The Central Idea
Suppose we have vectors \(v_1,v_2,\ldots,v_k\). We can multiply each vector by a scalar and then add the results:
\[ c_1v_1+c_2v_2+\cdots+c_kv_k. \]Any vector produced this way is a linear combination of \(v_1,v_2,\ldots,v_k\).
The set of every possible linear combination is their span. If none of the vectors can be recreated from the others, the collection is linearly independent.
What Is a Linear Combination?
A linear combination of vectors \(v_1,\ldots,v_k\) has the form
\[ \boxed{c_1v_1+c_2v_2+\cdots+c_kv_k} \]where \(c_1,\ldots,c_k\) are scalars.
The word linear matters. We are allowed to scale vectors and add them. We are not introducing products such as \(v_1v_2\), powers such as \(v_1^2\), or other nonlinear operations.
Example: A Linear Combination in \(\mathbb{R}^2\)
Let
\[ v_1= \begin{bmatrix} 1\\ 2 \end{bmatrix}, \qquad v_2= \begin{bmatrix} 3\\ -1 \end{bmatrix}. \]The vector
\[ 2v_1-3v_2 \]is a linear combination of \(v_1\) and \(v_2\).
\[ 2 \begin{bmatrix} 1\\ 2 \end{bmatrix} - 3 \begin{bmatrix} 3\\ -1 \end{bmatrix} = \begin{bmatrix} 2\\ 4 \end{bmatrix} + \begin{bmatrix} -9\\ 3 \end{bmatrix} = \boxed{ \begin{bmatrix} -7\\ 7 \end{bmatrix}}. \]Coefficients Control the Combination
In
\[ c_1v_1+c_2v_2, \]the scalars \(c_1\) and \(c_2\) determine how much of each vector is used.
Positive coefficients preserve direction, negative coefficients reverse direction, zero coefficients remove a vector from that particular combination, and coefficients with magnitude greater than \(1\) stretch the corresponding vector.
Linear Combinations Are Not Limited to Coordinate Vectors
The same idea works in every vector space.
For polynomials, if
\[ p_1(x)=1+x,\qquad p_2(x)=x^2, \]then
\[ 3p_1(x)-2p_2(x) \]is the linear combination
\[ \boxed{3+3x-2x^2}. \]For matrices, linear combinations are formed by scalar multiplication and matrix addition. For functions, they are formed by scalar multiplication and pointwise addition.
What Is Span?
The span of vectors \(v_1,\ldots,v_k\) is the set of all their linear combinations.
\[ \boxed{ \operatorname{span}\{v_1,\ldots,v_k\} = \left\{ c_1v_1+\cdots+c_kv_k: c_1,\ldots,c_k\in\mathbb{R} \right\}} \]Span answers a generation question:
What vectors can be built from this collection?
The Span of One Nonzero Vector
Let
\[ v= \begin{bmatrix} 2\\ 1 \end{bmatrix}. \]Every vector in \(\operatorname{span}\{v\}\) has the form
\[ c \begin{bmatrix} 2\\ 1 \end{bmatrix} = \begin{bmatrix} 2c\\ c \end{bmatrix}. \]Geometrically, these vectors form the line through the origin pointing in the direction of \(v\).
The Span of Two Vectors in \(\mathbb{R}^2\)
Consider
\[ e_1= \begin{bmatrix} 1\\ 0 \end{bmatrix}, \qquad e_2= \begin{bmatrix} 0\\ 1 \end{bmatrix}. \]Every vector
\[ \begin{bmatrix} a\\ b \end{bmatrix} \]can be written as
\[ a e_1+b e_2. \]Therefore,
\[ \boxed{\operatorname{span}\{e_1,e_2\}=\mathbb{R}^2}. \]Two Vectors Do Not Always Span the Plane
Let
\[ v_1= \begin{bmatrix} 1\\ 2 \end{bmatrix}, \qquad v_2= \begin{bmatrix} 2\\ 4 \end{bmatrix}. \]Since
\[ v_2=2v_1, \]both vectors lie on the same line through the origin.
