Learning Outcomes
- Determine if a set of Euclidean vectors spans \(\IR^m\) by solving appropriate vector equations.
Section 19.2 Spanning Sets (EV2)
Subsection 19.2.1 Warm Up
Activity 19.16.
Given a set of ingredients and a meal, a recipe is a list of amounts of each ingredient required to prepare the given meal.
(a)
Use the words vector and linear combination to create a new statement that is analogous to one above.
(b)
Building on your analogy, what role might the word span play?
Subsection 19.2.2 Class Activities
Observation 19.17.
Any single non-zero vector/number \(x\) in \(\IR^1\) spans \(\IR^1\text{,}\) since \(\IR^1=\setBuilder{cx}{c\in\IR}\text{.}\)
Activity 19.18.
How many vectors are required to span \(\IR^2\text{?}\) Sketch a drawing in the \(xy\) plane to support your answer.
- \(\displaystyle 1\)
- \(\displaystyle 2\)
- \(\displaystyle 3\)
- \(\displaystyle 4\)
- Infinitely Many
Activity 19.19.
How many vectors are required to span \(\IR^3\text{?}\)
- \(\displaystyle 1\)
- \(\displaystyle 2\)
- \(\displaystyle 3\)
- \(\displaystyle 4\)
- Infinitely Many
Fact 19.20.
At least \(m\) vectors are required to span \(\IR^m\text{.}\)
Activity 19.21.
Consider the question: Does every vector in \(\IR^3\) belong to \(\vspan\left\{\left[\begin{array}{c}1\\-1\\0\end{array}\right],
\left[\begin{array}{c}-2\\0\\1\end{array}\right],\left[\begin{array}{c}-2\\-2\\2\end{array}\right]\right\}\text{?}\)
(a)
Determine if \(\left[\begin{array}{c} 7 \\ -3 \\ -2 \end{array}\right]\) belongs to \(\vspan\left\{\left[\begin{array}{c}1\\-1\\0\end{array}\right],
\left[\begin{array}{c}-2\\0\\1\end{array}\right],\left[\begin{array}{c}-2\\-2\\2\end{array}\right]\right\}\text{.}\)
Answer.
The vector belongs to the span.
(b)
Given this result, do we now know whether every vector in \(\mathbb R^3\) belongs to \(\vspan\left\{\left[\begin{array}{c}1\\-1\\0\end{array}\right],
\left[\begin{array}{c}-2\\0\\1\end{array}\right],\left[\begin{array}{c}-2\\-2\\2\end{array}\right]\right\}\text{?}\)
Answer.
No; we’ve checked one vector, but that doesn’t mean some other vector can’t fail to belong.
(c)
Determine if \(\left[\begin{array}{c} 0 \\ -4 \\ 3 \end{array}\right]\) belongs to \(\vspan\left\{\left[\begin{array}{c}1\\-1\\0\end{array}\right],
\left[\begin{array}{c}-2\\0\\1\end{array}\right],\left[\begin{array}{c}-2\\-2\\2\end{array}\right]\right\}\text{.}\)
Answer.
The vector belongs to the span.
(d)
Determine if \(\left[\begin{array}{c} 2 \\ 5 \\ 7 \end{array}\right]\) belongs to \(\vspan\left\{\left[\begin{array}{c}1\\-1\\0\end{array}\right],
\left[\begin{array}{c}-2\\0\\1\end{array}\right],\left[\begin{array}{c}-2\\-2\\2\end{array}\right]\right\}\text{.}\)
Answer.
The vector does not belong to the span.
(e)
Fix the SageMath code below to visualize \(\vspan\left\{\left[\begin{array}{c}1\\-1\\0\end{array}\right],
\left[\begin{array}{c}-2\\0\\1\end{array}\right],\left[\begin{array}{c}-2\\-2\\2\end{array}\right]\right\}\text{.}\)
Answer.
v2 = vector([-2,0,1])
v3 = vector([-2,-2,2])
# ...
linear_combo = a*v1 + b*v2 + c*v3
Activity 19.22.
We’d prefer a more methodical method to decide if every vector in \(\IR^n\) belongs to some spanning set, compared to the guess-and-check methods we used in Activity 19.21.
(a)
An arbitrary vector \(\left[\begin{array}{c}\unknown\\\unknown\\\unknown\end{array}\right]\) belongs to \(\vspan\left\{\left[\begin{array}{c}1\\-1\\0\end{array}\right],
\left[\begin{array}{c}-2\\0\\1\end{array}\right],\left[\begin{array}{c}-2\\-2\\2\end{array}\right]\right\}\) provided the equation
\begin{equation*}
x_1\left[\begin{array}{c}1\\-1\\0\end{array}\right]+
x_2\left[\begin{array}{c}-2\\0\\1\end{array}\right]+
x_3\left[\begin{array}{c}-2\\-2\\2\end{array}\right]=\left[\begin{array}{c}\unknown\\\unknown\\\unknown\end{array}\right]
\end{equation*}
has...
- no solutions.
- exactly one solution.
- at least one solution.
- infinitely-many solutions.
Answer.
A.
(b)
We’re guaranteed at least one solution if the RREF of the corresponding augmented matrix has no contradictions; likewise, we have no solutions if the RREF corresponds to the contradiction \(0=1\text{.}\) Given
\begin{equation*}
\left[\begin{array}{ccc|c}1&-2&-2&\unknown\\-1&0&-2&\unknown\\0&1&2&\unknown\end{array}\right]\sim
\left[\begin{array}{ccc|c}1&0&2&\unknown\\0&1&2&\unknown\\0&0&0&\unknown\end{array}\right]
\end{equation*}
we may conclude that the set does not span all of \(\IR^3\) because...
