Vector Equations and the Span
Before Deciding whether a vector is in a span means row reducing an augmented matrix. Builds on: Systems of Linear Equations and Row Reduction
Next Span leads to the matrix equation Ax = b and, later, to dot products and orthogonality. Leads to: The Matrix Equation Ax = b · Inner Products, Length, and Orthogonality
Before You Start: Prerequisite Check
Can you do these? (Click to reveal self-test)
Test yourself on these prerequisite skills:
Row reduction: What does it mean for an augmented matrix to be consistent?
Check
A system is consistent when the augmented column is not a pivot column, equivalently when no row reduces to the form $[0 \; 0 \; \cdots \; 0 \mid c]$ with $c \neq 0$.
Vector arithmetic: Compute $3\begin{pmatrix}2\\-1\end{pmatrix} - 2\begin{pmatrix}0\\4\end{pmatrix}$.
Check
\[ 3\begin{pmatrix}2\\-1\end{pmatrix} - 2\begin{pmatrix}0\\4\end{pmatrix} = \begin{pmatrix}6\\-3\end{pmatrix} + \begin{pmatrix}0\\-8\end{pmatrix} = \begin{pmatrix}6\\-11\end{pmatrix} \]
Connection: How does the augmented matrix for a system encode the system’s information?
Check
The coefficient of each variable in each equation fills one column, and the constant fills the augmented column. Position replaces variable names.
If you struggled: Review Systems of Linear Equations and Row Reduction before continuing.
Try This First
Before any definitions, try this.
Write the vector $\begin{pmatrix}4\\6\end{pmatrix}$ as a sum of a scalar multiple of $\begin{pmatrix}1\\0\end{pmatrix}$ and a scalar multiple of $\begin{pmatrix}0\\1\end{pmatrix}$. Find the two scalars.
Show Solution
\[ 4 \begin{pmatrix}1\\0\end{pmatrix} + 6 \begin{pmatrix}0\\1\end{pmatrix} = \begin{pmatrix}4\\0\end{pmatrix} + \begin{pmatrix}0\\6\end{pmatrix} = \begin{pmatrix}4\\6\end{pmatrix} \]
The scalars are $4$ and $6$. You just wrote $\begin{pmatrix}4\\6\end{pmatrix}$ as a linear combination of the two given vectors.
What if the target were $\begin{pmatrix}5\\-3\end{pmatrix}$? The scalars would be $5$ and $-3$: any vector in $\mathbb{R}^2$ can be reached this way, because $\begin{pmatrix}1\\0\end{pmatrix}$ and $\begin{pmatrix}0\\1\end{pmatrix}$ cover all directions.
Now the harder question: what if the two given vectors were $\begin{pmatrix}1\\2\end{pmatrix}$ and $\begin{pmatrix}2\\4\end{pmatrix}$? Could you still reach $\begin{pmatrix}4\\6\end{pmatrix}$? Section 1.3 takes on exactly this question.
Reaching Targets
The central question of this section: given a collection of vectors, which target vectors can be built from them by scaling and adding?
This is not merely an abstract question. A system of linear equations asks exactly this: “is the right-hand-side vector $\mathbf{b}$ buildable from the columns of the coefficient matrix?” Reframing the system as a vector equation reveals this geometric structure and unifies the algebra you already know with a spatial picture.
The vocabulary:
- A linear combination of vectors $\mathbf{v}_1, \ldots, \mathbf{v}_p$ with scalars $c_1, \ldots, c_p$ is the vector $c_1 \mathbf{v}_1 + \cdots + c_p \mathbf{v}_p$.
- The span of $\{\mathbf{v}_1, \ldots, \mathbf{v}_p\}$ is the set of all linear combinations: every target reachable by choosing some scalars.
Prerequisite Map
Quick Reference
| Property | Value |
|---|---|
| Course | MATH 231 |
| Difficulty | Beginner |
| Time | ~30 minutes |
| Source | Lay, Linear Algebra and Its Applications, Section 1.3 |
Key Terms at a Glance
| Term | Definition |
|---|---|
| Linear combination | $c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + \cdots + c_p\mathbf{v}_p$ for any scalars $c_i$ |
| Span | The set of all linear combinations of a given list of vectors |
| Vector equation | $x_1\mathbf{v}_1 + \cdots + x_p\mathbf{v}_p = \mathbf{b}$ with unknowns $x_1, \ldots, x_p$ |
| $\mathbf{b} \in \text{Span}\{\mathbf{v}_1,\ldots,\mathbf{v}_p\}$ | The vector equation has at least one solution |
Testing Span Membership
To decide whether $\mathbf{b}$ is in $\text{Span}\{\mathbf{v}_1, \ldots, \mathbf{v}_p\}$:
Form the augmented matrix $[\mathbf{v}_1 \; \cdots \; \mathbf{v}_p \mid \mathbf{b}]$ and row-reduce. If the system is consistent, then $\mathbf{b}$ is in the span. If not, it is not.
