The Central Idea
Vector spaces allow two basic operations:
- vector addition;
- scalar multiplication.
By combining these operations, we can form expressions such as
\[ a_1\mathbf{v}_1+a_2\mathbf{v}_2+\cdots+a_k\mathbf{v}_k. \]Such expressions are called linear combinations.
The vectors obtainable from all possible scalar choices form the span of the original vectors.
Linear independence asks whether any vector in the collection is algebraically redundant.
What Is a Linear Combination?
Let
\[ \mathbf{v}_1,\mathbf{v}_2,\ldots,\mathbf{v}_k \]be vectors in a vector space \(V\) over a field \(F\).
A linear combination of these vectors is any expression of the form
\[ \boxed{ a_1\mathbf{v}_1 + a_2\mathbf{v}_2 + \cdots + a_k\mathbf{v}_k } \]where
\[ a_1,a_2,\ldots,a_k\in F. \]The Scalars Are Coefficients
In a linear combination,
\[ a_1,\ldots,a_k \]are scalar coefficients.
Each scalar tells us how much of the corresponding vector contributes to the final result.
The scalar may be positive, negative, zero, fractional, real, complex, or otherwise appropriate for the chosen field.
Example: A Linear Combination in \(\mathbb{R}^2\)
Let
\[ \mathbf{u}=(1,2) \]and
\[ \mathbf{v}=(3,-1). \]Consider
\[ 2\mathbf{u}-3\mathbf{v}. \]Then
\[ 2\mathbf{u} = (2,4), \]and
\[ -3\mathbf{v} = (-9,3). \]Therefore,
\[ 2\mathbf{u}-3\mathbf{v} = (-7,7). \]Every Individual Vector Is a Linear Combination
A vector is always a linear combination of a collection containing that vector.
For example,
\[ \mathbf{v}_1 = 1\mathbf{v}_1 + 0\mathbf{v}_2 + \cdots + 0\mathbf{v}_k. \]Thus each generator belongs to the span generated by the collection.
The Zero Vector Is Always a Linear Combination
Choosing every scalar coefficient to be zero gives
\[ 0\mathbf{v}_1+\cdots+0\mathbf{v}_k = \mathbf{0}. \]Therefore the zero vector belongs to every span:
\[ \boxed{ \mathbf{0}\in \operatorname{span} \{\mathbf{v}_1,\ldots,\mathbf{v}_k\} }. \]What Is Span?
The span of vectors
\[ \mathbf{v}_1,\ldots,\mathbf{v}_k \]is the set of all their linear combinations.
Symbolically,
\[ \boxed{ \operatorname{span} \{\mathbf{v}_1,\ldots,\mathbf{v}_k\} = \left\{ a_1\mathbf{v}_1+\cdots+a_k\mathbf{v}_k : a_i\in F \right\} }. \]Span Means “Everything You Can Build”
The span answers the question:
What vectors can be constructed using these vectors together with vector addition and scalar multiplication?
The original vectors act as generators.
Their span is the complete linear world generated by those vectors.
Span Is a Subspace
The span of any collection of vectors in \(V\) is always a subspace of \(V\).
It contains the zero vector.
It is closed under vector addition because the sum of two linear combinations is another linear combination.
It is closed under scalar multiplication because scaling a linear combination simply changes its coefficients.
The Smallest Subspace Containing the Generators
The span of a collection is the smallest subspace containing every vector in that collection.
If \(W\) is any subspace containing
\[ \mathbf{v}_1,\ldots,\mathbf{v}_k, \]then \(W\) must also contain every linear combination of them.
Therefore,
\[ \boxed{ \operatorname{span} \{\mathbf{v}_1,\ldots,\mathbf{v}_k\} \subseteq W }. \]Span of One Nonzero Vector in \(\mathbb{R}^2\)
Let
\[ \mathbf{v}=(2,1). \]Its span consists of all scalar multiples:
\[ \operatorname{span}\{\mathbf{v}\} = \{t(2,1):t\in\mathbb{R}\}. \]Equivalently,
\[ \boxed{ \operatorname{span}\{(2,1)\} = \{(2t,t):t\in\mathbb{R}\} }. \]Geometrically, this is a line through the origin.
