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1 We need to find coefficients \(a_1\) and \(a_2\) such that \(\vec {u}=a_1\vec {v}_1+a_2\vec {v}_2\). To do this we need to solve the vector equation:
This shows that \(a_1=\frac {1}{2}\) and \(a_2=\frac {3}{2}\), and we can express \(\vec {u}\) as a linear combination of \(\vec {v}_1\) and \(\vec {v}_2\) as follows:
Observe that because vector \(\vec {u}\) is a linear combination of \(\vec {v}_1\) and \(\vec {v}_2\), \(\vec {u}\) is the diagonal of a parallelogram whose sides are scalar
multiples of \(\vec {v}_1\) and \(\vec {v}_2\). As such, \(\vec {u}\) lies in the same plane as \(\vec {v}_1\) and \(\vec {v}_2\), as illustrated below.
We conclude that there are no solutions, and \(\vec {w}\) is not a linear combination of \(\vec {v}_1\) and \(\vec {v}_2\).
Geometrically, this means that \(\vec {w}\) is not the diagonal of any parallelogram whose sides are scalar multiples of \(\vec {v}_1\) and \(\vec {v}_2\). Thus, \(\vec {w}\) does
not lie in the plane determined by \(\vec {v}_1\) and \(\vec {v}_2\).
In part 1 of Example 1 we expressed \(\vec {u}\) as a linear combination of \(\vec {v}_1\) and \(\vec {v}_2\), and concluded that \(\vec {u}\) lies in the plane determined by \(\vec {v}_1\)
and \(\vec {v}_2\). We say that \(\vec {u}\) is in the span of \(\vec {v}_1\) and \(\vec {v}_2\). In fact, every vector in the plane determined by \(\vec {v}_1\) and \(\vec {v}_2\) is in the span of \(\vec {v}_1\) and \(\vec {v}_2\). We say
that \(\vec {v}_1\) and \(\vec {v}_2\)span the plane.
In contrast, vector \(\vec {w}\) of part 2 of Example 1 is not a linear combination of \(\vec {v}_1\) and \(\vec {v}_2\). We say that \(\vec {w}\) is not in the span of \(\vec {v}_1\) and
\(\vec {v}_2\).
The following video takes another look at Example 1 using our new vocabulary.
Definition of Span
Let \(\vec {v}_1, \vec {v}_2,\ldots ,\vec {v}_p\) be vectors in \(\RR ^n\). The set \(S\) of all linear combinations of \(\vec {v}_1, \vec {v}_2,\ldots ,\vec {v}_p\) is called the span of \(\vec {v}_1, \vec {v}_2,\ldots ,\vec {v}_p\). We write
and we say that vectors \(\vec {v}_1, \vec {v}_2,\ldots ,\vec {v}_p\)span\(S\). Any vector in \(S\) is said to be in the span of \(\vec {v}_1, \vec {v}_2,\ldots ,\vec {v}_p\). The set \(\{\vec {v}_1, \vec {v}_2,\ldots ,\vec {v}_p\}\) is called a spanning set for \(S\).
The span of \(\begin{bmatrix}-3\\1\end{bmatrix}\) is the set of all linear combinations of \(\begin{bmatrix}-3\\1\end{bmatrix}\). Since we are looking for linear combinations of only one
vector, we are really looking for all of its scalar multiples. So, the span will be the set of all vectors of the form \(\vec {v}=a\begin{bmatrix}-3\\1\end{bmatrix}\). All such
vectors lie on the line determined by \(\begin{bmatrix}-3\\1\end{bmatrix}\).
First, observe that \(\begin{bmatrix}2\\2\end{bmatrix}\) and \(\begin{bmatrix}-1\\0\end{bmatrix}\) are not scalar multiples of each other.
Geometrically, we can use Procedure ?? to express any vector of \(\RR ^2\) as a linear combination of \(\begin{bmatrix}2\\2\end{bmatrix}\) and \(\begin{bmatrix}-1\\0\end{bmatrix}\), indicating that the two
vectors span all of \(\RR ^2\).
