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Mathematical Expression Editor
Introduction to Bases
Coordinate Vectors
When we first introduced vectors we learned to represent them using component
notation. If \(\vec {u}=\begin{bmatrix}2\\5\end{bmatrix}\) we know that the head of \(\vec {u}\) is located at the point \((2, 5)\). But there is another
way to look at the component form. Observe that \(\vec {u}\) can be expressed as a linear
combination of the standard unit vectors \(\vec {i}\) and \(\vec {j}\):
In fact, any vector \(\vec {v}=\begin{bmatrix}a\\b\end{bmatrix}\) of \(\RR ^2\) can be written as a linear combination of \(\vec {i}\) and \(\vec {j}\):
This gives us an alternative way of interpreting the component notation:
\[\left [\begin{array}{c} a\\b \end{array}\right ] \begin{array}{c} \longleftarrow \\ \longleftarrow \end{array} \begin{array}{c} \mbox {coefficient in front of $\vec {i}$}\\\mbox {coefficient in front of $\vec {j}$} \end{array}\]
We say that \(a\) and \(b\) are coordinates of \(\vec {v}\) with respect to \(\{\vec {i}, \vec {j}\}\), and \(\begin{bmatrix}a\\b\end{bmatrix}\) is said to be
the coordinate vector for \(\vec {v}\) with respect to \(\{\vec {i}, \vec {j}\}\). Every vector \(\vec {v}\) of \(\RR ^2\) can be thus
represented using \(\vec {i}\) and \(\vec {j}\). Moreover, such representation in terms of \(\vec {i}\) and \(\vec {j}\)
is unique for each vector, meaning that we will never have two different
coordinate vectors representing the same vector. We will refer to \(\{\vec {i}, \vec {j}\}\) as a basis of
\(\RR ^2\).
The order in which the basis elements are written matters. For example, \(\vec {u}\) is
represented by the coordinate vector \(\begin{bmatrix}2\\5\end{bmatrix}\) with respect to \(\{\vec {i}, \vec {j}\}\), but changing the basis to \(\{\vec {j}, \vec {i}\}\)
would change the coordinate vector to \(\begin{bmatrix}5\\2\end{bmatrix}\).
\[\left [\begin{array}{c} 5\\2 \end{array}\right ] \begin{array}{c} \longleftarrow \\ \longleftarrow \end{array} \begin{array}{c} \mbox {coefficient in front of the first basis element }\\\mbox {coefficient in front of the second basis element} \end{array}\]
Clearly, standard unit vectors \(\vec {i}\) and \(\vec {j}\) are very convenient, but other vectors can also
be used in place of \(\vec {i}\) and \(\vec {j}\) to represent \(\vec {u}\).
The diagram below shows \(\vec {u}\) together with vectors \(\vec {w}_1\) and \(\vec {w}_2\).
It is easy to see that
\[\vec {u}=2\vec {w}_1+\vec {w}_2\]
as shown below.
If we declare \(\{\vec {w}_1, \vec {w}_2\}\) to be a basis of \(\RR ^2\), then we can say that the coordinate vector for \(\vec {u}\) with
respect to \(\{\vec {w}_1, \vec {w}_2\}\) is \(\begin{bmatrix}2\\1\end{bmatrix}\) .
\[\left [\begin{array}{c} 2\\1 \end{array}\right ] \begin{array}{c} \longleftarrow \\ \longleftarrow \end{array} \begin{array}{c} \mbox {coefficient in front of the first basis element }\\\mbox {coefficient in front of the second basis element} \end{array}\]
What Constitutes a Basis?
In the previous section we had used the term basis without defining it. Now is the
time to pause and think about what we want a basis to do. Let’s focus on \(\RR ^n\) and
subspaces of \(\RR ^n\). What we establish here will easily generalize to other vector
spaces.
Based on our previous discussion, given any vector \(\vec {v}\) of \(\RR ^n\) (or a subspace \(V\) of \(\RR ^n\)), we want
to be able to write a coordinate vector for \(\vec {v}\) with respect to the given basis of \(\RR ^n\)
(or \(V\)). Based on this condition, we will require that basis vectors span \(\RR ^n\) (or
\(V\)).
For example, consider \(\vec {w}_1\) and \(\vec {w}_2\) shown below.
The set \(\{\vec {w}_1, \vec {w}_2\}\) cannot be a basis for \(\RR ^3\) because \(\vec {w}_1\) and \(\vec {w}_2\) span a plane in \(\RR ^3\), and any
vector not in that plane cannot be written as a linear combination of \(\vec {w}_1\) and
\(\vec {w}_2\).