Span of the Zero Vector
If the only generator is
\[ \mathbf{0}, \]then every scalar multiple is still
\[ \mathbf{0}. \]Therefore,
\[ \boxed{ \operatorname{span}\{\mathbf{0}\} = \{\mathbf{0}\} }. \]Two Vectors Can Span the Plane
Consider
\[ \mathbf{e}_1=(1,0) \]and
\[ \mathbf{e}_2=(0,1). \]Any vector
\[ (x,y)\in\mathbb{R}^2 \]can be written as
\[ (x,y) = x(1,0)+y(0,1). \]Therefore,
\[ \boxed{ \operatorname{span}\{\mathbf{e}_1,\mathbf{e}_2\} = \mathbb{R}^2 }. \]Two Vectors May Span Only a Line
Consider
\[ \mathbf{u}=(1,2) \]and
\[ \mathbf{v}=(2,4). \]Since
\[ \mathbf{v}=2\mathbf{u}, \]every linear combination
\[ a\mathbf{u}+b\mathbf{v} \]becomes
\[ (a+2b)\mathbf{u}. \]Thus the two vectors generate only the same line as \(\mathbf{u}\).
Testing Whether a Vector Lies in a Span
To determine whether
\[ \mathbf{w} \]belongs to
\[ \operatorname{span} \{\mathbf{v}_1,\ldots,\mathbf{v}_k\}, \]ask whether scalars
\[ a_1,\ldots,a_k \]exist such that
\[ \boxed{ a_1\mathbf{v}_1+\cdots+a_k\mathbf{v}_k = \mathbf{w} }. \]Example: Membership in a Span
Determine whether
\[ (5,1) \]lies in the span of
\[ (1,1) \]and
\[ (2,-1). \]We seek \(a,b\) such that
\[ a(1,1)+b(2,-1)=(5,1). \]Comparing components gives
\[ a+2b=5 \]and
\[ a-b=1. \]Solving the Span-Membership System
From
\[ a-b=1, \]we have
\[ a=1+b. \]Substitute into
\[ a+2b=5: \] \[ 1+b+2b=5. \]Thus
\[ 3b=4, \]so
\[ b=\frac43. \]Then
\[ a=\frac73. \]Therefore,
\[ \boxed{ (5,1) = \frac73(1,1) + \frac43(2,-1) }. \]So \((5,1)\) belongs to the span.
Span Questions Become Systems of Equations
In coordinate spaces, testing span membership often becomes a system of linear equations.
Each coordinate produces one equation in the unknown scalar coefficients.
Therefore matrix methods, elimination, and row reduction naturally connect to span.
What Is Linear Independence?
Vectors
\[ \mathbf{v}_1,\ldots,\mathbf{v}_k \]are linearly independent if the equation
\[ a_1\mathbf{v}_1+\cdots+a_k\mathbf{v}_k = \mathbf{0} \]has only the solution
\[ \boxed{ a_1=a_2=\cdots=a_k=0 }. \]The Trivial Linear Combination
The equation
\[ a_1\mathbf{v}_1+\cdots+a_k\mathbf{v}_k = \mathbf{0} \]always has the solution
\[ a_1=\cdots=a_k=0. \]This is called the trivial solution.
Independence means that no other solution exists.
What Is Linear Dependence?
A collection of vectors is linearly dependent if
\[ a_1\mathbf{v}_1+\cdots+a_k\mathbf{v}_k = \mathbf{0} \]has some solution in which not all coefficients are zero.
Such a relation is called a nontrivial linear relation.
Dependence Means Redundancy
If a collection is linearly dependent, then at least one vector can be written as a linear combination of the others.
For example, if
\[ 2\mathbf{u}-3\mathbf{v}+\mathbf{w} = \mathbf{0}, \]then
\[ \mathbf{w} = -2\mathbf{u}+3\mathbf{v}. \]Thus \(\mathbf{w}\) adds no new span beyond \(\mathbf{u}\) and \(\mathbf{v}\).
Example: Dependent Vectors
Consider
\[ \mathbf{u}=(1,2) \]and
\[ \mathbf{v}=(2,4). \]Since
\[ 2\mathbf{u}-\mathbf{v} = \mathbf{0}, \]with coefficients \(2\) and \(-1\) not both zero, the vectors are linearly dependent.
