Span of Vectors in Linear Algebra

In linear algebra, the span of a set of vectors is the collection of all possible linear combinations that those vectors can create. This guide explores the definition of vector spans and demonstrates how to determine whether a set of vectors spans a given vector space.

Table of Contents

Space Spanned by Vectors

If vectors \( \textbf{u}_1, \textbf{u}_2, \dots, \textbf{u}_n \) are in a vector space \( V \), then the set \( W \) of all linear combinations of these vectors is a subspace of \( V \), called the span of \( \textbf{u}_1, \textbf{u}_2, \dots, \textbf{u}_n \).

We also say that the vectors span \( W \), denoted as:

\[ W = \text{span}\{\textbf{u}_1, \textbf{u}_2, \dots, \textbf{u}_n\} \]

Properties of Span

The span of a set of vectors has several important properties:

  1. Always a subspace: The span of any set of vectors is always a subspace of the ambient vector space.
  2. Contains the zero vector: The zero vector is always in the span (set all scalars to zero).
  3. Smallest subspace containing the vectors: \( \text{span}\{\textbf{u}_1, \dots, \textbf{u}_n\} \) is the smallest subspace that contains all the vectors \( \textbf{u}_1, \dots, \textbf{u}_n \).
  4. Dimension: The dimension of \( \text{span}\{\textbf{u}_1, \dots, \textbf{u}_n\} \) equals the number of linearly independent vectors in the set (i.e., the rank of the matrix with these vectors as columns).
  5. Redundancy: Adding a vector that is already in the span does not change the span. Removing a linearly dependent vector does not change the span either.

Key Fact: In \( \mathbb{R}^n \), a set of vectors spans \( \mathbb{R}^n \) if and only if the matrix formed by these vectors (as columns) has a pivot position in every row when row-reduced. Equivalently, for a square \( n \times n \) matrix, the vectors span \( \mathbb{R}^n \) if and only if the determinant is non-zero.

Examples with Solutions

Example 1: Proving that Two Vectors Span \( \mathbb{R}^2 \)

Show Solution

Show that the vectors \( \textbf{u}_1 = \begin{bmatrix} 1 \\ 5 \end{bmatrix} \) and \( \textbf{u}_2 = \begin{bmatrix} -1 \\ 2 \end{bmatrix} \) span the vector space \( \mathbb{R}^2 \).

Solution

A general vector in \( \mathbb{R}^2 \) is of the form \( \begin{bmatrix} a \\ b \end{bmatrix} \), where \( a \) and \( b \) can take any real numbers in \( \mathbb{R} \).

To show that \( \textbf{u}_1 \) and \( \textbf{u}_2 \) span \( \mathbb{R}^2 \), we must demonstrate that any arbitrary vector \( \begin{bmatrix} a \\ b \end{bmatrix} \) can be written as a linear combination of \( \textbf{u}_1 \) and \( \textbf{u}_2 \). That is, we need to find scalars \( r_1 \) and \( r_2 \) such that:

\[ \begin{bmatrix} a \\ b \end{bmatrix} = r_1 \begin{bmatrix} 1 \\ 5 \end{bmatrix} + r_2 \begin{bmatrix} -1 \\ 2 \end{bmatrix} \]

Performing scalar multiplication and vector addition on the right side:

\[ \begin{bmatrix} a \\ b \end{bmatrix} = \begin{bmatrix} r_1 - r_2 \\ 5r_1 + 2r_2 \end{bmatrix} \quad \text{(I)} \]

This gives us a system of linear equations:

\[ \begin{cases} r_1 - r_2 = a \\ 5r_1 + 2r_2 = b \end{cases} \]

We solve this system using the method of elimination. Multiply the top equation by \( 2 \):

\[ 2r_1 - 2r_2 = 2a \]

Add this result to the second equation to eliminate \( r_2 \):

\[ (2r_1 - 2r_2) + (5r_1 + 2r_2) = 2a + b \implies 7r_1 = 2a + b \implies r_1 = \frac{2a + b}{7} \]

From the first equation, express \( r_2 \):

\[ r_2 = r_1 - a = \frac{2a + b}{7} - a = \frac{2a + b - 7a}{7} = \frac{b - 5a}{7} \]

Since we can always find valid scalars \( r_1 \) and \( r_2 \) for any values of \( a \) and \( b \), any vector in \( \mathbb{R}^2 \) can be expressed as a linear combination of \( \textbf{u}_1 \) and \( \textbf{u}_2 \). Therefore, these vectors span \( \mathbb{R}^2 \).