Their span is therefore the same as the span of \(v_1\) alone:
\[ \boxed{ \operatorname{span}\{v_1,v_2\} = \operatorname{span}\{v_1\}}. \]Span Is Always a Subspace
The span of any collection of vectors in a vector space is a subspace of that vector space.
Why? Because linear combinations are automatically closed under vector addition and scalar multiplication, and choosing every coefficient equal to zero produces the zero vector.
This creates an important connection with the previous lesson:
\[ \boxed{\text{vectors} \longrightarrow \text{their span} \longrightarrow \text{a subspace}}. \]The Smallest Subspace Containing the Vectors
The span of \(v_1,\ldots,v_k\) is not merely some subspace containing those vectors. It is the smallest subspace that contains all of them.
Any subspace containing \(v_1,\ldots,v_k\) must also contain every scalar multiple and every sum of those vectors. Therefore it must contain every linear combination.
Testing Whether a Vector Lies in a Span
To determine whether a vector \(b\) lies in
\[ \operatorname{span}\{v_1,\ldots,v_k\}, \]ask whether scalars \(c_1,\ldots,c_k\) exist such that
\[ \boxed{c_1v_1+\cdots+c_kv_k=b}. \]This converts a span question into a system of linear equations.
Example: Membership in a Span
Determine whether
\[ b= \begin{bmatrix} 7\\ 1 \end{bmatrix} \]belongs to the span of
\[ v_1= \begin{bmatrix} 1\\ 1 \end{bmatrix}, \qquad v_2= \begin{bmatrix} 2\\ -1 \end{bmatrix}. \]We seek \(c_1,c_2\) satisfying
\[ c_1 \begin{bmatrix} 1\\ 1 \end{bmatrix} + c_2 \begin{bmatrix} 2\\ -1 \end{bmatrix} = \begin{bmatrix} 7\\ 1 \end{bmatrix}. \]Comparing coordinates gives
\[ c_1+2c_2=7, \] \[ c_1-c_2=1. \]Subtracting the second equation from the first gives
\[ 3c_2=6, \]so \(c_2=2\). Then \(c_1=3\).
Therefore,
\[ \boxed{ \begin{bmatrix} 7\\ 1 \end{bmatrix} = 3 \begin{bmatrix} 1\\ 1 \end{bmatrix} + 2 \begin{bmatrix} 2\\ -1 \end{bmatrix}}. \]When a Vector Is Not in the Span
If the corresponding system
\[ c_1v_1+\cdots+c_kv_k=b \]is inconsistent, then \(b\) is not in the span.
Thus:
\[ \boxed{ b\in\operatorname{span}\{v_1,\ldots,v_k\} \iff c_1v_1+\cdots+c_kv_k=b \text{ has a solution}.} \]Span and Matrix Equations
Place the vectors \(v_1,\ldots,v_k\) into the columns of a matrix:
\[ A= \begin{bmatrix} \vert & \vert & & \vert\\ v_1 & v_2 & \cdots & v_k\\ \vert & \vert & & \vert \end{bmatrix}. \]Then
\[ c_1v_1+\cdots+c_kv_k=b \]becomes
\[ \boxed{Ax=b}. \]Span questions and systems of linear equations are therefore two views of the same underlying structure.
What Is Linear Independence?
Vectors \(v_1,\ldots,v_k\) are linearly independent if the equation
\[ c_1v_1+\cdots+c_kv_k=0 \]has only the trivial solution
\[ \boxed{c_1=c_2=\cdots=c_k=0}. \]In an independent collection, no vector is unnecessary.
What Is Linear Dependence?
A collection is linearly dependent if there is a nontrivial solution to
\[ c_1v_1+\cdots+c_kv_k=0. \]Nontrivial means that at least one coefficient is not zero.
Such a relation shows that one or more vectors can be recreated from the others.