Example: Independent Vectors
Consider
\[ \mathbf{e}_1=(1,0) \]and
\[ \mathbf{e}_2=(0,1). \]Suppose
\[ a\mathbf{e}_1+b\mathbf{e}_2 = \mathbf{0}. \]Then
\[ (a,b)=(0,0), \]so
\[ a=0,\qquad b=0. \]Therefore \(\mathbf{e}_1\) and \(\mathbf{e}_2\) are linearly independent.
A Single Vector
A set containing one vector
\[ \{\mathbf{v}\} \]is linearly independent exactly when
\[ \mathbf{v}\ne\mathbf{0}. \]If
\[ a\mathbf{v}=\mathbf{0} \]and \(\mathbf{v}\ne\mathbf{0}\), then field arithmetic forces
\[ a=0. \]Any Collection Containing the Zero Vector Is Dependent
Suppose a collection contains
\[ \mathbf{0}. \]Then we can assign coefficient \(1\) to the zero vector and coefficient \(0\) to every other vector.
This gives a nontrivial relation:
\[ 1\mathbf{0} + 0\mathbf{v}_1 + \cdots = \mathbf{0}. \]Therefore the collection is linearly dependent.
Scalar Multiples Create Dependence
If
\[ \mathbf{v}=c\mathbf{u} \]for some scalar \(c\), then
\[ c\mathbf{u}-\mathbf{v} = \mathbf{0}. \]Unless both vectors are absent from the relation in a trivial way, this gives a nontrivial dependence relation.
Thus two nonzero scalar multiples are linearly dependent.
Two Vectors in \(\mathbb{R}^2\)
Two nonzero vectors in \(\mathbb{R}^2\) are linearly dependent exactly when one is a scalar multiple of the other.
Geometrically, dependent vectors point along the same line through the origin.
Independent vectors point in genuinely different directions.
Three Vectors in \(\mathbb{R}^2\) Must Be Dependent
The plane has only two independent coordinate directions.
Therefore any collection of three vectors in
\[ \mathbb{R}^2 \]is linearly dependent.
This fact anticipates the idea of dimension.
Four Vectors in \(\mathbb{R}^3\) Must Be Dependent
Similarly, no more than three vectors can be linearly independent in
\[ \mathbb{R}^3. \]Therefore every collection of four vectors in \(\mathbb{R}^3\) is dependent.
Testing Independence Algebraically
To test
\[ \mathbf{v}_1,\ldots,\mathbf{v}_k, \]solve
\[ \boxed{ a_1\mathbf{v}_1+\cdots+a_k\mathbf{v}_k = \mathbf{0} }. \]If the only solution is
\[ a_1=\cdots=a_k=0, \]the vectors are independent.
If a nonzero coefficient solution exists, they are dependent.
Example: Independence in \(\mathbb{R}^2\)
Test
\[ \mathbf{u}=(1,2) \]and
\[ \mathbf{v}=(3,1). \]Set
\[ a(1,2)+b(3,1)=(0,0). \]This gives
\[ a+3b=0 \]and
\[ 2a+b=0. \]Solving the Independence System
From
\[ a+3b=0, \]we get
\[ a=-3b. \]Substitute into
\[ 2a+b=0: \] \[ -6b+b=0. \]Thus
\[ -5b=0, \]so
\[ b=0. \]Then
\[ a=0. \]Therefore the vectors are linearly independent.
Matrix Form of a Linear Combination
Coordinate vectors can be placed as columns of a matrix.
Then
\[ a_1\mathbf{v}_1+\cdots+a_k\mathbf{v}_k = \mathbf{w} \]becomes
\[ \boxed{ A\mathbf{x}=\mathbf{w} }. \]Here the columns of \(A\) are the vectors
\[ \mathbf{v}_1,\ldots,\mathbf{v}_k, \]and
\[ \mathbf{x} = \begin{pmatrix} a_1\\ \vdots\\ a_k \end{pmatrix}. \]Span and the Column Space
The span of the columns of a matrix is called its column space.
Therefore
\[ \operatorname{Col}(A) \]consists of every vector that can be written as a linear combination of the columns of \(A\).
This connects generalized vector-space structure to matrix methods.
