Understanding and Computing Vector Spans

Defining Span through Linear Combinations

The fundamental building block for the concept of span is the linear combination. You have likely encountered this idea before, but we will formalize it here.

A vector $\mathbf{w}$ is a linear combination of a set of vectors $\{{\mathbf{v}}_1,{\mathbf{v}}_2,\dots ,{\mathbf{v}}_k\}$ if it can be expressed as a sum of scalar multiples of those vectors:

where $c_1,c_2,\dots ,c_k$ are scalars.

The span is simply the set of all possible linear combinations of a set of vectors.

As defined in the text, for a set of vectors $S=\{{\mathbf{v}}_1,\dots ,{\mathbf{v}}_p\}$, their span is:

The Span of a Set is Always a Subspace

In the previous lesson, we used the Subspace Test to verify if a set defined by certain rules (e.g., solutions to $A\mathbf{x}=0$) formed a subspace. The concept of span gives us a direct way to construct a subspace.

Theorem: For any non-empty set of vectors $S$ in a vector space $V$, the span of $S$ is a subspace of $V$.

This is a fundamental result. It guarantees that any set we generate by taking all possible linear combinations of some starting vectors will automatically satisfy the three conditions of the Subspace Test.

The proof confirms that $\text{span}(S)$ is a subspace by showing:

1. Contains $0$: The zero vector can be formed by choosing all scalars to be zero.
2. Closure under Addition: The sum of any two linear combinations is itself a linear combination.
3. Closure under Scalar Multiplication: A scalar multiple of a linear combination is also a linear combination.

The Geometry of Span

Visualizing the span can build powerful intuition. In ${\mathbb{R}}^2$ or ${\mathbb{R}}^3$, the span of a set of vectors represents a geometric object that always passes through the origin.

To summarize the geometric view in ${\mathbb{R}}^3$:

• The span of a single non-zero vector $\mathbf{v}$ is the set of all scalar multiples $c\mathbf{v}$. This is the line passing through the origin and the point defined by $\mathbf{v}$.
• The span of two non-collinear vectors $\mathbf{u}$ and $\mathbf{v}$ is the set of all combinations $c_1\mathbf{u}+c_2\mathbf{v}$. This forms the plane passing through the origin and the points defined by $\mathbf{u}$ and $\mathbf{v}$.
• If we have three vectors that do not all lie in the same plane, their span can encompass all of ${\mathbb{R}}^3$.

This animation from Dr. Trefor Bazett provides another excellent visualization of a plane being generated by the span of two vectors.

How to Compute the Span

The most important practical skill related to span is determining whether a given vector $\mathbf{w}$ belongs to the span of a set of vectors $S=\{{\mathbf{v}}_1,\dots ,{\mathbf{v}}_k\}$.

Based on the definition, this question is equivalent to asking:
"Do there exist scalars $c_1,\dots ,c_k$ such that $c_1{\mathbf{v}}_1+c_2{\mathbf{v}}_2+\cdots +c_k{\mathbf{v}}_k=\mathbf{w}$?"

This vector equation can be rewritten as a system of linear equations. Let's see how.

If we are in ${\mathbb{R}}^m$, we can write the vectors ${\mathbf{v}}_i$ and $\mathbf{w}$ as columns. The vector equation becomes a matrix equation $A\mathbf{c}=\mathbf{w}$, where the columns of matrix $A$ are the vectors ${\mathbf{v}}_i$, $\mathbf{c}$ is the vector of unknown scalars, and $\mathbf{w}$ is the target vector.

The question of whether $\mathbf{w}$ is in the span of $S$ is now a question of whether the system $A\mathbf{c}=\mathbf{w}$ has a solution.

The process is purely algorithmic:

1. Construct an augmented matrix $[A|\mathbf{w}]$, where the columns of $A$ are the vectors in your set $S$ and $\mathbf{w}$ is the vector you are testing.
2. Perform row reduction to bring the matrix into row echelon form.
3. Check for consistency.
   • If the system is consistent (has at least one solution), then $\mathbf{w}$ is in the span of $S$. A solution provides the specific scalars for the linear combination.
   • If the system is inconsistent (e.g., a row like $[00\dots |1]$), then $\mathbf{w}$ is not in the span of $S$.

Practice Problem

Let's walk through an example from the exercises in the Runestone text.

Problem:
Suppose that S=\set{\colvector{2\\-1\\3\\4},\,\colvector{3\\2\\-2\\1}}. Let $W=\text{spn}S$. Is the vector $$\begin{gathered} \mathbf{x}=\text{colvector}5 \\ 8 \\ -12 \\ -5 \end{gathered}$$ in $W$?

Solution:
We are looking for scalars ${\alpha }_1$ and ${\alpha }_2$ such that:

This corresponds to the augmented matrix:

Row reducing this matrix yields:

This system is consistent and gives a unique solution: ${\alpha }_1=-2$ and ${\alpha }_2=3$. Since a solution exists, yes, $\mathbf{x}$ is in the span of $S$. We can verify this:

Now, try this one yourself. Using the same set $S$, is the vector $$\begin{gathered} \mathbf{y}=\text{colvector}5 \\ 1 \\ 3 \\ 5 \end{gathered}$$ in $\text{spn}S$? (You can find the detailed solution in Exercise C41 in the "SS Spanning Sets" resource LINK).

A Deeper Look at Span (Optional)

There is a more abstract, "top-down" way to define span that you may find interesting. Instead of building it from the "bottom-up" with linear combinations, we can define it as the intersection of all subspaces that contain the set $S$.

This perspective is powerful in more abstract proofs and highlights the robust, fundamental nature of the span in linear algebra.

Conclusion

This lesson introduced a core constructive tool in linear algebra. You now have two primary ways of thinking about subspaces: as sets that satisfy the three-part Subspace Test, and as sets that can be generated as the span of a collection of vectors.

Key Takeaways:

• A linear combination is a sum of scaled vectors.
• The span of a set of vectors $S$ is the set of all possible linear combinations of vectors from $S$.
• The span of any set of vectors is always a subspace.
• To determine if a vector $\mathbf{w}$ is in $\text{span}(S)$, you set up a vector equation that becomes a system of linear equations. If the system is consistent, the vector is in the span.
• Given your background in computer science, it's worth noting that checking for membership in a span is a purely algorithmic process. We will explore the implementation of algorithms like Gaussian elimination later in the course.

Preview of the next lesson:
We now know how to generate a subspace using a set of vectors. A natural question follows: are all the vectors in our spanning set necessary? For example, the span of $\{\left(\begin{array}{c}1\\ 0\end{array}\right),\left(\begin{array}{c}2\\ 0\end{array}\right)\}$ is just the x-axis, the same as the span of just $\{\left(\begin{array}{c}1\\ 0\end{array}\right)\}$. The second vector was redundant. This idea of redundancy, or lack thereof, leads directly to our next topic: linear independence.