To verify this claim algebraically we will show that an arbitrary vector \(\begin{bmatrix}s\\t\end{bmatrix}\) of \(\RR ^2\) can be written as a linear combination of \(\begin{bmatrix}2\\2\end{bmatrix}\) and
\(\begin{bmatrix}-1\\0\end{bmatrix}\).
This shows that every vector of \(\RR ^2\) can be written as a linear combination of \(\begin{bmatrix}2\\2\end{bmatrix}\) and \(\begin{bmatrix}-1\\0\end{bmatrix}\):
Geometrically, we can interpret all such linear combinations as diagonals of parallelograms determined by scalar multiples of \(\begin{bmatrix}5\\0\\4\end{bmatrix}\)
and \(\begin{bmatrix}0\\4\\2\end{bmatrix}\). All such diagonals will lie in the plane determined by \(\begin{bmatrix}5\\0\\4\end{bmatrix}\) and \(\begin{bmatrix}0\\4\\2\end{bmatrix}\). Let this plane be called \(p\). A portion of \(p\) is shown below.
Because Procedure ?? can be applied to vectors that lie in \(p\) just as easily as it can be applied to vectors of \(\RR ^2\), we conclude that
every vector in \(p\) can be expressed as a linear combination of \(\begin{bmatrix}5\\0\\4\end{bmatrix}\) and \(\begin{bmatrix}0\\4\\2\end{bmatrix}\). Thus,
If a friend told you that they have a line spanned by \(\begin{bmatrix}1\\1\end{bmatrix}\) and \(\begin{bmatrix}2\\2\end{bmatrix}\) and \(\begin{bmatrix}3\\3\end{bmatrix}\), 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."
What is the span of these vectors? A line, \(\RR ^2\), A parallelogram, A parallelepiped
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 \(\begin{bmatrix}2\\1\end{bmatrix}\) from \(\left \{\begin{bmatrix}2\\-1\end{bmatrix}, \begin{bmatrix}-4\\2\end{bmatrix}, \begin{bmatrix}2\\1\end{bmatrix}\right \}\).
Which of the following is true?
\(\mbox {span}\left (\begin{bmatrix}2\\-1\end{bmatrix}, \begin{bmatrix}-4\\2\end{bmatrix}\right )=\mbox {span}\left (\begin{bmatrix}2\\-1\end{bmatrix}, \begin{bmatrix}-4\\2\end{bmatrix}, \begin{bmatrix}2\\1\end{bmatrix}\right )\)\(\mbox {span}\left (\begin{bmatrix}2\\-1\end{bmatrix}, \begin{bmatrix}-4\\2\end{bmatrix}\right )\) is a line\(\mbox {span}\left (\begin{bmatrix}2\\-1\end{bmatrix}, \begin{bmatrix}-4\\2\end{bmatrix}\right )=\RR ^2\)\(\mbox {span}\left (\begin{bmatrix}2\\-1\end{bmatrix}, \begin{bmatrix}-4\\2\end{bmatrix}\right )\) is a parallelogram.
Removing \(\begin{bmatrix}2\\1\end{bmatrix}\) from \(\left \{\begin{bmatrix}2\\-1\end{bmatrix}, \begin{bmatrix}-4\\2\end{bmatrix}, \begin{bmatrix}2\\1\end{bmatrix}\right \}\) changed, did not change the span.
Now let’s remove \(\begin{bmatrix}-4\\2\end{bmatrix}\) from the original collection of vectors.
Which of the following is true?