Key Concepts
Vectors in $\mathbb{R}^n$
A vector in $\mathbb{R}^n$ is a column of $n$ real numbers:
\[ \mathbf{u} = \begin{pmatrix} u_1 \\ u_2 \\ \vdots \\ u_n \end{pmatrix} \]
Two operations define the algebra of $\mathbb{R}^n$:
- Addition: $\mathbf{u} + \mathbf{v} = \begin{pmatrix}u_1 + v_1 \\ \vdots \\ u_n + v_n\end{pmatrix}$ (add entry by entry)
- Scalar multiplication: $c\mathbf{u} = \begin{pmatrix}cu_1 \\ \vdots \\ cu_n\end{pmatrix}$ (multiply every entry by $c$)
Geometric picture in $\mathbb{R}^2$: a vector $\begin{pmatrix}a\\b\end{pmatrix}$ is an arrow from the origin to the point $(a, b)$.
Linear Combinations
A linear combination of vectors $\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_p$ in $\mathbb{R}^n$ with scalars $c_1, c_2, \ldots, c_p$ is:
\[ c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + \cdots + c_p\mathbf{v}_p \]
The scalars are the “amounts” of each vector you mix in. Negative scalars are permitted; they reverse direction.
Example: With $\mathbf{v}_1 = \begin{pmatrix}1\\-1\\2\end{pmatrix}$ and $\mathbf{v}_2 = \begin{pmatrix}0\\3\\1\end{pmatrix}$, the combination $2\mathbf{v}_1 - \mathbf{v}_2$ gives:
\[ 2\begin{pmatrix}1\\-1\\2\end{pmatrix} - \begin{pmatrix}0\\3\\1\end{pmatrix} = \begin{pmatrix}2\\-2\\4\end{pmatrix} + \begin{pmatrix}0\\-3\\-1\end{pmatrix} = \begin{pmatrix}2\\-5\\3\end{pmatrix} \]
Vector Equations
The equation
\[ x_1\mathbf{v}_1 + x_2\mathbf{v}_2 + \cdots + x_p\mathbf{v}_p = \mathbf{b} \]
is a vector equation with unknowns $x_1, x_2, \ldots, x_p$. A solution is a list of scalars making the left side equal the right side.
Key connection: A vector equation is equivalent to a system of linear equations. Each component of the vector equation gives one scalar equation. This means vector equations can be solved by row reduction.
Source: Lay, Linear Algebra and Its Applications, Section 1.3: Vector Equations.
The Span
Definition: The span of a set of vectors $\{\mathbf{v}_1, \ldots, \mathbf{v}_p\}$ in $\mathbb{R}^n$ is the collection of all possible linear combinations:
\[ \text{Span}\{\mathbf{v}_1, \ldots, \mathbf{v}_p\} = \{c_1\mathbf{v}_1 + \cdots + c_p\mathbf{v}_p \mid c_1, \ldots, c_p \in \mathbb{R}\} \]
The span is a set of vectors (a subset of $\mathbb{R}^n$), not a single vector.
Geometric cases in $\mathbb{R}^3$:
| Vectors | Their Span |
|---|---|
| One nonzero vector $\{\mathbf{v}\}$ | A line through the origin in the direction of $\mathbf{v}$ |
| Two vectors, not parallel, not zero | A plane through the origin |
| Two parallel (or one zero) vectors | A line (same as one vector) |
| Three vectors, none a combo of others | All of $\mathbb{R}^3$ |
Span always contains the zero vector (take all scalars equal to $0$).
The Row Reduction Test
Question: Is $\mathbf{b}$ in $\text{Span}\{\mathbf{v}_1, \ldots, \mathbf{v}_p\}$?
Answer: Asking whether $\mathbf{b}$ is a linear combination of $\mathbf{v}_1, \ldots, \mathbf{v}_p$ is exactly asking whether the vector equation $x_1\mathbf{v}_1 + \cdots + x_p\mathbf{v}_p = \mathbf{b}$ has a solution. This is the same as asking whether the augmented matrix
\[ [\mathbf{v}_1 \; \mathbf{v}_2 \; \cdots \; \mathbf{v}_p \mid \mathbf{b}] \]
is consistent.