Independence and the Homogeneous System
If the columns of \(A\) are
\[ \mathbf{v}_1,\ldots,\mathbf{v}_k, \]then their independence is tested by
\[ \boxed{ A\mathbf{x}=\mathbf{0} }. \]The columns are linearly independent exactly when this homogeneous system has only the trivial solution
\[ \mathbf{x}=\mathbf{0}. \]Pivot Columns and Independence Preview
Row reduction can identify which columns contribute independent directions.
Pivot positions reveal which variables are constrained and whether free variables exist.
A free variable in
\[ A\mathbf{x}=\mathbf{0} \]indicates a nontrivial solution and therefore linear dependence.
Span and Independence Are Different Questions
Span asks:
What can these vectors build?
Independence asks:
Is any vector unnecessary because it can already be built from the others?
A collection may span a large space while still containing redundant vectors.
Example: Spanning but Dependent
Consider
\[ (1,0),\qquad (0,1),\qquad (1,1). \]The first two vectors already span
\[ \mathbb{R}^2. \]The third vector satisfies
\[ (1,1) = (1,0)+(0,1). \]Therefore all three vectors still span \(\mathbb{R}^2\), but the collection is linearly dependent.
Removing a Redundant Vector Does Not Change the Span
If one vector is a linear combination of the others, removing it does not change the span.
In the previous example,
\[ \operatorname{span} \{(1,0),(0,1),(1,1)\} \]equals
\[ \operatorname{span} \{(1,0),(0,1)\}. \]This is the bridge from spanning sets to bases.
Adding a Vector Already in the Span Creates Dependence
Suppose
\[ \mathbf{v}_{k+1} \in \operatorname{span} \{\mathbf{v}_1,\ldots,\mathbf{v}_k\}. \]Then
\[ \mathbf{v}_{k+1} = a_1\mathbf{v}_1+\cdots+a_k\mathbf{v}_k. \]Rearranging,
\[ a_1\mathbf{v}_1+\cdots+a_k\mathbf{v}_k - \mathbf{v}_{k+1} = \mathbf{0}. \]This is a nontrivial linear relation, so the enlarged collection is dependent.
Adding a Vector Outside the Span Preserves Independence
Suppose
\[ \mathbf{v}_1,\ldots,\mathbf{v}_k \]are linearly independent and
\[ \mathbf{w} \notin \operatorname{span} \{\mathbf{v}_1,\ldots,\mathbf{v}_k\}. \]Then adding \(\mathbf{w}\) produces a larger linearly independent collection.
The new vector contributes a genuinely new direction.
Polynomials as Linear Combinations
In
\[ P_2(\mathbb{R}), \]every polynomial
\[ a+bx+cx^2 \]can be written as
\[ a(1)+b(x)+c(x^2). \]Therefore,
\[ \boxed{ \operatorname{span}\{1,x,x^2\} = P_2(\mathbb{R}) }. \]Independence of \(1,x,x^2\)
Suppose
\[ a+bx+cx^2=0 \]as the zero polynomial.
A polynomial is identically zero only when every coefficient is zero.
Therefore,
\[ a=0,\qquad b=0,\qquad c=0. \]Thus
\[ \boxed{\{1,x,x^2\}} \]is linearly independent.
Matrices as Linear Combinations
In the vector space of \(2\times2\) real matrices, consider
\[ E_{11} = \begin{pmatrix} 1&0\\ 0&0 \end{pmatrix}, \quad E_{12} = \begin{pmatrix} 0&1\\ 0&0 \end{pmatrix}, \] \[ E_{21} = \begin{pmatrix} 0&0\\ 1&0 \end{pmatrix}, \quad E_{22} = \begin{pmatrix} 0&0\\ 0&1 \end{pmatrix}. \]Any matrix
\[ \begin{pmatrix} a&b\\ c&d \end{pmatrix} \]can be written as
\[ aE_{11}+bE_{12}+cE_{21}+dE_{22}. \]These four matrices therefore span \(M_{2\times2}(\mathbb{R})\).
Functions as Linear Combinations
In a function space, linear combinations are formed pointwise.
If
\[ h=af+bg, \]then
\[ \boxed{ h(x)=a f(x)+b g(x) }. \]Thus span and independence apply to functions just as they do to coordinate vectors.
Example: Functions That Are Dependent
Consider
\[ f(x)=x \]and
\[ g(x)=3x. \]Since
\[ g=3f, \]the functions are linearly dependent.