Dependence Means Redundancy
Suppose
\[ 2v_1-v_2+3v_3=0. \]Solving for \(v_2\) gives
\[ v_2=2v_1+3v_3. \]So \(v_2\) adds no new direction to the span:
\[ \boxed{ \operatorname{span}\{v_1,v_2,v_3\} = \operatorname{span}\{v_1,v_3\}}. \]Example: A Dependent Pair
Let
\[ v_1= \begin{bmatrix} 1\\ 3 \end{bmatrix}, \qquad v_2= \begin{bmatrix} 2\\ 6 \end{bmatrix}. \]Since
\[ v_2=2v_1, \]we have the nontrivial relation
\[ 2v_1-v_2=0. \]Therefore the vectors are linearly dependent.
Example: An Independent Pair
Consider
\[ v_1= \begin{bmatrix} 1\\ 0 \end{bmatrix}, \qquad v_2= \begin{bmatrix} 0\\ 1 \end{bmatrix}. \]If
\[ c_1v_1+c_2v_2=0, \]then
\[ \begin{bmatrix} c_1\\ c_2 \end{bmatrix} = \begin{bmatrix} 0\\ 0 \end{bmatrix}. \]Hence \(c_1=c_2=0\), so the vectors are linearly independent.
The Zero Vector Forces Dependence
Any collection containing the zero vector is linearly dependent.
For example, if \(v_2=0\), then
\[ 0v_1+1v_2+0v_3+\cdots+0v_k=0 \]is already a nontrivial linear relation.
A Single Vector
A collection containing one vector \(\{v\}\) is linearly independent exactly when
\[ v\neq 0. \]If \(v\neq0\), the equation \(cv=0\) forces \(c=0\). If \(v=0\), then every scalar \(c\) satisfies \(cv=0\).
Two Vectors
Two nonzero vectors are linearly dependent precisely when one is a scalar multiple of the other.
In \(\mathbb{R}^2\), this means the vectors point along the same line through the origin.
If they are not scalar multiples, they are independent and span the entire plane.
Three Vectors in \(\mathbb{R}^3\)
Three vectors in \(\mathbb{R}^3\) may be:
- independent and able to span all of \(\mathbb{R}^3\);
- dependent while spanning a plane;
- dependent while spanning only a line;
- dependent because one or more vectors are zero.
The number of vectors alone does not determine the span. Their relationships matter.
More Vectors Than Dimensions
Any collection of more than \(n\) vectors in \(\mathbb{R}^n\) must be linearly dependent.
For example, any four vectors in \(\mathbb{R}^3\) are dependent.
There are only three independent coordinate directions available in \(\mathbb{R}^3\), so a fourth vector cannot contribute an entirely new independent direction.
Fewer Vectors Than Dimensions
Fewer than \(n\) vectors cannot span all of \(\mathbb{R}^n\).
For example, two vectors cannot span all of \(\mathbb{R}^3\).
They may span a plane, a line, or only the zero vector, depending on the vectors chosen.
Independence Does Not Mean Perpendicular
Linear independence is not the same as perpendicularity.
For example,
\[ \begin{bmatrix} 1\\ 0 \end{bmatrix} \qquad\text{and}\qquad \begin{bmatrix} 1\\ 1 \end{bmatrix} \]are not perpendicular, but they are linearly independent.
Orthogonality is a stronger geometric relationship that will be developed later in this section.
Independence Does Not Mean Unit Length
Independent vectors do not need length \(1\).
For example,
\[ \begin{bmatrix} 5\\ 0 \end{bmatrix}, \qquad \begin{bmatrix} 0\\ -8 \end{bmatrix} \]are independent even though neither is a unit vector.
Testing Independence with a Homogeneous System
To test whether \(v_1,\ldots,v_k\) are independent, solve
\[ c_1v_1+\cdots+c_kv_k=0. \]Place the vectors in the columns of a matrix \(A\). Then solve
\[ \boxed{Ax=0}. \]If the homogeneous system has only \(x=0\), the vectors are independent. If it has a nonzero solution, they are dependent.
