The span of a set of vectors

Section 2.3 The span of a set of vectors

Activity 2.3.0.1.

If the matrix equation \(A\xvec = \bvec\) is inconsistent, give the statement that best describes the pivots of the augmented matrix \(\left[\begin{array}{r|r} A \amp \bvec \end{array}\right]\text{.}\)

There must be a pivot in the last row of the augmented matrix.
There must be a pivot in the last column of the augmented matrix.
The augmented matrix must be the identity matrix.
We can't determine anything about the pivots without the specific matrix.

Activity 2.3.0.2.

Consider the matrix \(A\)

\begin{equation*} A = \left[ \begin{array}{rrr} 1 \amp 0 \amp -2 \\ -2 \amp 2 \amp 2 \\ 1 \amp 1 \amp -3 \end{array}\right]\text{.} \end{equation*}

(a)

If \(\bvec=\threevec{2}{2}{5}\text{,}\) is the equation \(A\xvec = \bvec\) consistent? If so, find a solution.

(b)

If \(\bvec=\threevec{2}{2}{6}\text{,}\) is the equation \(A\xvec = \bvec\) consistent? If so, find a solution.

(c)

Identify the pivot positions of \(A\text{.}\)

(d)

For our two choices of the vector \(\bvec\text{,}\) one equation \(A\xvec = \bvec\) has a solution and the other does not. What feature of the pivot positions of the matrix \(A\) tells us to expect this?

There is a row of the coefficient matrix with no pivot position.
There is column of the coefficient matrix with no pivot position.
The coefficient matrix is not the identity matrix.
We can't determine anything about the pivot positions without the specific matrix.

Definition 2.3.0.1.

The span of a set of vectors \(\vvec_1,\vvec_2,\ldots,\vvec_n\) is the set of all linear combinations of the vectors.

In other words, the span of \(\vvec_1,\vvec_2,\ldots,\vvec_n\) consists of all the vectors \(\bvec\) for which the equation

\begin{equation*} \left[\begin{array}{rrrr} \vvec_1\amp\vvec_2\amp\ldots\amp\vvec_n \end{array}\right] \xvec = \bvec \end{equation*}

is consistent.

Let's look at two examples to develop some intuition for the concept of span.

Activity 2.3.0.3.

Consider the set of vectors

\begin{equation*} \vvec = \twovec{1}{2}, \wvec = \twovec{-2}{-4}\text{.} \end{equation*}

Go to the Applet in Activity 2.3.2 ³, part (a) in Understanding Linear Algebra.

(a)

What vector is the linear combination of \(\vvec\) and \(\wvec\) with weights:

\(a = 2\) and \(b=0\text{?}\)
\(a = 1\) and \(b=1\text{?}\)
\(a = 0\) and \(b=-1\text{?}\)

(b)

Can the vector \(\twovec{2}{4}\) be expressed as a linear combination of \(\vvec\) and \(\wvec\text{?}\) Is the vector \(\twovec{2}{4}\) in the span of \(\vvec\) and \(\wvec\text{?}\)

(c)

Can the vector \(\twovec{3}{0}\) be expressed as a linear combination of \(\vvec\) and \(\wvec\text{?}\) Is the vector \(\twovec{3}{0}\) in the span of \(\vvec\) and \(\wvec\text{?}\)

(d)

Describe the set of vectors in the span of \(\vvec\) and \(\wvec\text{.}\)

(e)

For what vectors \(\bvec\) does the equation

\begin{equation*} \left[\begin{array}{rr} 1 \amp -2 \\ 2 \amp -4 \end{array}\right] \xvec = \bvec \end{equation*}

have a solution?

Activity 2.3.0.4.

We will now look at an example where

\begin{equation*} \vvec = \twovec{2}{1}, \wvec = \twovec{1}{2}\text{.} \end{equation*}

Go to the Applet in Activity 2.3.2 ⁴, part (b) in Understanding Linear Algebra.

(a)

What vector is the linear combination of \(\vvec\) and \(\wvec\) with weights:

\(a = 2\) and \(b=0\text{?}\)
\(a = 1\) and \(b=1\text{?}\)
\(a = 0\) and \(b=-1\text{?}\)

(b)

Can the vector \(\twovec{-2}{2}\) be expressed as a linear combination of \(\vvec\) and \(\wvec\text{?}\) Is the vector \(\twovec{-2}{2}\) in the span of \(\vvec\) and \(\wvec\text{?}\)

(c)

Can the vector \(\twovec{3}{0}\) be expressed as a linear combination of \(\vvec\) and \(\wvec\text{?}\) Is the vector \(\twovec{3}{0}\) in the span of \(\vvec\) and \(\wvec\text{?}\)

(d)

Describe the set of vectors in the span of \(\vvec\) and \(\wvec\text{.}\)

(e)

For what vectors \(\bvec\) does the equation

\begin{equation*} \left[\begin{array}{rr} 2 \amp 1 \\ 1 \amp 2 \end{array}\right] \xvec = \bvec \end{equation*}

have a solution?

In the next couple of activities, we will look at the span of sets of vectors in \(\real^3\text{.}\)

Activity 2.3.0.5.

In this activity, we will look at the span of a set of vectors in \(\real^3\text{.}\)

(a)

Suppose \(\vvec=\threevec{1}{2}{1}\text{.}\) Give a written description of \(\laspan{\vvec}\) and a rough sketch of it.

