If a friend told you that they have a line spanned by and and , you would probably think that your friend’s description is a little excessive. Isn’t one of the above vectors sufficient to describe the line? A line can be described as a span of one vector, but it can also be described as a span of two or more vectors. There are many advantages, however, to using the most efficient description possible. In this section we will begin to explore what makes a description “more efficient.”
In this Exploration we will examine what can happen to the span of a collection of vectors when a vector is removed from the collection.
First, let’s remove from .
Which of the following is true?
Removing from changed, did not change the span.
Now let’s remove from the original collection of vectors.
Which of the following is true?
As you just discovered, removing a vector from a collection of vectors may or may not affect the span of the collection. We will refer to vectors that can be removed from a collection without changing the span as redundant. In Exploration exp:redundantVecs1, is redundant, while is not.
Our next goal is to see what causes of Exploration exp:redundantVecs1 to be redundant. The answer lies not in the vector itself, but in its relationship to the other vectors in the collection. Observe that . In other words, is a scalar multiple of another vector in the set. To see why this matters, let’s pick an arbitrary vector in . Vector is in the span because it can be written as a linear combination of the three vectors as follows
But is not essential to this linear combination because it can be replaced with , as shown below.
Regardless of what vector we write as a linear combination of, and , we will always be able to replace with , placing into the span of and , and making redundant. (Note that we can just as easily write , and argue that is redundant.) We conclude that only one of and is needed to maintain the span of the original three vectors. We have The left-most collection in this expression contains redundant vectors; the other two collections do not.
In Exploration exp:redundantVecs1 we found one vector to be redundant because we could replace it with a scalar multiple of another vector in the set. The following Exploration delves into what happens when a vector in a given set is a linear combination of the other vectors.
is A line, A plane, , A parallelepiped
Can we remove one of the vectors from the set without changing the span? Observe that we can write as a linear combination of the other two vectors \begin{equation} \label{eq:redundant}\begin{bmatrix}4\\4\\-1\end{bmatrix}=2\begin{bmatrix}1\\2\\-1\end{bmatrix}+1\begin{bmatrix}2\\0\\1\end{bmatrix} \end{equation}
This means that we can write any vector in as a linear combination of only and by replacing with the expression in (eq:redundant). For example,
We have
We conclude that vector is redundant. Can each of the other two vectors in the set be considered redundant? You will address this question in Practice Problem prob:redundant1.
Collections of vectors that do not contain redundant vectors are very important in linear algebra. We will refer to such collections as linearly independent. Collections of vectors that contain redundant vectors will be called linearly dependent. The following section offers a definition that will allow us to easily determine linear dependence and independence of vectors.
If, in addition to the trivial solution, a non-trivial solution (not all are zero) exists, then we say that the set is linearly dependent.
It turns out that these questions are equivalent. In other words, if the answer to one of them is “YES”, the answer to the other two is also “YES”. Conversely, if the answer to one of them is “NO”, then the answer to the other two is also “NO”. We will start by illustrating this idea with an example, then conclude this section by formally proving the equivalency.
Clearly is a solution to the equation. The question is whether another solution exists.
The vector equation translates into the following system:
Writing the system in augmented matrix form and applying elementary row operations gives us the following reduced row-echelon form: This shows that (eq:linrelationpart1) has infinitely many solutions: Letting , we obtain the following:
\begin{equation} \label{eq:ex1} -6\begin{bmatrix}2\\-3\end{bmatrix}+0 \begin{bmatrix}0\\3\end{bmatrix}+6\begin{bmatrix}1\\-1\end{bmatrix}+6\begin{bmatrix}1\\-2\end{bmatrix}=\vec{0} \end{equation} We conclude that the vectors are linearly dependent.
Observe that (eq:ex1) allows us to solve for one of the vectors and express it as a linear combination of the others. For example, \begin{equation} \label{eq:ex1lincomb} \begin{bmatrix}2\\-3\end{bmatrix}=0 \begin{bmatrix}0\\3\end{bmatrix}+\begin{bmatrix}1\\-1\end{bmatrix}+\begin{bmatrix}1\\-2\end{bmatrix} \end{equation} This would not be possible if a nontrivial solution to the equation did not exist.