Example: Testing Three Vectors
Consider
\[ v_1= \begin{bmatrix} 1\\ 0\\ 1 \end{bmatrix}, \quad v_2= \begin{bmatrix} 0\\ 1\\ 1 \end{bmatrix}, \quad v_3= \begin{bmatrix} 1\\ 1\\ 2 \end{bmatrix}. \]Notice that
\[ v_3=v_1+v_2. \]Therefore,
\[ v_1+v_2-v_3=0 \]is a nontrivial relation, and the vectors are linearly dependent.
Span Can Stay the Same After Removing a Vector
If a collection is linearly dependent, at least one vector can be removed without changing the span.
This is one of the most useful practical consequences of dependence: redundant data can be discarded while preserving the generated subspace.
Span Can Grow When an Independent Vector Is Added
Suppose \(v\) is not already in
\[ \operatorname{span}\{v_1,\ldots,v_k\}. \]Then adding \(v\) enlarges the span:
\[ \operatorname{span}\{v_1,\ldots,v_k\} \subsetneq \operatorname{span}\{v_1,\ldots,v_k,v\}. \]The new vector contributes a genuinely new direction.
Generating Sets
A collection of vectors that spans a vector space is called a spanning set or generating set.
For example,
\[ \left\{ \begin{bmatrix}1\\0\end{bmatrix}, \begin{bmatrix}0\\1\end{bmatrix} \right\} \]generates \(\mathbb{R}^2\).
A spanning set may contain redundant vectors. Independence tells us whether it does.
A Spanning Set Can Be Redundant
The set
\[ \left\{ \begin{bmatrix}1\\0\end{bmatrix}, \begin{bmatrix}0\\1\end{bmatrix}, \begin{bmatrix}1\\1\end{bmatrix} \right\} \]spans \(\mathbb{R}^2\), but it is linearly dependent because
\[ \begin{bmatrix} 1\\ 1 \end{bmatrix} = \begin{bmatrix} 1\\ 0 \end{bmatrix} + \begin{bmatrix} 0\\ 1 \end{bmatrix}. \]The third vector is unnecessary for spanning the plane.
Independent Does Not Automatically Mean Spanning
The single vector
\[ \begin{bmatrix} 1\\ 2 \end{bmatrix} \]is linearly independent by itself, but it does not span all of \(\mathbb{R}^2\).
Independence asks whether the vectors are redundant. Spanning asks whether they generate the whole target space.
Spanning Does Not Automatically Mean Independent
A collection can span a space and still contain redundant vectors.
To obtain a minimal, nonredundant generating system, we want both properties at once:
\[ \boxed{\text{spanning}+\text{independence}}. \]That combination leads directly to the concept of a basis in the next lesson.
Linear Combinations of Polynomials
Consider the polynomial space \(P_2\), consisting of polynomials of degree at most \(2\).
Let
\[ p_1(x)=1,\qquad p_2(x)=x,\qquad p_3(x)=x^2. \]Every polynomial
\[ a+bx+cx^2 \]can be written
\[ ap_1+bp_2+cp_3. \]Therefore,
\[ \boxed{\operatorname{span}\{1,x,x^2\}=P_2}. \]Independence of \(1,x,x^2\)
Suppose
\[ c_1+c_2x+c_3x^2=0 \]as the zero polynomial.
Every coefficient must be zero:
\[ c_1=c_2=c_3=0. \]Hence
\[ \boxed{\{1,x,x^2\}\text{ is linearly independent in }P_2}. \]A Dependent Polynomial Collection
Consider
\[ 1,\qquad x,\qquad 1+x. \]Since
\[ (1+x)-1-x=0, \]the collection is linearly dependent.
The polynomial \(1+x\) adds no new direction beyond those already supplied by \(1\) and \(x\).