\(\mbox {span}\left (\begin{bmatrix}2\\-1\end{bmatrix}, \begin{bmatrix}2\\1\end{bmatrix}\right )=\mbox {span}\left (\begin{bmatrix}2\\-1\end{bmatrix}, \begin{bmatrix}-4\\2\end{bmatrix}, \begin{bmatrix}2\\1\end{bmatrix}\right )\)\(\mbox {span}\left (\begin{bmatrix}2\\-1\end{bmatrix}, \begin{bmatrix}2\\1\end{bmatrix}\right )\) is a line\(\mbox {span}\left (\begin{bmatrix}2\\-1\end{bmatrix}, \begin{bmatrix}2\\1\end{bmatrix}\right )\) is the right side of the coordinate plane.\(\mbox {span}\left (\begin{bmatrix}2\\-1\end{bmatrix}, \begin{bmatrix}2\\1\end{bmatrix}\right )\) is a parallelogram.
Removing \(\begin{bmatrix}-4\\2\end{bmatrix}\) from \(\left \{\begin{bmatrix}2\\-1\end{bmatrix}, \begin{bmatrix}-4\\2\end{bmatrix}, \begin{bmatrix}2\\1\end{bmatrix}\right \}\) changed, did not change the span.
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 , \(\begin{bmatrix}-4\\2\end{bmatrix}\) is redundant,
while \(\begin{bmatrix}2\\1\end{bmatrix}\) is not.
Let \(\{\vec {v}_1,\vec {v}_2,\dots ,\vec {v}_k\}\) be a set of vectors in \(\RR ^n\). If we can remove one vector without changing the span of this set, then that vector is redundant.
In other words, if
we say that \(\vec {v}_j\) is a redundant element of \(\{\vec {v}_1,\vec {v}_2,\dots ,\vec {v}_k\}\), or simply redundant.
Our next goal is to see what causes \(\begin{bmatrix}-4\\2\end{bmatrix}\) of Exploration to be redundant. The answer lies not in the vector itself, but in its
relationship to the other vectors in the collection. Observe that \(\begin{bmatrix}-4\\2\end{bmatrix}=-2\begin{bmatrix}2\\-1\end{bmatrix}\). In other words, \(\begin{bmatrix}-4\\2\end{bmatrix}\) is a scalar multiple of another vector in the
set. To see why this matters, let’s pick an arbitrary vector \(\vec {w}=\begin{bmatrix}0\\2\end{bmatrix}\) in \(\mbox {span}\left (\begin{bmatrix}2\\-1\end{bmatrix}, \begin{bmatrix}-4\\2\end{bmatrix}, \begin{bmatrix}2\\1\end{bmatrix}\right )\). Vector \(\vec {w}\) is in the span because it can be written as a linear
combination of the three vectors as follows
But \(\begin{bmatrix}-4\\2\end{bmatrix}\) is not essential to this linear combination because it can be replaced with \(-2\begin{bmatrix}2\\-1\end{bmatrix}\), as shown below.
Regardless of what vector \(\vec {w}\) we write as a linear combination of\(\begin{bmatrix}2\\-1\end{bmatrix}\),\( \begin{bmatrix}-4\\2\end{bmatrix}\) and \(\begin{bmatrix}2\\1\end{bmatrix}\), we will always be able to replace \(\begin{bmatrix}-4\\2\end{bmatrix}\) with \(-2\begin{bmatrix}2\\-1\end{bmatrix}\),
placing \(\vec {w}\) into the span of \(\begin{bmatrix}2\\-1\end{bmatrix}\) and \(\begin{bmatrix}2\\1\end{bmatrix}\), and making \(\begin{bmatrix}-4\\2\end{bmatrix}\) redundant. (Note that we can just as easily write \(\begin{bmatrix}2\\-1\end{bmatrix}=-\frac {1}{2}\begin{bmatrix}-4\\2\end{bmatrix}\), and argue that \(\begin{bmatrix}2\\-1\end{bmatrix}\) is
redundant.) We conclude that only one of \(\begin{bmatrix}-4\\2\end{bmatrix}\) and \(\begin{bmatrix}2\\-1\end{bmatrix}\) 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 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.
The three vectors are shown below. RIGHT-CLICK and DRAG to rotate the interactive graph.
