Learning Outcomes
- Determine if a set of Euclidean vectors is linearly dependent or independent by solving an appropriate vector equation.
Section 2.4 Linear Independence (EV4)
Subsection 2.4.1 Warm Up
Activity 2.4.1.
Consider the vector equation
\begin{equation*}
x_1\left[\begin{array}{c}1\\1\\1\end{array}\right]+x_2\left[\begin{array}{c}2\\0\\-1\end{array}\right]+x_3\left[\begin{array}{c}-1\\2\\0\end{array}\right]=\left[\begin{array}{c}-1\\7\\4\end{array}\right].
\end{equation*}
(a)
Decide which of \(\left[\begin{array}{c}3\\-1\\2\end{array}\right]\) or \(\left[\begin{array}{c}1\\1\\1\end{array}\right]\) is a solution vector.
(b)
Consider now the following vector equation:
\begin{equation*}
y_1\left[\begin{array}{c}1\\1\\1\end{array}\right]+y_2\left[\begin{array}{c}2\\0\\-1\end{array}\right]+y_3\left[\begin{array}{c}-1\\2\\0\end{array}\right]+y_4\left[\begin{array}{c}-1\\7\\4\end{array}\right]=\vec{0}.
\end{equation*}
How is this vector equation related to the original one?
(c)
Use the solution vector you found in part (a) to construct a solution vector to this new equation.
Subsection 2.4.2 Class Activities
Activity 2.4.2.
Consider the two sets
\begin{equation*}
S=\left\{
\left[\begin{array}{c}2\\3\\1\end{array}\right],
\left[\begin{array}{c}1\\1\\4\end{array}\right]
\right\} \hspace{3em}
T=\left\{
\left[\begin{array}{c}2\\3\\1\end{array}\right],
\left[\begin{array}{c}1\\1\\4\end{array}\right],
\left[\begin{array}{c}-1\\0\\-11\end{array}\right]
\right\}
\end{equation*}
where \(T\) contains a vector missing from \(S\text{.}\) Which of the following is true?
- \(\vspan S\) contains a vector missing from \(\vspan T\text{.}\)
- \(\vspan T\) contains a vector missing from \(\vspan S\text{.}\)
- \(\vspan S\) and \(\vspan T\) contain the same vectors.
Definition 2.4.3.
We say that a set of vectors is linearly dependent if one vector in the set belongs to the span of the others. Otherwise, we say the set is linearly independent.
You can think of linearly dependent sets as containing a redundant vector, in the sense that you can drop a vector out without reducing the span of the set. In the above image, all three vectors lay in the same planar subspace, but only two vectors are needed to span the plane, so the set is linearly dependent.
Remark 2.4.4.
In Activity 2.4.2 we had
\begin{equation*}
S=\left\{
\left[\begin{array}{c}2\\3\\1\end{array}\right],
\left[\begin{array}{c}1\\1\\4\end{array}\right]
\right\} \hspace{1em} \not= \hspace{1em}
T=\left\{
\left[\begin{array}{c}2\\3\\1\end{array}\right],
\left[\begin{array}{c}1\\1\\4\end{array}\right],
\left[\begin{array}{c}-1\\0\\-11\end{array}\right]
\right\}
\end{equation*}
different, while
\begin{equation*}
\vspan S=\left\{
a\left[\begin{array}{c}2\\3\\1\end{array}\right]+
b\left[\begin{array}{c}1\\1\\4\end{array}\right] \middle|
a,b\in\mathbb R
\right\} =
\end{equation*}
\begin{equation*}
\vspan T=\left\{
a\left[\begin{array}{c}2\\3\\1\end{array}\right]+
b\left[\begin{array}{c}1\\1\\4\end{array}\right]+
c\left[\begin{array}{c}-1\\0\\-11\end{array}\right]\middle|
a,b,c\in\mathbb R
\right\}
\end{equation*}
were the same. This is possible because while \(S\) is linearly independent, \(T\)’s third vector made it linearly dependent:
\begin{equation*}
1\left[\begin{array}{c}2\\3\\1\end{array}\right]
-3\left[\begin{array}{c}1\\1\\4\end{array}\right]=
\left[\begin{array}{c}-1\\0\\-11\end{array}\right]
\end{equation*}
Activity 2.4.5.
Consider the following three vectors in \(\IR^3\text{:}\)
\begin{equation*}
\vec v_1=\left[\begin{array}{c}-2 \\ 0 \\ 3\end{array}\right],
\vec v_2=\left[\begin{array}{c}1 \\ 3 \\ 0\end{array}\right],
\text{ and }
\vec v_3=\left[\begin{array}{c}-2 \\ 6 \\ 6\end{array}\right]\text{.}
\end{equation*}
(a)
Let \(\vec v_4 = 3\vec v_1 - \vec v_2 - 2 \vec v_3 =
\left[\begin{array}{c}-3 \\ -15 \\ -3\end{array}\right]\text{.}\) The set \(\{\vec v_1,\vec v_2,\vec v_3,\vec v_4\}\) is...
