations.
A basis is a specific independent collection whose span is the whole vector space.
These are related but distinct ideas.
Common Error: Assuming Every Spanning Set Is a Basis
A spanning set may contain redundant vectors.
For example,
\[ (1,0),\quad (0,1),\quad (1,1) \]span \(\mathbb{R}^2\), but the collection is dependent.
Therefore it is not a basis.
Common Error: Assuming Every Independent Set Is a Basis
An independent set may fail to span the whole space.
For example,
\[ \{(1,0)\} \]is independent in \(\mathbb{R}^2\), but it spans only a line.
Therefore it is not a basis for \(\mathbb{R}^2\).
Common Error: Confusing Dimension with Number of Coordinates Shown
Dimension is an intrinsic property of the vector space.
A subspace of \(\mathbb{R}^3\) may consist of triples but still have dimension \(1\) or \(2\).
The number of ambient coordinates does not automatically equal the dimension of a subspace.
Common Error: Counting a Redundant Spanning Set
Dimension is the number of vectors in a basis, not the number of vectors in an arbitrary spanning set.
Three vectors may span \(\mathbb{R}^2\), but
\[ \dim(\mathbb{R}^2)=2. \]A Reliable Basis-Test Procedure
- Identify the vector space \(V\).
- Identify the proposed basis vectors.
- Test whether they span \(V\).
- Test whether they are linearly independent.
- If both conditions hold, the collection is a basis.
- If the dimension is already known and the collection has exactly \(\dim(V)\) vectors, proving either spanning or independence is enough.
A Reliable Dimension Procedure
- Find a spanning set for the vector space or subspace.
- Remove redundant vectors until the set is independent.
- The remaining independent spanning set is a basis.
- Count the basis vectors.
- That count is the dimension.
Example: Full Basis and Dimension Analysis
Consider
\[ W= \operatorname{span} \{ (1,0,1), (0,1,1), (1,1,2) \}. \]Observe that
\[ (1,1,2) = (1,0,1)+(0,1,1). \]Therefore the third vector is redundant.
So
\[ W= \operatorname{span} \{(1,0,1),(0,1,1)\}. \]The remaining two vectors are not scalar multiples, so they are independent.
Therefore,
\[ \boxed{ \{(1,0,1),(0,1,1)\} } \]is a basis for \(W\), and
\[ \boxed{\dim(W)=2}. \]Practice 1: Standard Basis
State the standard basis of
\[ \mathbb{R}^3. \]Practice 2: Dimension of Coordinate Space
Find
\[ \dim(\mathbb{R}^5). \]Practice 3: Basis or Not?
Determine whether
\[ \{(1,0),(0,1)\} \]is a basis for
\[ \mathbb{R}^2. \]Practice 4: Dependent Spanning Set
Is
\[ \{(1,0),(0,1),(1,1)\} \]a basis for \(\mathbb{R}^2\)?
Practice 5: Polynomial Dimension
Find
\[ \dim P_4(\mathbb{R}). \]Practice 6: Matrix Dimension
Find
\[ \dim M_{3\times2}(\mathbb{R}). \]Practice 7: Basis Coordinates
Let
\[ B=\{(1,1),(1,-1)\}. \]Find the coordinates of
\[ (4,2) \]relative to \(B\).
Practice 8: Dimension of a Line
Find the dimension of
\[ L= \{(t,-3t):t\in\mathbb{R}\}. \]Practice 9: Remove Redundancy
Reduce
\[ \{(1,0),(0,1),(2,2)\} \]to a basis for its span.
Practice 10: Identify the Error
A student says:
“\(\mathbb{R}^2\) has dimension \(3\) because I found three vectors that span it.”
Explain the error.
Practice Answers
Practice 1
The standard basis is
\[ \boxed{ \{(1,0,0),(0,1,0),(0,0,1)\} }. \]Practice 2
\[ \boxed{ \dim(\mathbb{R}^5)=5 }. \]Practice 3
Yes. The vectors span \(\mathbb{R}^2\) and are linearly independent.
Therefore they form a basis.
Practice 4
No. Although the set spans \(\mathbb{R}^2\),
\[ (1,1) = (1,0)+(0,1), \]so the collection is dependent.
Practice 5
A basis is
\[ \{1,x,x^2,x^3,x^4\}. \]Therefore,
\[ \boxed{ \dim P_4(\mathbb{R})=5 }. \]Practice 6
A \(3\times2\) matrix has
\[ 3\cdot2=6 \]independent entry positions.
Therefore,
\[ \boxed{ \dim M_{3\times2}(\mathbb{R})=6 }. \]Practice 7
Solve
\[ a(1,1)+b(1,-1)=(4,2). \]This gives
\[ a+b=4 \]and
\[ a-b=2. \]Therefore,
\[ a=3,\qquad b=1. \]Hence
\[ \boxed{ [(4,2)]_B = \begin{pmatrix} 3\\ 1 \end{pmatrix} }. \]Practice 8
The space is
\[ L= \operatorname{span}\{(1,-3)\}. \]Since the spanning vector is nonzero,
\[ \boxed{\dim(L)=1}. \]Practice 9
Since
\[ (2,2) = 2(1,0)+2(0,1), \]the third vector is redundant.
A basis is
\[ \boxed{ \{(1,0),(0,1)\} }. \]Practice 10
Dimension is the number of vectors in a basis, not an arbitrary spanning set.
A spanning set can contain redundant vectors.
Every basis of \(\mathbb{R}^2\) contains exactly
\[ \boxed{2} \]vectors.
What to Remember
A basis is a linearly independent spanning set.
\[ \boxed{ \text{basis} = \text{independent} + \text{spanning} }. \]Every vector has a unique coordinate representation relative to a basis:
\[ \mathbf{v} = a_1\mathbf{v}_1+\cdots+a_n\mathbf{v}_n. \]If a finite-dimensional vector space has a basis with \(n\) vectors, then
\[ \boxed{\dim(V)=n}. \]Important examples include:
- \(\dim(\mathbb{R}^n)=n\);
- \(\dim P_n(\mathbb{R})=n+1\);
- \(\dim M_{m\times n}(\mathbb{R})=mn\);
- a line through the origin has dimension \(1\);
- a plane through the origin has dimension \(2\).
Bases provide coordinates, and dimension measures the number of independent directions needed to describe the vector space.
Where We Go Next
A basis gives coordinates to vectors. The next natural question is how functions between vector spaces interact with vector addition and scalar multiplication.
The next lesson, Linear Transformations, studies structure-preserving maps between vector spaces and shows how bases allow those transformations to be represented by matrices.