Linear Combinations of Matrices
Let
\[ A= \begin{bmatrix} 1&0\\ 0&0 \end{bmatrix}, \qquad B= \begin{bmatrix} 0&0\\ 0&1 \end{bmatrix}. \]Then any combination
\[ c_1A+c_2B \]has the form
\[ \begin{bmatrix} c_1&0\\ 0&c_2 \end{bmatrix}. \]Thus \(A\) and \(B\) span the subspace of diagonal \(2\times2\) matrices.
Linear Combinations of Functions
If
\[ f(x)=\sin x,\qquad g(x)=\cos x, \]then
\[ a\sin x+b\cos x \]is a linear combination of \(f\) and \(g\).
Their span consists of all functions of that form.
This same structure appears in differential equations, Fourier methods, signal models, and oscillatory systems.
Linear Combinations as Building Instructions
It is useful to think of a linear combination as a set of building instructions.
The vectors are the available building blocks. The coefficients tell us how much of each block to use. Span describes every object that can be assembled from those blocks.
Independence asks whether any block can already be assembled from the others.
Geometric Interpretation in \(\mathbb{R}^3\)
In three-dimensional space:
- the span of one nonzero vector is a line through the origin;
- the span of two independent vectors is a plane through the origin;
- the span of three independent vectors is all of \(\mathbb{R}^3\).
Dependence occurs when one of the proposed directions fails to add a new dimension.
Why the Origin Keeps Appearing
Every span contains the zero vector because setting all coefficients to zero gives
\[ 0v_1+\cdots+0v_k=0. \]This is why linear spans in coordinate space pass through the origin.
A shifted line or plane that does not pass through the origin may be affine, but it cannot be the span of vectors in the usual linear sense.
Common Error: Confusing Span with the Original Set
The set
\[ \{v_1,v_2\} \]contains only two vectors.
The span
\[ \operatorname{span}\{v_1,v_2\} \]usually contains infinitely many vectors because the coefficients can vary continuously.
Common Error: Using Only Positive Coefficients
A span uses all allowed scalars, not merely positive numbers.
In a real vector space, coefficients may be positive, negative, or zero. Restricting coefficients changes the generated set and generally destroys the vector-space structure.
Common Error: Testing Independence with \(Ax=b\)
Span membership uses
\[ Ax=b. \]Independence uses the homogeneous system
\[ \boxed{Ax=0}. \]These questions are related, but they are not identical.
Common Error: Thinking Any Nonzero Coefficients Prove Dependence
Dependence requires a nontrivial combination that equals the zero vector.
The statement
\[ 2v_1+3v_2=w \]does not by itself prove that \(v_1\) and \(v_2\) are dependent.
Common Error: Assuming More Vectors Produce a Larger Span
Adding a vector already contained in the existing span does not enlarge the span.
If
\[ v_3=2v_1-v_2, \]then
\[ \operatorname{span}\{v_1,v_2,v_3\} = \operatorname{span}\{v_1,v_2\}. \]Common Error: Assuming a Large Set Is Automatically Useful
More vectors can create more redundancy rather than more information.
In computational work, redundant columns may increase storage, obscure structure, and complicate interpretation. Detecting dependence is therefore not only theoretical; it is part of simplifying multidimensional systems.
A Reliable Procedure for Span Membership
- Write the target vector \(b\).
- Write the generating vectors \(v_1,\ldots,v_k\).
- Set \(c_1v_1+\cdots+c_kv_k=b\).
- Translate the vector equation into scalar equations or \(Ax=b\).
- Solve for the coefficients.
- If a solution exists, \(b\) lies in the span.
- If the system is inconsistent, \(b\) does not lie in the span.
A Reliable Procedure for Linear Independence
- Place the vectors into the equation \(c_1v_1+\cdots+c_kv_k=0\).
- Translate the equation into a homogeneous linear system.
- Solve for the coefficients.
- If the only solution is \(c_1=\cdots=c_k=0\), the vectors are independent.
- If any nontrivial solution exists, the vectors are dependent.
- When useful, identify which vector is redundant by solving the relation for it.