- linearly dependent: at least one vector is a linear combination of others
- linearly independent: no vector is a linear combination of others
Answer.
A. We explicitly constructed \(\vec v_4\) as a linear combination of the others. (It turns out \(\vec v_3\) is a linear combo as well.)
(b)
Find
\begin{equation*}
\RREF \left[\begin{array}{ccc|c}
\vec v_1 & \vec v_2 & \vec v_3 & \vec v_4 \\
\end{array}\right]=
\RREF \left[\begin{array}{ccc|c}
-2 & 1 &-2 & -3 \\
0 & 3 & 6 & -15 \\
3 &0 &6 & -3
\end{array}\right]= \unknown .
\end{equation*}
What does this guarantee about the solution set for the vector equation \(x_1\vec{v}_1+x_2\vec{v}_2+x_3\vec{v}_3 =\vec v_4\text{?}\)
- It is inconsistent.
- It is consistent with one solution.
- It is consistent with infinitely many solutions.
- None of these.
Answer.
C. Since
\begin{equation*}
\RREF \left[\begin{array}{ccc|c}
-2 & 1 &-2 & -3 \\
0 & 3 & 6 & -15 \\
3 &0 &6 & -3
\end{array}\right] =
\RREF \left[\begin{array}{ccc|c}
1 & 0 &2 & -1 \\
0 & 1 & 2 & -5 \\
0 &0 &0 & 0
\end{array}\right]
\end{equation*}
has no contradiction and a free variable, it has infinitely-many solutions.
(c)
Let \(\vec w=\left[\begin{array}{c}\unknown \\ \unknown \\ \unknown\end{array}\right]\) be some arbitrary vector in \(\mathbb R^3\text{.}\) Consider
\begin{equation*}
\RREF \left[\begin{array}{cccc}
\vec v_1 & \vec v_2 & \vec v_3 & \vec v_4 \\
\end{array}\right]=
\RREF \left[\begin{array}{cccc}
-2 & 1 &-2 & -3 \\
0 & 3 & 6 & -15 \\
3 &0 &6 & -3
\end{array}\right]= \unknown .
\end{equation*}
What does this guarantee about the solution set for the vector equation \(x_1\vec{v}_1+x_2\vec{v}_2+x_3\vec{v}_3 + x_4\vec v_4=\vec w\text{?}\)
- It is inconsistent.
- It is consistent with exactly one solution.
- It is not consistent with exactly one solution.
- It is consistent with infinitely many solutions.
Answer.
D. Due to the row of zeros, it might be inconsistent. But if it is consistent, the free variable column guarantees it will have infinitely-many solutions.
(d)
Which of the following describes any solution to the equation \(x_1\vec{v}_1+x_2\vec{v}_2+x_3\vec{v}_3 + x_4\vec v_4=\vec{w}\) for any linearly dependent set \(\{\vec v_1,\vec v_2,\vec v_3,\vec v_4\}\) and some vector \(\vec w\in \mathbb R^n\text{?}\)
- No such solution can exist.
- It cannot be unique.
- It must be unique.
Answer.
B. If we assume the solution exists, the previous answer shows us that it cannot be the only one, due to the free variable column.
(e)
Which of the following describes any solution to the equation \(x_1\vec{v}_1+x_2\vec{v}_2+x_3\vec{v}_3 + x_4\vec v_4=\vec{w}\) for any linearly independent set \(\{\vec v_1,\vec v_2,\vec v_3,\vec v_4\}\) and some vector \(\vec w\in \mathbb R^n\text{?}\)
- No such solution can exist.
- It cannot be unique.
- It must be unique.
Answer.
C. If we assume the solution exists, it must be the only one, due to the lack of a free variable column.
Fact 2.4.6.
For any vector space, the set \(\{\vec v_1,\dots\vec v_n\}\) is linearly dependent if and only if the vector equation \(x_1\vec v_1+ x_2 \vec v_2+\dots+x_n\vec v_n=\vec w\) cannot have a unique solution for any \(\vec w\text{.}\)
On the other hand, the set of vectors \(\{\vec v_1,\dots\vec v_n\}\) is linearly independent if and only if any solution to the vector equation
\begin{equation*}
x_1\vec v_1+ x_2 \vec v_2 + \cdots + x_n\vec v_n = \vec w
\end{equation*}
must be unique for any \(\vec w\text{.}\)
Activity 2.4.7.