- the row \([0\,1\,2\,|\,\unknown]\) prevents a contradiction.
- the row \([0\,1\,2\,|\,\unknown]\) allows a contradiction.
- the row \([0\,0\,0\,|\,\unknown]\) prevents a contradiction.
- the row \([0\,0\,0\,|\,\unknown]\) allows a contradiction.
Answer.
D.
Fact 19.23.
The set \(\{\vec v_1,\dots,\vec v_n\}\) spans all of \(\IR^m\) exactly when the vector equation
\begin{equation*}
x_1 \vec{v}_1 + \cdots + x_n\vec{v}_n = \vec{w}
\end{equation*}
is consistent for every vector \(\vec{w}\in\IR^m\text{.}\)
Likewise, the set \(\{\vec v_1,\dots,\vec v_n\}\) fails to span all of \(\IR^m\) exactly when the vector equation
\begin{equation*}
x_1 \vec{v}_1 + \cdots + x_n\vec{v}_n = \vec{w}
\end{equation*}
is inconsistent for some vector \(\vec{w}\in\IR^m\text{.}\)
Note these two possibilities are decided based on whether or not the RREF of the vector equation’s coefficient matrix (that is, \(\RREF[\vec v_1\,\dots\,\vec v_n]\)) has either all pivot rows, or at least one non-pivot row (a row of zeroes):
\begin{equation*}
\left[\begin{array}{ccc|c}1&-2&-2\\-1&0&-2\\0&1&2\end{array}\right]\sim
\left[\begin{array}{ccc|c}1&0&2\\0&1&2\\0&0&0\end{array}\right]\text{.}
\end{equation*}
Activity 19.24.
Consider the set of vectors \(S=\left\{
\left[\begin{array}{c}2\\3\\0\\-1\end{array}\right],
\left[\begin{array}{c}1\\-4\\3\\0\end{array}\right],
\left[\begin{array}{c}1\\7\\-3\\-1\end{array}\right],
\left[\begin{array}{c}0\\3\\5\\7\end{array}\right],
\left[\begin{array}{c}3\\13\\7\\16\end{array}\right]
\right\}\) and the question “Does \(\IR^4=\vspan S\text{?}\)”
(a)
Rewrite this question in terms of the solutions to a vector equation.
(b)
Answer your new question, and use this to answer the original question.
Activity 19.25.
Let \(\vec{v}_1, \vec{v}_2, \vec{v}_3 \in \IR^7\) be three Euclidean vectors, and suppose \(\vec{w}\) is another vector with \(\vec{w} \in \vspan \left\{ \vec{v}_1, \vec{v}_2, \vec{v}_3 \right\}\text{.}\) What can you conclude about \(\vspan \left\{ \vec{w}, \vec{v}_1, \vec{v}_2, \vec{v}_3 \right\} \text{?}\)
- \(\vspan \left\{ \vec{w}, \vec{v}_1, \vec{v}_2, \vec{v}_3 \right\} \) is larger than \(\vspan \left\{ \vec{v}_1, \vec{v}_2, \vec{v}_3 \right\} \text{.}\)
- \(\vspan \left\{ \vec{w}, \vec{v}_1, \vec{v}_2, \vec{v}_3 \right\} \) is the same as \(\vspan \left\{ \vec{v}_1, \vec{v}_2, \vec{v}_3 \right\} \text{.}\)
- \(\vspan \left\{ \vec{w}, \vec{v}_1, \vec{v}_2, \vec{v}_3 \right\} \) is smaller than \(\vspan \left\{ \vec{v}_1, \vec{v}_2, \vec{v}_3 \right\} \text{.}\)
Subsection 19.2.3 Individual Practice
Activity 19.26.
One of our important results in this lesson is Fact 19.20, which states that a set of \(n\) vectors is required to span \(\IR^n\text{.}\) While we developed some geometric intuition for why this true, we did not prove it in class. Before coming to class next time, follow the steps outlined below to convince yourself of this fact using the concepts we learned in this lesson.
(a)
Let \(\{\vec{v}_1,\dots, \vec{v}_m\}\) be a set of vectors living in \(\IR^n\) and assume that \(m <n\text{.}\) How many rows and how many columns will the matrix \([\vec{v}_1\cdots \vec{v}_m]\) have?
(b)
Given no additional information about the vectors \(\vec{v}_1,\dots, \vec{v}_m\text{,}\) what is the maximum possible number of pivots in \(\RREF[\vec v_1\,\dots\,\vec v_m]\text{?}\)
(c)
Conclude that our given set of vector cannot span all of \(\IR^n\text{.}\)
Subsection 19.2.4 Videos
Subsection 19.2.5 Exercises
Exercises available at
https://tbil.org/preview/linear-algebra/exercises/#/bank/EV2/.Subsection 19.2.6 Mathematical Writing Explorations
Exploration 19.27.
Construct each of the following, or show that it is impossible:
- A set of 2 vectors that spans \(\mathbb{R}^3\)
- A set of 3 vectors that spans \(\mathbb{R}^3\)
- A set of 3 vectors that does not span \(\mathbb{R}^3\)
- A set of 4 vectors that spans \(\mathbb{R}^3\)
For any of the sets you constructed that did span the required vector space, are any of the vectors a linear combination of the others in your set?
Exploration 19.28.
Based on these results, generalize this a conjecture about how a set of \(n-1, n\) and \(n+1\) vectors would or would not span \(\mathbb{R}^n\text{.}\)
Subsection 19.2.7 Sample Problem and Solution
Sample problem Example C.6.