Example: Full Span Workflow
Determine whether
\[ b= \begin{bmatrix} 4\\ 5\\ 1 \end{bmatrix} \]lies in the span of
\[ v_1= \begin{bmatrix} 1\\ 1\\ 0 \end{bmatrix}, \qquad v_2= \begin{bmatrix} 0\\ 1\\ 1 \end{bmatrix}, \qquad v_3= \begin{bmatrix} 1\\ 0\\ -1 \end{bmatrix}. \]Set
\[ c_1v_1+c_2v_2+c_3v_3=b. \]Coordinate comparison gives
\[ c_1+c_3=4, \] \[ c_1+c_2=5, \] \[ c_2-c_3=1. \]The third equation follows from the first two, so the system has infinitely many solutions. For example, choosing \(c_3=0\) gives
\[ c_1=4,\qquad c_2=1. \]Thus
\[ \boxed{b=4v_1+v_2}. \]Therefore \(b\) lies in the span.
Example: Full Independence Workflow
Test
\[ v_1= \begin{bmatrix} 1\\ 1\\ 0 \end{bmatrix}, \quad v_2= \begin{bmatrix} 0\\ 1\\ 1 \end{bmatrix}, \quad v_3= \begin{bmatrix} 1\\ 2\\ 1 \end{bmatrix}. \]Observe that
\[ v_3=v_1+v_2. \]Therefore,
\[ v_1+v_2-v_3=0 \]is a nontrivial linear relation.
Hence the collection is
\[ \boxed{\text{linearly dependent}}. \]Why These Ideas Matter in Multidimensional Systems
Many large systems are built from a limited number of underlying directions or patterns.
Linear combinations describe how those directions combine. Span identifies the reachable or representable states. Independence identifies which directions carry genuinely distinct information.
These ideas later support coordinate systems, matrix rank, eigenvector decompositions, least-squares models, singular-value methods, and dimensional reduction.
Practice 1: Compute a Linear Combination
Let
\[ v_1= \begin{bmatrix} 2\\ 1 \end{bmatrix}, \qquad v_2= \begin{bmatrix} -1\\ 3 \end{bmatrix}. \]Compute \(3v_1-2v_2\).
Practice 2: Recognize a Span
Describe geometrically
\[ \operatorname{span} \left\{ \begin{bmatrix} 4\\ -2 \end{bmatrix} \right\}. \]Practice 3: Span Membership
Determine whether
\[ \begin{bmatrix} 5\\ 4 \end{bmatrix} \]lies in the span of
\[ \begin{bmatrix} 1\\ 1 \end{bmatrix}, \qquad \begin{bmatrix} 2\\ 1 \end{bmatrix}. \]Practice 4: Dependent or Independent?
Determine whether
\[ \begin{bmatrix} 1\\ 2 \end{bmatrix}, \qquad \begin{bmatrix} 3\\ 6 \end{bmatrix} \]are linearly independent.
Practice 5: Independent Pair
Determine whether
\[ \begin{bmatrix} 1\\ 2 \end{bmatrix}, \qquad \begin{bmatrix} 2\\ 1 \end{bmatrix} \]are linearly independent.
Practice 6: Find the Redundant Vector
In the collection
\[ v_1= \begin{bmatrix} 1\\ 0\\ 1 \end{bmatrix}, \quad v_2= \begin{bmatrix} 0\\ 1\\ 1 \end{bmatrix}, \quad v_3= \begin{bmatrix} 2\\ 1\\ 3 \end{bmatrix}, \]identify a vector that can be written as a linear combination of the others.
Practice 7: Polynomial Span
Does
\[ p(x)=3-2x+5x^2 \]lie in \(\operatorname{span}\{1,x,x^2\}\)?
Practice 8: Polynomial Dependence
Determine whether
\[ 1,\qquad x,\qquad 2+3x \]are linearly independent.
Practice 9: Dimension Reasoning
Can four vectors in \(\mathbb{R}^3\) be linearly independent? Explain.