Example: Functions That Are Independent
Consider
\[ f(x)=1 \]and
\[ g(x)=x. \]If
\[ af+bg=0, \]then
\[ a+bx=0 \]for every \(x\).
Therefore
\[ a=0,\qquad b=0, \]so \(1\) and \(x\) are linearly independent.
Finite Spanning Sets
A vector space is called finite-dimensional if it can be spanned by a finite collection of vectors.
For example,
\[ \mathbb{R}^3 \]is spanned by
\[ (1,0,0),\quad (0,1,0),\quad (0,0,1). \]The precise meaning of dimension will be developed in the next lesson.
Infinite-Dimensional Spaces Preview
Some vector spaces cannot be spanned by any finite collection.
For example, the space of all real polynomials
\[ \mathbb{R}[x] \]requires arbitrarily high powers
\[ 1,x,x^2,x^3,\ldots. \]Such spaces are called infinite-dimensional.
Common Error: Confusing Span with the Original Set
The span usually contains far more vectors than the original collection.
For example,
\[ \{(1,0),(0,1)\} \]contains only two vectors.
But its span is all of
\[ \mathbb{R}^2, \]containing infinitely many vectors.
Common Error: Forgetting Arbitrary Scalars
Span includes all permitted scalar coefficients.
If the field is
\[ \mathbb{R}, \]coefficients may be any real numbers, not merely integers or positive values.
Common Error: Testing Independence Against a Nonzero Target
Independence is tested using the zero vector:
\[ a_1\mathbf{v}_1+\cdots+a_k\mathbf{v}_k = \mathbf{0}. \]Testing whether a particular nonzero vector is obtainable is a span question, not an independence test.
Common Error: Thinking Dependence Means Equal Vectors
Two equal nonzero vectors are certainly dependent, but dependence is more general.
For example,
\[ (1,0),\quad (0,1),\quad (1,1) \]are three distinct vectors, yet
\[ (1,0)+(0,1)-(1,1)=\mathbf{0}. \]Therefore they are dependent.
Common Error: Assuming More Vectors Produce a Larger Span
Adding a vector already in the existing span does not enlarge the span.
If
\[ \mathbf{w} \in \operatorname{span}\{\mathbf{u},\mathbf{v}\}, \]then
\[ \operatorname{span}\{\mathbf{u},\mathbf{v},\mathbf{w}\} = \operatorname{span}\{\mathbf{u},\mathbf{v}\}. \]Common Error: Assuming a Spanning Set Is Independent
A spanning set may contain redundant vectors.
For example,
\[ (1,0),\quad (0,1),\quad (1,1) \]span \(\mathbb{R}^2\), but they are dependent.
A collection that both spans and is independent receives a special name: a basis.
A Reliable Linear-Combination Procedure
- Write the target vector as a scalar-weighted sum of the generators.
- Introduce one unknown scalar for each generator.
- Expand the vector equation.
- Compare coordinates, coefficients, entries, or function values as appropriate.
- Solve the resulting system.
- If a solution exists, the target is in the span.
- If no solution exists, the target lies outside the span.
A Reliable Independence Procedure
- Set a general linear combination equal to the zero vector.
- Introduce scalar coefficients \(a_1,\ldots,a_k\).
- Translate the vector equation into scalar equations.
- Solve the homogeneous system.
- If only the zero coefficients work, the vectors are independent.
- If any nontrivial solution exists, the vectors are dependent.
- When dependent, identify which vector can be expressed using the others.
Example: Full Span and Independence Analysis
Consider
\[ \mathbf{v}_1=(1,1,0), \quad \mathbf{v}_2=(0,1,1), \quad \mathbf{v}_3=(1,2,1). \]Observe that
\[ \mathbf{v}_1+\mathbf{v}_2 = (1,2,1) = \mathbf{v}_3. \]Therefore,
\[ \mathbf{v}_1+\mathbf{v}_2-\mathbf{v}_3 = \mathbf{0}. \]The coefficients
\[ 1,\quad1,\quad-1 \]are not all zero, so the three vectors are linearly dependent.
Also,
\[ \operatorname{span} \{\mathbf{v}_1,\mathbf{v}_2,\mathbf{v}_3\} = \operatorname{span} \{\mathbf{v}_1,\mathbf{v}_2\}. \]The third vector is redundant.