- Consistent: $\mathbf{b}$ is in the span.
- Inconsistent: $\mathbf{b}$ is not in the span.
Worked Example: Testing Span Membership
Is $\mathbf{b} = \begin{pmatrix}7\\4\\-5\end{pmatrix}$ in $\text{Span}\left\{\mathbf{v}_1 = \begin{pmatrix}1\\-2\\0\end{pmatrix},\; \mathbf{v}_2 = \begin{pmatrix}2\\1\\-3\end{pmatrix}\right\}$?
Form the augmented matrix: \[ \left[\begin{array}{rr|r} 1 & 2 & 7 \\ -2 & 1 & 4 \\ 0 & -3 & -5 \end{array}\right] \]
Row reduce:
$R_2 \to R_2 + 2R_1$: \[ \left[\begin{array}{rr|r} 1 & 2 & 7 \\ 0 & 5 & 18 \\ 0 & -3 & -5 \end{array}\right] \]
$R_2 \to \frac{1}{5}R_2$: \[ \left[\begin{array}{rr|r} 1 & 2 & 7 \\ 0 & 1 & 18/5 \\ 0 & -3 & -5 \end{array}\right] \]
$R_3 \to R_3 + 3R_2$: \[ \left[\begin{array}{rr|r} 1 & 2 & 7 \\ 0 & 1 & 18/5 \\ 0 & 0 & -5 + 54/5 \end{array}\right] = \left[\begin{array}{rr|r} 1 & 2 & 7 \\ 0 & 1 & 18/5 \\ 0 & 0 & 29/5 \end{array}\right] \]
Row 3 reads $0 = \frac{29}{5}$: inconsistent.
Conclusion: $\mathbf{b}$ is NOT in $\text{Span}\{\mathbf{v}_1, \mathbf{v}_2\}$.
Common Pitfalls
| Mistake | Why It Is Wrong | Correct Approach |
|---|---|---|
| Confusing the span with the list of given vectors | The span is infinitely large: all linear combinations, not just the original vectors | Span includes every scaling and every sum of the given vectors |
| Assuming two vectors in $\mathbb{R}^2$ always span all of $\mathbb{R}^2$ | If both vectors point in the same direction, their span is only a line | Check whether the augmented column becomes a pivot column |
| Treating the span as a single vector | The span is a set of vectors, the targets reachable from the given set | Write “the span contains...” or “the span is the set of all...” |
| Forgetting that the zero vector is always in the span | Setting all scalars to $0$ gives the zero vector | Any span includes $\mathbf{0}$ |
| Solving the vector equation without setting up the augmented matrix correctly | The columns of the matrix are the given vectors, in order | Place $\mathbf{v}_j$ as column $j$; place $\mathbf{b}$ as the augmented column |
When Two Vectors Span Only a Line
Consider $\mathbf{v}_1 = \begin{pmatrix}1\\2\end{pmatrix}$ and $\mathbf{v}_2 = \begin{pmatrix}2\\4\end{pmatrix}$.
The second vector is $2\mathbf{v}_1$. Every linear combination is:
\[ c_1\begin{pmatrix}1\\2\end{pmatrix} + c_2\begin{pmatrix}2\\4\end{pmatrix} = (c_1 + 2c_2)\begin{pmatrix}1\\2\end{pmatrix} \]
No matter what $c_1$ and $c_2$ you choose, the result is always a multiple of $\begin{pmatrix}1\\2\end{pmatrix}$. The span is the line through the origin with slope 2, not all of $\mathbb{R}^2$.
To reach $\begin{pmatrix}3\\1\end{pmatrix}$, you would need the augmented matrix $\left[\begin{array}{rr|r}1&2&3\\2&4&1\end{array}\right]$ to be consistent. After $R_2 \to R_2 - 2R_1$: row 2 becomes $[0, 0, -5]$. Inconsistent: $\begin{pmatrix}3\\1\end{pmatrix}$ is not in the span.
Practice Problems
Compute $3\mathbf{u} - 2\mathbf{v}$ where $\mathbf{u} = \begin{pmatrix}1\\-2\\0\end{pmatrix}$ and $\mathbf{v} = \begin{pmatrix}3\\1\\-4\end{pmatrix}$.
Is your result a linear combination of $\mathbf{u}$ and $\mathbf{v}$? Justify.