Find
\begin{equation*}
\RREF\left[\begin{array}{ccccc|c}
2&2&3&-1&4&0\\
3&0&13&10&3&0\\
0&0&7&7&0&0\\
-1&3&16&14&1&0
\end{array}\right]
\end{equation*}
and mark the part of the matrix that demonstrates that
\begin{equation*}
S=\left\{
\left[\begin{array}{c}2\\3\\0\\-1\end{array}\right],
\left[\begin{array}{c}2\\0\\0\\3\end{array}\right],
\left[\begin{array}{c}3\\13\\7\\16\end{array}\right],
\left[\begin{array}{c}-1\\10\\7\\14\end{array}\right],
\left[\begin{array}{c}4\\3\\0\\1\end{array}\right]
\right\}
\end{equation*}
is linearly dependent (the part that shows its linear system cannot have a unique solution).
Activity 2.4.8.
(a)
Write a statement involving the solutions of a vector equation that’s equivalent to each claim:
(i)
“The set of vectors \(\left\{ \left[\begin{array}{c}
1 \\
-1 \\
0 \\
-1
\end{array}\right] , \left[\begin{array}{c}
5 \\
5 \\
3 \\
1
\end{array}\right] , \left[\begin{array}{c}
9 \\
11 \\
6 \\
2
\end{array}\right] \right\}\) is linearly independent.”
(ii)
“The set of vectors \(\left\{ \left[\begin{array}{c}
1 \\
-1 \\
0 \\
-1
\end{array}\right] , \left[\begin{array}{c}
5 \\
5 \\
3 \\
1
\end{array}\right] , \left[\begin{array}{c}
9 \\
11 \\
6 \\
2
\end{array}\right] \right\}\) is linearly dependent.”
(b)
Explain how to determine which of these statements is true.
Observation 2.4.9.
Compare the following results:
- A set of \(\IR^m\) vectors \(\{\vec v_1,\dots\vec v_n\}\) is linearly independent if and only if \(\RREF\left[\begin{array}{ccc}\vec v_1&\dots&\vec v_n\end{array}\right]\) has all pivot columns.
- A set of \(\IR^m\) vectors \(\{\vec v_1,\dots\vec v_n\}\) is linearly dependent if and only if \(\RREF\left[\begin{array}{ccc}\vec v_1&\dots&\vec v_n\end{array}\right]\) has at least one non-pivot column.
- A set of \(\IR^m\) vectors \(\{\vec v_1,\dots\vec v_n\}\) spans \(\IR^m\) if and only if \(\RREF\left[\begin{array}{ccc}\vec v_1&\dots&\vec v_n\end{array}\right]\) has all pivot rows.
- A set of \(\IR^m\) vectors \(\{\vec v_1,\dots\vec v_n\}\) fails to span \(\IR^m\) if and only if \(\RREF\left[\begin{array}{ccc}\vec v_1&\dots&\vec v_n\end{array}\right]\) has at least one non-pivot row.
Activity 2.4.10.
What is the largest number of \(\IR^4\) vectors that can form a linearly independent set?
- \(\displaystyle 3\)
- \(\displaystyle 4\)
- \(\displaystyle 5\)
- You can have infinitely many vectors and still be linearly independent.
Activity 2.4.11.
Is it possible for the set of Euclidean vectors \(\{\vec v_1, \vec v_2,\ldots, \vec v_n, \vec 0\}\) to be linearly independent?
- Yes
- No
Subsection 2.4.3 Individual Practice
Remark 2.4.12.
Recall that in Activity 2.2.1 we used the words vector, linear combination, and span to make an analogy with recipes, ingredients, and meals. In this analogy, a recipe was defined to be a list of amounts of each ingredient to build a particular meal.
Activity 2.4.13.
Consider the statement: The set of vectors \(\left\{\vec{v}_1,\vec{v}_2,\vec{v}_3\right\}\) is linearly dependent because the vector \(\vec{v}_3\) is a linear combination of \(\vec{v_1}\) and \(\vec{v}_2\text{.}\) Construct an analogous statement involving ingredients, meals, and recipes, using the terms linearly (in)dependent and linear combination.
Activity 2.4.14.
The following exercises are designed to help develop your geometric intuition around linear dependence.
(a)
Draw sketches that depict the following:
- Three linearly independent vectors in \(\IR^3\text{.}\)
- Three linearly dependent vectors in \(\IR^3\text{.}\)
(b)
If you have three linearly dependent vectors, is it necessarily the case that one of the vectors is a multiple of the other?
Subsection 2.4.4 Videos
Subsection 2.4.5 Exercises
Exercises available at
https://tbil.org/preview/linear-algebra/exercises/#/bank/EV4/.Subsection 2.4.6 Mathematical Writing Explorations
Exploration 2.4.15.
Prove the result of Observation 2.4.9, by showing that, given a set \(S = \{\vec{v}_1,\vec{v}_2,\ldots,\vec{v}_n\}\) of vectors, \(S\) is linearly independent iff the equation \(x_1\vec{v}_1 + x_2\vec{v}_2 + \ldots\ + x_n\vec{v}_n = \vec{0}\) is only true when \(x_1 = x_2 = \cdots = x_n = 0\text{.}\)
Subsection 2.4.7 Sample Problem and Solution
Sample problem Example B.1.8.