Practice 10: Spanning Reasoning
Can two vectors span all of \(\mathbb{R}^3\)? Explain.
Practice 11: Zero Vector
Is the collection
\[ \left\{ \begin{bmatrix}1\\2\end{bmatrix}, \begin{bmatrix}0\\0\end{bmatrix} \right\} \]linearly independent?
Practice 12: Same Span?
Let
\[ v_1= \begin{bmatrix} 1\\ 0 \end{bmatrix}, \qquad v_2= \begin{bmatrix} 0\\ 1 \end{bmatrix}, \qquad v_3= \begin{bmatrix} 1\\ 1 \end{bmatrix}. \]Compare
\[ \operatorname{span}\{v_1,v_2\} \]with
\[ \operatorname{span}\{v_1,v_2,v_3\}. \]Practice Answers
Practice 1
\[ 3 \begin{bmatrix} 2\\ 1 \end{bmatrix} - 2 \begin{bmatrix} -1\\ 3 \end{bmatrix} = \begin{bmatrix} 6\\ 3 \end{bmatrix} + \begin{bmatrix} 2\\ -6 \end{bmatrix} = \boxed{ \begin{bmatrix} 8\\ -3 \end{bmatrix}}. \]Practice 2
The span is the line through the origin in the direction
\[ \begin{bmatrix} 4\\ -2 \end{bmatrix}. \]Practice 3
Solve
\[ c_1 \begin{bmatrix} 1\\ 1 \end{bmatrix} + c_2 \begin{bmatrix} 2\\ 1 \end{bmatrix} = \begin{bmatrix} 5\\ 4 \end{bmatrix}. \]This gives
\[ c_1+2c_2=5,\qquad c_1+c_2=4. \]Hence \(c_2=1\) and \(c_1=3\). Therefore the target vector is in the span.
\h3>Practice 4The second vector is three times the first:
\[ \begin{bmatrix} 3\\ 6 \end{bmatrix} = 3 \begin{bmatrix} 1\\ 2 \end{bmatrix}. \]Therefore the vectors are linearly dependent.
Practice 5
Neither vector is a scalar multiple of the other, so the pair is linearly independent.
Practice 6
We have
\[ v_3=2v_1+v_2. \]Therefore \(v_3\) is redundant.
Practice 7
Yes:
\[ \boxed{p(x)=3(1)-2(x)+5(x^2)}. \]Practice 8
Since
\[ 2+3x=2(1)+3x, \]the third polynomial is a linear combination of the first two. The collection is linearly dependent.
Practice 9
No. Any set containing more than three vectors in \(\mathbb{R}^3\) must be linearly dependent.
Practice 10
No. Two vectors can span at most a two-dimensional subspace of \(\mathbb{R}^3\).
Practice 11
No. Any collection containing the zero vector is linearly dependent.
Practice 12
Since
\[ v_3=v_1+v_2, \]adding \(v_3\) does not enlarge the span:
\[ \boxed{ \operatorname{span}\{v_1,v_2\} = \operatorname{span}\{v_1,v_2,v_3\} = \mathbb{R}^2}. \]What to Remember
A linear combination has the form
\[ c_1v_1+\cdots+c_kv_k. \]The span of a collection is the set of every vector obtainable from such combinations.
- Span describes what a collection can generate.
- Every span is a subspace.
- Span membership is tested by solving \(Ax=b\).
- Independence is tested by solving \(Ax=0\).
- A nontrivial solution to \(Ax=0\) means dependence.
- Dependence means at least one vector is redundant.
- An independent vector added outside an existing span enlarges that span.
The key distinction is:
\[ \boxed{ \text{span asks what can be generated; independence asks what is redundant}.} \]Where We Go Next
A spanning set may contain too many vectors, while an independent set may contain too few to generate the entire space.
The next lesson, Bases and Dimension, brings these two requirements together. A basis is a linearly independent spanning set, and the number of vectors in a basis defines the dimension of the space.