\(\mbox {span}\left (\begin{bmatrix}1\\2\\-1\end{bmatrix},\begin{bmatrix}2\\0\\1\end{bmatrix},\begin{bmatrix}4\\4\\-1\end{bmatrix}\right )\) is A line, A plane, \(\RR ^3\), A parallelepiped
Can we remove one of the vectors from the set without changing the span? Observe that we can write \(\begin{bmatrix}4\\4\\-1\end{bmatrix}\) as a linear
combination of the other two vectors
This means that we can write any vector in \(\mbox {span}\left (\begin{bmatrix}1\\2\\-1\end{bmatrix},\begin{bmatrix}2\\0\\1\end{bmatrix},\begin{bmatrix}4\\4\\-1\end{bmatrix}\right )\) as a linear combination of only \(\begin{bmatrix}1\\2\\-1\end{bmatrix}\) and \(\begin{bmatrix}2\\0\\1\end{bmatrix}\) by replacing \(\begin{bmatrix}4\\4\\-1\end{bmatrix}\) with the expression in (1). For
example,
We conclude that vector \(\begin{bmatrix}4\\4\\-1\end{bmatrix}\) is redundant. Can each of the other two vectors in the set \(\left \{\begin{bmatrix}1\\2\\-1\end{bmatrix},\begin{bmatrix}2\\0\\1\end{bmatrix},\begin{bmatrix}4\\4\\-1\end{bmatrix}\right \}\) be considered redundant? You will
address this question in the problem set.
Collections of vectors that do not contain redundant vectors are very important in linear algebra. In subsequent sections, we
will formally introduce such collections as linearly independent. Collections of vectors that contain redundant vectors will be
called linearly dependent.
Let \(\vec {v}=\begin{bmatrix}3\\4\\5\end{bmatrix}\). Give an example of at least one vector \(\vec {w}\) such that \(\vec {v}\), \(\vec {w}\) do NOT span a plane in \(\RR ^3\). Describe \(\mbox {span}(\vec {v}, \vec {w})\).
Prove or disprove. The zero vector of \(\RR ^n\) is contained in the span of any collection of vectors of \(\RR ^n\).
In Exploration we considered the following set of vectors
and demonstrated that \(\begin{bmatrix}4\\4\\-1\end{bmatrix}\) is redundant by using the fact that it is a linear combination of the other two vectors.
1.
Express each of \(\begin{bmatrix}1\\2\\-1\end{bmatrix}\) and \(\begin{bmatrix}2\\0\\1\end{bmatrix}\) as a linear combination of the remaining vectors.
If \(\vec {w}\) is in \(\mbox {span}\left (\begin{bmatrix}1\\2\\-1\end{bmatrix},\begin{bmatrix}2\\0\\1\end{bmatrix},\begin{bmatrix}4\\4\\-1\end{bmatrix}\right )\), then \(\vec {w}\) is in \(\mbox {span}\left (\begin{bmatrix}2\\0\\1\end{bmatrix},\begin{bmatrix}4\\4\\-1\end{bmatrix}\right )\).Both \(\begin{bmatrix}1\\2\\-1\end{bmatrix}\) and \(\begin{bmatrix}2\\0\\1\end{bmatrix}\) are redundant in \(\left \{\begin{bmatrix}1\\2\\-1\end{bmatrix},\begin{bmatrix}2\\0\\1\end{bmatrix},\begin{bmatrix}4\\4\\-1\end{bmatrix}\right \}\).We can remove \(\begin{bmatrix}1\\2\\-1\end{bmatrix}\) and \(\begin{bmatrix}2\\0\\1\end{bmatrix}\) from \(\left \{\begin{bmatrix}1\\2\\-1\end{bmatrix},\begin{bmatrix}2\\0\\1\end{bmatrix},\begin{bmatrix}4\\4\\-1\end{bmatrix}\right \}\)
at the same time without affecting the span.
Show that if the zero vector is part of a collection of two or more vectors, the zero vector is redundant.
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