Rewrite the following system as a single vector equation $x_1\mathbf{v}_1 + x_2\mathbf{v}_2 = \mathbf{b}$:
\[ \begin{cases} 3x_1 - x_2 = 5 \\ x_1 + 2x_2 = -1 \end{cases} \]
Also write the augmented matrix you would row-reduce to test whether $\mathbf{b}$ is in $\text{Span}\{\mathbf{v}_1, \mathbf{v}_2\}$.
Is $\mathbf{b} = \begin{pmatrix}3\\-1\\0\end{pmatrix}$ in $\text{Span}\left\{\begin{pmatrix}1\\0\\-1\end{pmatrix}, \begin{pmatrix}0\\1\\2\end{pmatrix}\right\}$?
Set up the augmented matrix, row-reduce, and state your conclusion with a complete justification.
Let $\mathbf{v}_1 = \begin{pmatrix}1\\2\\3\end{pmatrix}$ and $\mathbf{v}_2 = \begin{pmatrix}-2\\-4\\-6\end{pmatrix}$.
(a) Describe $\text{Span}\{\mathbf{v}_1, \mathbf{v}_2\}$ geometrically. What is this set?
(b) Is $\begin{pmatrix}2\\4\\6\end{pmatrix}$ in the span? Is $\begin{pmatrix}1\\2\\4\end{pmatrix}$ in the span? For each, justify your answer without fully row-reducing.
For which value(s) of $h$ is $\mathbf{b} = \begin{pmatrix}-2\\h\\4\end{pmatrix}$ in $\text{Span}\left\{\mathbf{v}_1 = \begin{pmatrix}1\\-1\\2\end{pmatrix},\; \mathbf{v}_2 = \begin{pmatrix}0\\1\\3\end{pmatrix}\right\}$?
Set up and fully row-reduce the augmented matrix, treating $h$ as a symbol, and state the condition on $h$.
Common Misconceptions
the span of a set of vectors is just the set of those vectors. The span is the set of all linear combinations, which is in general an infinite collection. Even from a single nonzero vector, the span contains infinitely many vectors: every scalar multiple. The original vectors are in the span, but they are far from the only members.
adding more vectors to a set always makes the span larger. If the new vector is already a linear combination of the existing ones, it adds nothing to the span. For example, adding $\begin{pmatrix}2\\4\end{pmatrix}$ to the set $\left\{\begin{pmatrix}1\\2\end{pmatrix}\right\}$ does not enlarge the span because $\begin{pmatrix}2\\4\end{pmatrix} = 2\begin{pmatrix}1\\2\end{pmatrix}$ is already reachable.
$n$ vectors in $\mathbb{R}^n$ must span all of $\mathbb{R}^n$. Counting vectors does not guarantee span coverage. If several vectors point in the same or linearly related directions, they cover a lower-dimensional subset. For example, any number of multiples of a single vector in $\mathbb{R}^3$ span only a line, not $\mathbb{R}^3$.
Mastery Checklist
Novice (Level 1-2):
Competent (Level 3-4):
Proficient (Level 5):
Mental Model
The Mixing Console
Imagine each vector as a “channel” on a mixing console, pointing in some direction in space. The scalars are the volume knobs. The span is everything you can produce by turning the knobs: every blend, every level, positive and negative.
If two channels point in the same direction, turning both up or down just makes the sound louder or softer in that one direction. You cannot add any new directions no matter how you adjust the knobs. That is why two parallel vectors span only a line.
If two channels point in genuinely different directions, you can blend them to reach any direction in the plane they define. And if three vectors cover three independent directions in $\mathbb{R}^3$, you can reach any point in $\mathbb{R}^3$.
The row reduction test is the precise way to answer: “does this particular target fall within reach of my current set of channels?”
Connections
Looking ahead in MATH 231:
- The Matrix Equation $Ax = \mathbf{b}$: multiplying $A$ by $\mathbf{x}$ is exactly a linear combination of the columns of $A$; the column space of $A$ is the span of those columns
- Solution Sets and Free Variables: the parametric form of a solution set is described using span
- Null Space, Column Space, and Basis: the column space of a matrix is the span of its columns, a central object in the theory
Real-world connections:
- In data science, a data matrix’s column space is the span of the feature vectors; regression asks whether the response is in that span.
- In signal processing, a basis for a signal space is a spanning set; decomposing a signal means writing it as a linear combination of basis elements.
Source: Lay, Linear Algebra and Its Applications, Section 1.3: Vector Equations.
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|---|---|---|
| Systems and Row Reduction | MATH231 Skills Index | Matrix Equation Ax = b |
Last updated: 2026-06-16
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