Using the linear combination in (eq:ex1lincomb) and the argument of Exploration exp:redundantVecs2, we conclude that is redundant in We find that the answer to all questions in Remark remark:LinIndEquiv is “YES”.
item:linindpart2 To address linear independence, we need to solve the equation
Converting the equation to augmented matrix form and performing row reduction gives us This shows that is the only solution. Therefore the two vectors are linearly independent.
Furthermore, we cannot write one of the vectors as a linear combination of the other. (Do you see that the only way this would be possible with a set of two vectors is if they were scalar multiples of each other?)
Finally, we observe that removing either vector would change the span from a plane in to a line in , so the answer to all three questions in Remark remark:LinIndEquiv is “NO”.
has a non-trivial solution. In other words at least one of the constants, say , does not equal zero. This allows us to solve for :
\begin{align*} -c_j\vec{v}_j&=c_1\vec{v}_1+c_2\vec{v}_2+\ldots +c_k\vec{v}_k \\ \\ \vec{v}_j&=-\frac{c_1}{c_j}\vec{v}_1-\frac{c_2}{c_j}\vec{v}_2-\ldots -\frac{c_k}{c_j}\vec{v}_k \end{align*}
(Do you see why it was important to have one of the constants nonzero?) This shows that may be expressed as a linear combination of the other vectors.
th:lindeplincombofother_b th:lindeplincombofother_c First, suppose is a linear combination of . We will show that is redundant by showing that To show equality of the two spans we will pick a vector in the left span and show that it is also an element of the span on the right. Then, we will pick a vector in the right span and show that it is also an element of the span on the left, and we will conclude that the sets are equal.
Observe that if is in , then it has to be in . (Why?)
Now suppose is in . We need to show that is also in .
By assumption, we can write as \begin{equation} \label{eq:vj} \vec{v}_j=a_1\vec{v}_1+a_2\vec{v}_2+\dots +a_{j-1}\vec{v}_{j-1}+a_{j+1}\vec{v}_{j+1}+\dots +a_k\vec{v}_k. \end{equation}
Since is in , we have Substituting the expression in (eq:vj) for and simplifying, we obtain the following
th:lindeplincombofother_c th:lindeplincombofother_a Suppose that is redundant, so that Consider a vector in \begin{equation} \label{eq:w1} \vec{w}=a_1\vec{v}_1+a_2\vec{v}_2+\dots +a_j\vec{v}_j+\dots +a_k\vec{v}_k \end{equation} Since the span contains ALL possible linear combinations of , we may choose such that .
By assumption, is also in . Therefore, we can express as a linear combination \begin{equation} \label{eq:w2} \vec{w}=b_1\vec{v}_1+b_2\vec{v}_2+\dots +b_{j-1}\vec{v}_{j-1}+b_{j+1}\vec{v}_{j+1}+\dots +b_k\vec{v}_k. \end{equation}
We complete the proof by showing there exists a non-trivial solution to \begin{equation} \label{eq:LinIndepDefRepeated} c_1\vec{v}_1+c_2\vec{v}_2+\ldots +c_j\vec{v}_j+\ldots +c_k\vec{v}_k=\vec{0}. \end{equation} Subtracting expression (eq:w2) from (eq:w1) we obtain
These three parts of the proof show that if one of the conditions is true, all three must be true. It is a logical consequence that if one of the three conditions is false, all three must be false.
Theorem th:lindeplincombofother gives us a convenient ways of looking at linear dependence/independence geometrically. When looking at two or more vectors, we ask, “can one of the vectors be written as a linear combination of the others?” We can also ask, “is one of the vectors redundant?” If the answer to either of these questions is “YES”, then the vectors are linearly dependent.
Two vectors are linearly dependent if and only if one is a scalar multiple of the other. Two nonzero linearly dependent vectors may look like this:
or like this:
Two linearly independent vectors will look like this:
Given a set of three nonzero vectors, we have the following possibilities:
and demonstrated that is redundant by using the fact that it is a linear combination of the other two vectors.
Problems prob:linindmultchoice1-prob:linindmultchoice4
Are the given vectors linearly independent?
Problems prob:TFlinind1-prob:TFlinind2
True or False?
Problems prob:linindmultchoice5-prob:linindmultchoice7
Each problem below provides information about vectors . If possible, determine whether the vectors are linearly dependent or independent.
Problems prob:linindmultchoice9-prob:linindmultchoice10
Each diagram below shows a collection of vectors. Are the vectors linearly dependent or independent?