Alternative Check: The determinant of the matrix with these vectors as columns is \( (1)(2) - (-1)(5) = 2 + 5 = 7 \neq 0 \), confirming they span \( \mathbb{R}^2 \).

Example 2: Showing Vectors Do NOT Span \( \mathbb{R}^2 \)

Show Solution

Show that the vectors \( \textbf{u}_1 = \begin{bmatrix} 1 \\ -2 \end{bmatrix} \) and \( \textbf{u}_2 = \begin{bmatrix} 2 \\ -4 \end{bmatrix} \) DO NOT span the vector space \( \mathbb{R}^2 \).

Solution

We test whether an arbitrary vector \( \begin{bmatrix} a \\ b \end{bmatrix} \) can be written as a linear combination of \( \textbf{u}_1 \) and \( \textbf{u}_2 \):

\[ \begin{bmatrix} a \\ b \end{bmatrix} = r_1 \begin{bmatrix} 1 \\ -2 \end{bmatrix} + r_2 \begin{bmatrix} 2 \\ -4 \end{bmatrix} = \begin{bmatrix} r_1 + 2r_2 \\ -2r_1 - 4r_2 \end{bmatrix} \]

This gives the system:

\[ \begin{cases} r_1 + 2r_2 = a \\ -2r_1 - 4r_2 = b \end{cases} \]

Multiply the top equation by \( 2 \):

\[ 2a = 2r_1 + 4r_2 \]

Add the equations together:

\[ 2a + b = (2r_1 + 4r_2) + (-2r_1 - 4r_2) \implies 2a + b = 0 \]

This reveals that a solution only exists if \( 2a + b = 0 \). If \( a \) and \( b \) are chosen such that \( 2a + b \neq 0 \), the system has no solution. Hence, the given vectors do not span \( \mathbb{R}^2 \).

Geometric Interpretation: Since \( \textbf{u}_2 = 2\textbf{u}_1 \), both vectors lie on the same line through the origin. Their span is only that line (a 1-dimensional subspace), not all of \( \mathbb{R}^2 \).

Example 3: Standard Basis Vectors Spanning \( \mathbb{R}^3 \)

Show Solution

Show that the vectors \( \textbf{u}_1 = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} \), \( \textbf{u}_2 = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} \), and \( \textbf{u}_3 = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \) span the vector space \( \mathbb{R}^3 \).

Solution

We set up the linear combination equation for a general vector in \( \mathbb{R}^3 \):

\[ \begin{bmatrix} a \\ b \\ c \end{bmatrix} = r_1 \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} + r_2 \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} + r_3 \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} r_1 \\ r_2 \\ r_3 \end{bmatrix} \]

Direct comparison yields:

\[ r_1 = a, \quad r_2 = b, \quad r_3 = c \]

Since unique scalars can always be found for any real numbers \( a, b, c \), the standard unit vectors span \( \mathbb{R}^3 \).

Example 4: Proving Three Non-Standard Vectors Span \( \mathbb{R}^3 \)

Show Solution

Show that the vectors \( \textbf{u}_1 = \begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix} \), \( \textbf{u}_2 = \begin{bmatrix} 0 \\ 1 \\ -1 \end{bmatrix} \), and \( \textbf{u}_3 = \begin{bmatrix} 0 \\ 2 \\ 1 \end{bmatrix} \) span the vector space \( \mathbb{R}^3 \).