Practice 1: Linear Combination
Let
\[ \mathbf{u}=(1,2), \qquad \mathbf{v}=(3,-1). \]Compute
\[ 3\mathbf{u}+2\mathbf{v}. \]Practice 2: Span of One Vector
Describe
\[ \operatorname{span}\{(1,3)\}. \]Practice 3: Standard Vectors
What is
\[ \operatorname{span}\{(1,0),(0,1)\}? \]Practice 4: Span Membership
Determine whether
\[ (6,9) \]belongs to
\[ \operatorname{span}\{(2,3)\}. \]Practice 5: Dependence by Scalar Multiple
Are
\[ (2,5) \]and
\[ (6,15) \]linearly independent?
Practice 6: Zero Vector
Is the collection
\[ \{(1,2),(0,0)\} \]linearly independent?
Practice 7: Independence Test
Determine whether
\[ (1,0) \]and
\[ (1,1) \]are linearly independent.
Practice 8: Polynomial Span
Show that
\[ 4-2x+7x^2 \]belongs to
\[ \operatorname{span}\{1,x,x^2\}. \]Practice 9: Redundant Vector
In the collection
\[ (1,0),\quad (0,1),\quad (2,3), \]express the third vector as a linear combination of the first two.
Practice 10: Identify the Error
A student says:
“Three vectors must span a larger space than two vectors because there are more of them.”
Explain the error.
Practice Answers
Practice 1
\[ 3(1,2)+2(3,-1) = (3,6)+(6,-2) = \boxed{(9,4)}. \]Practice 2
The span consists of all real scalar multiples:
\[ \boxed{ \operatorname{span}\{(1,3)\} = \{(t,3t):t\in\mathbb{R}\} }. \]Practice 3
Every \((x,y)\in\mathbb{R}^2\) satisfies
\[ (x,y)=x(1,0)+y(0,1). \]Therefore,
\[ \boxed{ \operatorname{span}\{(1,0),(0,1)\} = \mathbb{R}^2 }. \]Practice 4
Since
\[ (6,9)=3(2,3), \]the vector belongs to the span.
Practice 5
No. Since
\[ (6,15)=3(2,5), \]one vector is a scalar multiple of the other.
Therefore they are linearly dependent.
Practice 6
No. Any collection containing the zero vector is linearly dependent.
Practice 7
Suppose
\[ a(1,0)+b(1,1)=(0,0). \]Then
\[ (a+b,b)=(0,0). \]Thus
\[ b=0 \]and then
\[ a=0. \]Therefore the vectors are linearly independent.
Practice 8
\[ 4-2x+7x^2 = 4(1)-2(x)+7(x^2). \]Therefore it belongs to
\[ \operatorname{span}\{1,x,x^2\}. \]Practice 9
\[ \boxed{ (2,3) = 2(1,0)+3(0,1) }. \]Practice 10
Additional vectors may be redundant.
If a new vector is already in the span of the existing vectors, adding it does not enlarge the span.
For example,
\[ (1,1) \]is already in the span of
\[ (1,0) \]and
\[ (0,1). \]What to Remember
A linear combination has the form
\[ \boxed{ a_1\mathbf{v}_1+\cdots+a_k\mathbf{v}_k }. \]The span is the set of all such combinations:
\[ \boxed{ \operatorname{span} \{\mathbf{v}_1,\ldots,\mathbf{v}_k\} }. \]A collection is linearly independent when
\[ a_1\mathbf{v}_1+\cdots+a_k\mathbf{v}_k = \mathbf{0} \]forces
\[ \boxed{ a_1=\cdots=a_k=0 }. \]A nontrivial solution means the collection is dependent.
The key distinction is:
- span measures what the vectors can build;
- independence measures whether any vectors are redundant.
A collection that spans a vector space without redundancy leads directly to the idea of a basis.
Where We Go Next
Spanning sets tell us how to build a vector space, while independent sets tell us which vectors contribute genuinely new directions. Combining those two properties gives one of the central objects of linear algebra.
The next lesson, Bases and Dimension, develops bases as independent spanning sets and defines dimension as the number of vectors needed in a basis for a finite-dimensional vector space.