(b)

Consider now the two vectors

\begin{equation*} \evec_1 = \threevec{1}{0}{0}, \evec_2 = \threevec{0}{1}{0}\text{.} \end{equation*}

Sketch the vectors. Then give a written description of \(\laspan{\evec_1,\evec_2}\) and a rough sketch of it.

(c)

Let's now look at this algebraically by writing write \(\bvec = \threevec{b_1}{b_2}{b_3}\text{.}\) Determine the conditions on \(b_1\text{,}\) \(b_2\text{,}\) and \(b_3\) so that \(\bvec\) is in \(\laspan{\evec_1,\evec_2}\) by considering the linear system

\begin{equation*} \left[\begin{array}{rr} \evec_1 \amp \evec_2 \\ \end{array}\right] \xvec = \bvec \end{equation*}

\begin{equation*} \left[\begin{array}{rr} 1 \amp 0 \\ 0 \amp 1 \\ 0 \amp 0 \\ \end{array}\right] \xvec = \threevec{b_1}{b_2}{b_3}\text{.} \end{equation*}

Be able to explain how this relates to your sketch of \(\laspan{\evec_1,\evec_2}\text{.}\)

Activity 2.3.0.6.

Consider the vectors

\begin{equation*} \vvec_1 = \threevec{1}{1}{-1}, \vvec_2 = \threevec{0}{2}{1}\text{.} \end{equation*}

(a)

Is the vector \(\bvec=\threevec{1}{-2}{4}\) in \(\laspan{\vvec_1,\vvec_2}\text{?}\)

(b)

Is the vector \(\bvec=\threevec{-2}{0}{3}\) in \(\laspan{\vvec_1,\vvec_2}\text{?}\)

(c)

Which of the following is the most appropriate description of \(\laspan{\vvec_1,\vvec_2}\text{?}\)

\(\laspan{\vvec_1,\vvec_2}\) is \(\real^2\text{.}\)
\(\laspan{\vvec_1,\vvec_2}\) is \(\real^3\text{.}\)
\(\laspan{\vvec_1,\vvec_2}\) is a plane in \(real^3\text{.}\)
\(\laspan{\vvec_1,\vvec_2}\) is a line in \(real^3\text{.}\)

Activity 2.3.0.7.

Consider the vectors

\begin{equation*} \vvec_1 = \threevec{1}{1}{-1}, \vvec_2 = \threevec{0}{2}{1}, \vvec_3 = \threevec{1}{-2}{4}\text{.} \end{equation*}

(a)

Form the matrix \(\left[\begin{array}{rrrr} \vvec_1 \amp \vvec_2 \amp \vvec_3 \end{array}\right]\) and find its reduced row echelon form.

(b)

What does the reduced row echelon form of the above matrix tell you about \(\laspan{\vvec_1,\vvec_2,\vvec_3}\text{?}\)

\(\laspan{\vvec_1,\vvec_2,\vvec_3}=\laspan{\vvec_1,\vvec_2}\text{.}\)
\(\laspan{\vvec_1,\vvec_2,\vvec_3}=\laspan{\vvec_2,\vvec_3}\text{.}\)
\(\laspan{\vvec_1,\vvec_2,\vvec_3}=\laspan{\vvec_1}\text{.}\)
\(\laspan{\vvec_1,\vvec_2,\vvec_3}=\real^3\text{.}\)

Activity 2.3.0.8.

What can we say more generally about when the span of a set of vectors is all of \(\real^3\text{?}\)

(a)

If a set of vectors \(\vvec_1,\vvec_2,\ldots,\vvec_n\) spans \(\real^3\text{,}\) what can you say about the pivots of the matrix \(\left[\begin{array}{rrrr} \vvec_1\amp\vvec_2\amp\ldots\amp\vvec_n \end{array}\right]\text{?}\)

(b)

What is the smallest number of vectors such that \(\laspan{\vvec_1,\vvec_2,\ldots,\vvec_n} = \real^3\text{?}\)

Activity 2.3.0.9.

The vector \(\left[\begin{array}{c}-1\\-6\\1\end{array}\right]\) belongs to \(\vspan\left\{\left[\begin{array}{c}1\\0\\-3\end{array}\right], \left[\begin{array}{c}-1\\-3\\2\end{array}\right]\right\}\) exactly when there exists a solution to the vector equation \(x_1\left[\begin{array}{c}1\\0\\-3\end{array}\right]+ x_2\left[\begin{array}{c}-1\\-3\\2\end{array}\right] =\left[\begin{array}{c}-1\\-6\\1\end{array}\right]\text{.}\)

(a)

Reinterpret this vector equation as a system of linear equations.

(b)

Find its solution set, using technology to find \(\RREF\) of its corresponding augmented matrix.

(c)

Given this solution set, does \(\left[\begin{array}{c}-1\\-6\\1\end{array}\right]\) belong to \(\vspan\left\{\left[\begin{array}{c}1\\0\\-3\end{array}\right], \left[\begin{array}{c}-1\\-3\\2\end{array}\right]\right\}\text{?}\)

Activity 2.3.0.10.

Determine if \(\left[\begin{array}{c}-1\\-9\\0\end{array}\right]\) belongs to \(\vspan\left\{\left[\begin{array}{c}1\\0\\-3\end{array}\right], \left[\begin{array}{c}-1\\-3\\2\end{array}\right]\right\}\) by solving an appropriate vector equation.

davidaustinm.github.io/ula/sec-span.html#activity-intro-span