On the other hand, the plane spanned by \(\vec {w}_1\) and \(\vec {w}_2\) is a subspace of \(\RR ^3\). Because every
vector in that plane can be written as a linear combination of \(\vec {w}_1\) and \(\vec {w}_2\), the set \(\{\vec {w}_1, \vec {w}_2\}\) can
potentially be a basis for the plane, provided that the set satisfies our second
requirement.
Our second requirement is that for a fixed basis of \(\RR ^n\) (or \(V\)), the coordinate vector for
each \(\vec {v}\) in \(\RR ^n\) (or \(V\)) should be unique. Uniqueness of representation in terms of the basis
elements will play an important role in our future study of functions that map vector
spaces to vector spaces.
The following theorem shows that the uniqueness requirement is equivalent to the
requirement that the basis vectors be linearly independent.
Suppose \(\{\vec {w}_1, \vec {w}_2,\ldots ,\vec {w}_p\}\) is a spanning set for a subspace \(V\) of \(\RR ^n\). Then every element \(\vec {v}\) of \(V\) has a unique
representation as a linear combination of \(\vec {w}_1, \vec {w}_2,\ldots ,\vec {w}_p\) if and only if the vectors \(\vec {w}_1, \vec {w}_2,\ldots ,\vec {w}_p\) are linearly
independent.
Suppose that every \(\vec {v}\) in \(V\) can be expressed as a unique linear combination of \(\vec {w}_1, \vec {w}_2,\ldots ,\vec {w}_p\).
This means that \(\vec {0}\) has a unique representation as a linear combination of \(\vec {w}_1, \vec {w}_2,\ldots ,\vec {w}_p\).
But
is a representation of \(\vec {0}\) in terms of \(\vec {w}_1, \vec {w}_2,\ldots ,\vec {w}_p\). Since we are assuming that such a representation
is unique, we conclude that there is no other. This means that the vectors \(\vec {w}_1, \vec {w}_2,\ldots ,\vec {w}_p\) are
linearly independent.
Conversely, suppose that vectors \(\vec {w}_1, \vec {w}_2,\ldots ,\vec {w}_p\) are linearly independent. An arbitrary element \(\vec {v}\) of \(V\)
can be expressed as a linear combination of \(\vec {w}_1, \vec {w}_2,\ldots ,\vec {w}_p\):
Because we assumed that \(\vec {w}_1, \vec {w}_2,\ldots ,\vec {w}_p\) are linearly independent, we must have
\[a_1-b_1=0,\, a_2-b_2=0,\,\ldots ,\,a_p-b_p=0\]
so that
\[a_1=b_1,\, a_2=b_2,\,\ldots ,\,a_p=b_p\]
This proves the representation of \(\vec {v}\) in terms of \(\vec {w}_1, \vec {w}_2,\ldots ,\vec {w}_p\) is unique.
Use \(V=\mbox {span}(\mathcal {S})\), where \(\mathcal {S}=\left \{\begin{bmatrix}5\\2\\4\end{bmatrix},\begin{bmatrix}4\\1\\1\end{bmatrix},\begin{bmatrix}-3\\0\\2\end{bmatrix}\right \}\) to illustrate why a set of linearly dependent vectors cannot be used as a
basis for a subspace by showing that linearly dependent vectors fail to ensure
uniqueness of coordinate vectors for vectors in \(V\).
We will first show that the elements
of \(\mathcal {S}\) are linearly dependent. Let \(A\) be a matrix whose columns are the vectors in
\(\mathcal {S}\).
This tells us that there are infinitely many nontrivial linear relations among the
elements of \(\mathcal {S}\). Letting \(t=1\) gives us one such nontrivial relation.
The set \(\mathcal {S}\) is linearly dependent. As a result, coordinate vectors for elements of \(V\) are
not unique and we do not want to use \(\mathcal {S}\) as a basis for \(V\).
Definition of a Basis
A set \(\mathcal {S}\) of vectors is called a basis of \(\RR ^n\) (or a basis of a subspace \(V\) of \(\RR ^n\)) provided
that
(a)
\(\mbox {span}(\mathcal {S})=\RR ^n\) (or \(V\))
(b)
\(\mathcal {S}\) is linearly independent.
