4.5: Spanning Sets in Rⁿ

Learning Objectives

• Determine the span of a set of vectors.
• Determine if a vector is contained in a specified span.

By generating all linear combinations of a set of vectors one can obtain various subsets of \(\mathbb{R}^{n}\) which we call subspaces. For example what set of vectors in \(\mathbb{R}^{3}\) generate the \(XY\)-plane? What is the smallest such set of vectors can you find? The tools of spanning, linear independence and basis are exactly what is needed to answer these and similar questions and are the focus of this section. The following definition is essential.

Definition \(\PageIndex{1}\): Subset

Let \(U\) and \(W\) be sets of vectors in \(\mathbb{R}^n\). If all vectors in \(U\) are also in \(W\), we say that \(U\) is a subset of \(W\), denoted \[U \subseteq W\nonumber \]

We begin this section with a definition.

Definition \(\PageIndex{2}\): Span of a Set of Vectors

The collection of all linear combinations of a set of vectors \(\{ \vec{u}_1, \cdots ,\vec{u}_k\}\) in \(\mathbb{R}^{n}\) is known as the span of these vectors and is written as \(\mathrm{span} \{\vec{u}_1, \cdots , \vec{u}_k\}\).

Consider the following example.

Example \(\PageIndex{1}\): Span of Vectors

Describe the span of the vectors \(\vec{u}=\left[ \begin{array}{rrr} 1 & 1 & 0 \end{array} \right]^T\) and \(\vec{v}=\left[ \begin{array}{rrr} 3 & 2 & 0 \end{array} \right]^T \in \mathbb{R}^{3}\).

Solution

You can see that any linear combination of the vectors \(\vec{u}\) and \(\vec{v}\) yields a vector of the form \(\left[ \begin{array}{rrr} x & y & 0 \end{array} \right]^T\) in the \(XY\)-plane.

Moreover every vector in the \(XY\)-plane is in fact such a linear combination of the vectors \(\vec{u}\) and \(\vec{v}\). That’s because \[\left[ \begin{array}{r} x \\ y \\ 0 \end{array} \right] = (-2x+3y) \left[ \begin{array}{r} 1 \\ 1 \\ 0 \end{array} \right] + (x-y)\left[ \begin{array}{r} 3 \\ 2 \\ 0 \end{array} \right]\nonumber \]

Thus \(\mathrm{span}\{\vec{u},\vec{v}\}\) is precisely the \(XY\)-plane. This concept is explained below in a interactive video format. Feel free to click on it and see the theory in action.

You can convince yourself that no single vector can span the \(XY\)-plane. In fact, take a moment to consider what is meant by the span of a single vector.

However you can make the set larger if you wish. For example consider the larger set of vectors \(\{ \vec{u}, \vec{v}, \vec{w}\}\) where \(\vec{w}=\left[ \begin{array}{rrr} 4 & 5 & 0 \end{array} \right]^T\). Since the first two vectors already span the entire \(XY\)-plane, the span is once again precisely the \(XY\)-plane and nothing has been gained. Of course if you add a new vector such as \(\vec{w}=\left[ \begin{array}{rrr} 0 & 0 & 1 \end{array} \right]^T\) then it does span a different space. What is the span of \(\vec{u}, \vec{v}, \vec{w}\) in this case?

The distinction between the sets \(\{ \vec{u}, \vec{v}\}\) and \(\{ \vec{u}, \vec{v}, \vec{w}\}\) will be made using the concept of linear independence.

Consider the vectors \(\vec{u}, \vec{v}\), and \(\vec{w}\) discussed above. In the next example, we will show how to formally demonstrate that \(\vec{w}\) is in the span of \(\vec{u}\) and \(\vec{v}\).

Example \(\PageIndex{3}\): Vector in a Span

Let \(\vec{u}=\left[ \begin{array}{rrr} 1 & 1 & 0 \end{array} \right]^T\) and \(\vec{v}=\left[ \begin{array}{rrr} 3 & 2 & 0 \end{array} \right]^T \in \mathbb{R}^{3}\). Show that \(\vec{w} = \left[ \begin{array}{rrr} 4 & 5 & 0 \end{array} \right]^{T}\) is in \(\mathrm{span} \left\{ \vec{u}, \vec{v} \right\}\).

Solution

For a vector to be in \(\mathrm{span} \left\{ \vec{u}, \vec{v} \right\}\), it must be a linear combination of these vectors. If \(\vec{w} \in \mathrm{span} \left\{ \vec{u}, \vec{v} \right\}\), we must be able to find scalars \(a,b\) such that\[\vec{w} = a \vec{u} +b \vec{v}\nonumber \]

We proceed as follows. \[\left[ \begin{array}{r} 4 \\ 5 \\ 0 \end{array} \right] = a \left[ \begin{array}{r} 1 \\ 1 \\ 0 \end{array} \right] + b \left[ \begin{array}{r} 3 \\ 2 \\ 0 \end{array} \right]\nonumber \] This is equivalent to the following system of equations \[\begin{aligned} a + 3b &= 4 \\ a + 2b &= 5\end{aligned}\]

We solving this system the usual way, constructing the augmented matrix and row reducing to find the reduced row-echelon form. \[\left[ \begin{array}{rr|r} 1 & 3 & 4 \\ 1 & 2 & 5 \end{array} \right] \rightarrow \cdots \rightarrow \left[ \begin{array}{rr|r} 1 & 0 & 7 \\ 0 & 1 & -1 \end{array} \right]\nonumber \] The solution is \(a=7, b=-1\). This means that \[\vec{w} = 7 \vec{u} - \vec{v}\nonumber \] Therefore we can say that \(\vec{w}\) is in \(\mathrm{span} \left\{ \vec{u}, \vec{v} \right\}\).

Example \(\PageIndex{4}\)

Let \(V = \big\{ \langle 5, -6, -1 \rangle, \langle 1, 4, 5 \rangle \big\} \). Which of the following vectors are in the space spanned by \(V\) ?

• \( \langle 5, -5, -3 \rangle \)
• \( \langle -1, -4, 5 \rangle \)
• \( \langle 2, 1, -4 \rangle \)
• \( \langle -5, -1, -1 \rangle \)
• \( \langle -5, 2, -3 \rangle \)

Solution

• Video Length: 7 minutes 41 seconds
• Context: This video demonstrates how to determine which given vectors lie in the span of a set of vectors by checking if they can be expressed as linear combinations of the spanning set.