Solution

We express an arbitrary vector in \( \mathbb{R}^3 \) as a linear combination:

\[ \begin{bmatrix} a \\ b \\ c \end{bmatrix} = r_1 \begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix} + r_2 \begin{bmatrix} 0 \\ 1 \\ -1 \end{bmatrix} + r_3 \begin{bmatrix} 0 \\ 2 \\ 1 \end{bmatrix} = \begin{bmatrix} r_1 \\ 2r_1 + r_2 + 2r_3 \\ -r_2 + r_3 \end{bmatrix} \]

This gives the system:

\[ \begin{cases} r_1 = a \\ 2r_1 + r_2 + 2r_3 = b \\ -r_2 + r_3 = c \end{cases} \]

Substituting \( r_1 = a \) into the second equation:

\[ 2a + r_2 + 2r_3 = b \implies r_2 + 2r_3 = b - 2a \]

Together with \( -r_2 + r_3 = c \), we add both equations:

\[ (r_2 + 2r_3) + (-r_2 + r_3) = (b - 2a) + c \implies 3r_3 = b + c - 2a \implies r_3 = \frac{b + c - 2a}{3} \]

Then from \( -r_2 + r_3 = c \):

\[ r_2 = r_3 - c = \frac{b + c - 2a}{3} - c = \frac{b + c - 2a - 3c}{3} = \frac{b - 2c - 2a}{3} \]

So the solution is:

\[ r_1 = a, \quad r_2 = \frac{b - 2c - 2a}{3}, \quad r_3 = \frac{b + c - 2a}{3} \]

Because solutions for the scalars exist for any vector in \( \mathbb{R}^3 \), these three vectors span \( \mathbb{R}^3 \).

Alternative Check: The determinant of the matrix \( \begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 2 \\ 0 & -1 & 1 \end{bmatrix} \) is \( 1 \cdot (1 \cdot 1 - 2 \cdot (-1)) = 1 \cdot (1 + 2) = 3 \neq 0 \), confirming they span \( \mathbb{R}^3 \).

Example 5: Using Matrix Row Operations to Test Spanning in \( \mathbb{R}^3 \)

Show Solution

Show that the vectors \( \textbf{u}_1 = \begin{bmatrix} -1 \\ 2 \\ 0 \end{bmatrix} \), \( \textbf{u}_2 = \begin{bmatrix} 0 \\ 2 \\ -1 \end{bmatrix} \), and \( \textbf{u}_3 = \begin{bmatrix} 1 \\ -4 \\ 1 \end{bmatrix} \) DO NOT span the vector space \( \mathbb{R}^3 \).

Solution

Set up the linear combination equation for a vector \( \begin{bmatrix} a \\ b \\ c \end{bmatrix} \):

\[ \begin{bmatrix} a \\ b \\ c \end{bmatrix} = r_1 \begin{bmatrix} -1 \\ 2 \\ 0 \end{bmatrix} + r_2 \begin{bmatrix} 0 \\ 2 \\ -1 \end{bmatrix} + r_3 \begin{bmatrix} 1 \\ -4 \\ 1 \end{bmatrix} \]

Write this system in matrix form:

\[ \begin{bmatrix} -1 & 0 & 1 \\ 2 & 2 & -4 \\ 0 & -1 & 1 \end{bmatrix} \begin{bmatrix} r_1 \\ r_2 \\ r_3 \end{bmatrix} = \begin{bmatrix} a \\ b \\ c \end{bmatrix} \]

Construct the augmented matrix and use elementary row operations to reduce it to row echelon form:

  1. Initial augmented matrix: \( \begin{bmatrix} -1 & 0 & 1 & | & a \\ 2 & 2 & -4 & | & b \\ 0 & -1 & 1 & | & c \end{bmatrix} \)
  2. Swap row 1 and row 2 (and adjust signs/rows): \( \begin{bmatrix} 2 & 2 & -4 & | & b \\ -1 & 0 & 1 & | & a \\ 0 & -1 & 1 & | & c \end{bmatrix} \)
  3. Add row 1 to twice row 2 to update row 2: \( \begin{bmatrix} 2 & 2 & -4 & | & b \\ 0 & 2 & -2 & | & 2a + b \\ 0 & -1 & 1 & | & c \end{bmatrix} \)
  4. Add row 2 to twice row 3 to update row 3: \( \begin{bmatrix} 2 & 2 & -4 & | & b \\ 0 & 2 & -2 & | & 2a + b \\ 0 & 0 & 0 & | & 2a + b + 2c \end{bmatrix} \)

The final row indicates that the system has no solution if \( 2a + b + 2c \neq 0 \). Therefore, since some vectors in \( \mathbb{R}^3 \) cannot be formed, the given vectors do not span \( \mathbb{R}^3 \).

Alternative Check: The determinant of the matrix \( \begin{bmatrix} -1 & 0 & 1 \\ 2 & 2 & -4 \\ 0 & -1 & 1 \end{bmatrix} \) is:

\[ -1 \cdot (2 \cdot 1 - (-4)(-1)) - 0 + 1 \cdot (2 \cdot (-1) - 2 \cdot 0) = -1 \cdot (2 - 4) + 1 \cdot (-2) = 2 - 2 = 0 \]

Since the determinant is zero, the vectors are linearly dependent and cannot span \( \mathbb{R}^3 \).

Observation: Notice that \( \textbf{u}_3 = -\textbf{u}_1 - 2\textbf{u}_2 \). Indeed: \( -\begin{bmatrix} -1 \\ 2 \\ 0 \end{bmatrix} - 2\begin{bmatrix} 0 \\ 2 \\ -1 \end{bmatrix} = \begin{bmatrix} 1 \\ -2 \\ 0 \end{bmatrix} + \begin{bmatrix} 0 \\ -4 \\ 2 \end{bmatrix} = \begin{bmatrix} 1 \\ -6 \\ 2 \end{bmatrix} \). This does not equal \( \textbf{u}_3 = \begin{bmatrix} 1 \\ -4 \\ 1 \end{bmatrix} \), so let me verify the linear dependence more carefully by solving for \( r_1, r_2, r_3 \):

Setting \( r_1 = 1, r_2 = 1, r_3 = 1 \): \( \begin{bmatrix} -1 \\ 2 \\ 0 \end{bmatrix} + \begin{bmatrix} 0 \\ 2 \\ -1 \end{bmatrix} + \begin{bmatrix} 1 \\ -4 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \). So \( \textbf{u}_1 + \textbf{u}_2 + \textbf{u}_3 = \textbf{0} \), confirming linear dependence.

Frequently Asked Questions

What is the span of a set of vectors?

The span of vectors \( \textbf{u}_1, \textbf{u}_2, \dots, \textbf{u}_n \) is the set of all possible linear combinations of these vectors. It forms a subspace of the vector space.

How do you determine if vectors span \( \mathbb{R}^n \)?

To determine if a set of vectors spans \( \mathbb{R}^n \), set up a linear combination equation equal to an arbitrary vector, form the augmented matrix, and check if the system is consistent for all possible vectors. If the coefficient matrix has a pivot in every row, the vectors span \( \mathbb{R}^n \).

Can two vectors span \( \mathbb{R}^3 \)?

No. At least three vectors are needed to span \( \mathbb{R}^3 \), because the dimension of \( \mathbb{R}^3 \) is 3. Two vectors can at most span a plane (2-dimensional subspace) in \( \mathbb{R}^3 \).

What is the difference between a spanning set and a basis?

A spanning set is any set of vectors whose span equals the vector space. A basis is a spanning set that is also linearly independent. A basis is the most efficient spanning set.

Does adding more vectors to a spanning set change the span?

No, adding more vectors to a spanning set does not change the span, as long as the added vectors are already in the span of the original set. The span remains the same subspace.

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