The standard unit vectors \(\vec {e}_1, \ldots ,\vec {e}_n\) are linearly independent and span \(\RR ^n\). Thus \(\{\vec {e}_1, \ldots ,\vec {e}_n\}\) is a basis of \(\RR ^n\).
The set \(\{\vec {e}_1, \ldots ,\vec {e}_n\}\) is called the standard basis of \(\RR ^n\).
Bases are not unique. For example, we know that vectors \(\vec {i}\) and \(\vec {j}\) form the standard
basis of \(\RR ^2\). But, as we discussed in Example ??, vectors
Any linearly independent spanning set in \(\RR ^n\) (or a subspace of \(\RR ^n\)) is a basis of \(\RR ^n\) (or the
subspace). The plural form of the word basis is bases. It is easy to see that \(\RR ^n\) and its
subspaces each has infinitely many bases.
Let \(V=\mbox {span}\left (\begin{bmatrix}-2\\1\\3\end{bmatrix},\begin{bmatrix}2\\-4\\1\end{bmatrix}\right )\). The set
is a basis for \(V\) because the two vectors in \(\mathcal {B}\) are linearly independent and span \(V\). Find
the coordinate vector for \(\vec {v}=\begin{bmatrix}2\\-10\\9\end{bmatrix}\) with respect to \(\mathcal {B}\).
We need to express \(\begin{bmatrix}2\\-10\\9\end{bmatrix}\) as a linear
combination of the elements of \(\mathcal {B}\). To this end, we need to solve the vector
equation:
The coefficient in front of the first basis vector is \(2\), the coefficient in front of the
second basis vector is \(3\). This means that the coordinate vector for \(\begin{bmatrix}2\\-10\\9\end{bmatrix}\) with respect to \(\mathcal {B}\) is
\(\begin{bmatrix}2\\3\end{bmatrix}\).
It may seem strange to you that the coordinate vector for a vector in \(\RR ^3\) only has two
components. But remember that subspace \(V\) is a plane. When viewed as a vector in the
plane, it makes sense that the coordinate vector for \(\begin{bmatrix}2\\-10\\9\end{bmatrix}\) only requires two components.
This issue is related to the question of dimension, which will be addressed in Bases
and Dimension.
To construct the coordinate vector for \(\begin{bmatrix}2\\-10\\9\end{bmatrix}\) with respect to \(\mathcal {B}\), we had to be mindful of the
order of the elements in \(\mathcal {B}\). Ordinarily, the order of elements in a set is irrelevant, and
the basis
When dealing with coordinate vectors, however, the order of the elements dictates
the order of the components of the coordinate vector coefficients. If we switch the
order of the elements in \(\mathcal {B}\), the coordinate vector becomes \(\begin{bmatrix}3\\2\end{bmatrix}\). For this reason, when we
come back to studying coordinate vectors in more detail, we will use the term ordered
basis to avoid confusion.
Practice Problems
Let \(\mathcal {B}=\left \{\begin{bmatrix}1\\1\end{bmatrix},\begin{bmatrix}-1\\2\end{bmatrix}\right \}\) be a basis for \(\RR ^2\). Find the coordinate vector for \(\vec {v}\) with respect to \(\mathcal {B}\).
Let \(\mathcal {B}=\left \{\begin{bmatrix}1\\1\end{bmatrix},\begin{bmatrix}-1\\2\end{bmatrix}\right \}\) be a basis for \(\RR ^2\). Find the coordinate vector for \(\vec {v}\) with respect to \(\mathcal {B}\).
Let \(\mathcal {B}=\left \{\begin{bmatrix}1\\-1\\3\end{bmatrix},\begin{bmatrix}2\\1\\-1\end{bmatrix}\right \}\) be a basis for \(\mbox {span}\left (\begin{bmatrix}1\\-1\\3\end{bmatrix},\begin{bmatrix}2\\1\\-1\end{bmatrix}\right )\). Find the coordinate vector for \(\begin{bmatrix}-4\\-2\\2\end{bmatrix}\) with respect to \(\mathcal {B}\).
Suppose \(\mathcal {B}=\left \{\begin{bmatrix}1\\1\\1\end{bmatrix},\begin{bmatrix}1\\0\\1\end{bmatrix}, \vec {w}\right \}\) is a basis for \(\RR ^3\). Find \(\vec {w}\) if the coordinate vector for \(\begin{bmatrix}-2\\-7\\4\end{bmatrix}\) is \(\begin{bmatrix}-1\\2\\-3\end{bmatrix}\).
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