Learning Outcomes
-
Compute a basis for the subspace spanned by a given set of Euclidean vectors, and determine the dimension of the subspace.
Section 2.6 Subspace Basis and Dimension (EV6)
Subsection 2.6.1 Warm Up
Activity 2.6.1.
Consider the set \(S\) of vectors in \(\IR^4\) given by
\begin{equation*}
S=\left\{\left[\begin{array}{c}2\\3\\0\\1\end{array}\right],\left[\begin{array}{c}2\\0\\1\\-1\end{array}\right]\right\}
\end{equation*}
(a)
Is the set \(S\) linearly independent or linearly dependent?
(b)
How would you describe the subspace \(\vspan{S}\) geometrically?
(c)
Subsection 2.6.2 Class Activities
Observation 2.6.2.
In Figure 16 we saw an example of two linearly independent vectors spanning a planar subspace of \(\IR^3\text{.}\)
Because these independent vectors fail to span \(\IR^3\text{,}\) they are not a basis for \(\IR^3\text{.}\) However, they still span a subspace of \(\IR^3\text{...}\)
Activity 2.6.3.
Consider the subspace of \(\IR^4\) given by \(W=\vspan\left\{
\left[\begin{array}{c}2\\3\\0\\1\end{array}\right],
\left[\begin{array}{c}2\\0\\1\\-1\end{array}\right],
\left[\begin{array}{c}2\\-3\\2\\-3\end{array}\right],
\left[\begin{array}{c}1\\5\\-1\\0\end{array}\right]
\right\}
\text{.}\)
(a)
Which feature of
\begin{equation*}
\RREF\left[\begin{array}{cccc}
2&2&2&1\\
3&0&-3&5\\
0&1&2&-1\\
1&-1&-3&0
\end{array}\right]=
\left[\begin{array}{cccc}
1&0&-1&0\\
0&1& 2&0\\
0&0& 0&1\\
0&0& 0&0
\end{array}\right]
\end{equation*}
that shows that \(W\)’s spanning set is linearly dependent?
-
The third column.
-
The fourth column.
-
The third row.
-
The fourth row.
(b)
If we removed the vector that causes this issue, what could we say about that set of three vectors?
-
The set spans the vector space \(\IR^4\text{,}\) but remains linearly dependent.
-
The set spans the subspace \(W\) of \(\IR^4\text{,}\) but remains linearly dependent.
-
The set spans the subspace \(W\) of \(\IR^4\text{,}\) and is now linearly independent.
-
The set no longer spans the subspace \(W\) of \(\IR^4\text{,}\) but is now linearly independent.
Definition 2.6.4.
Let \(W\) be a subspace of a vector space. A basis for \(W\) is a linearly independent set of vectors that spans \(W\) (but not necessarily the entire vector space).
Observation 2.6.5.
So given a set \(S=\{\vec v_1,\dots,\vec v_m\}\text{,}\) to compute a basis for the subspace \(\vspan S\text{,}\) simply remove the vectors corresponding to the non-pivot columns of \(\RREF[\vec v_1\,\dots\,\vec v_m]\text{.}\) For example, since
\begin{equation*}
\RREF
\left[\begin{array}{cccc}
1 & 2 & 0 & 1 \\
2 & 4 & -2 & 0 \\
3 & 6 & -2 & 1 \\
\end{array}\right]
=
\left[\begin{array}{cccc}
\markedPivot{1} & 2 & 0 & 1 \\
0 & 0 & \markedPivot{1} & 1 \\
0 & 0 & 0 & 0
\end{array}\right]
\end{equation*}
the subspace \(W=\vspan\setList{
\left[\begin{array}{c}1\\2\\3\end{array}\right],
\left[\begin{array}{c}2\\4\\6\end{array}\right],
\left[\begin{array}{c}0\\-2\\-2\end{array}\right],
\left[\begin{array}{c}1\\0\\1\end{array}\right]
}\) has \(\setList{
\left[\begin{array}{c}1\\2\\3\end{array}\right],
\left[\begin{array}{c}0\\-2\\-2\end{array}\right]
}\) as a basis.
Activity 2.6.6.
(a)
Find a basis for \(\vspan S\) where
\begin{equation*}
S=\left\{
\left[\begin{array}{c}2\\3\\0\\1\end{array}\right],
\left[\begin{array}{c}2\\0\\1\\-1\end{array}\right],
\left[\begin{array}{c}2\\-3\\2\\-3\end{array}\right],
\left[\begin{array}{c}1\\5\\-1\\0\end{array}\right]
\right\}\text{.}
\end{equation*}
(b)
Find a basis for \(\vspan T\) where
\begin{equation*}
T=\left\{
\left[\begin{array}{c}2\\0\\1\\-1\end{array}\right],
\left[\begin{array}{c}2\\-3\\2\\-3\end{array}\right],
\left[\begin{array}{c}1\\5\\-1\\0\end{array}\right],
\left[\begin{array}{c}2\\3\\0\\1\end{array}\right]
\right\}\text{.}
\end{equation*}
Observation 2.6.7.
Even though we found different bases for them, \(\vspan S\) and \(\vspan T\) are exactly the same subspace of \(\IR^4\text{,}\) since
\begin{equation*}
S=\left\{
\left[\begin{array}{c}2\\3\\0\\1\end{array}\right],
\left[\begin{array}{c}2\\0\\1\\-1\end{array}\right],
\left[\begin{array}{c}2\\-3\\2\\-3\end{array}\right],
\left[\begin{array}{c}1\\5\\-1\\0\end{array}\right]
\right\}
=
\left\{
\left[\begin{array}{c}2\\0\\1\\-1\end{array}\right],
\left[\begin{array}{c}2\\-3\\2\\-3\end{array}\right],
\left[\begin{array}{c}1\\5\\-1\\0\end{array}\right],
\left[\begin{array}{c}2\\3\\0\\1\end{array}\right]
\right\}=T\text{.}
\end{equation*}
Thus a given basis for a subspace need not be unique.
Definition 2.6.8.
Any non-trivial real vector space has infinitely-many different bases, but all the bases for a given vector space are exactly the same size. So we say the dimension of a vector space or subspace is equal to the size of any basis for the vector space.
As you’d expect, \(\IR^n\) has dimension \(n\text{.}\) For example, \(\IR^3\) has dimension \(3\) because any basis for \(\IR^3\) such as
\begin{equation*}
\setList{\vec e_1,\vec e_2,\vec e_3}
\text{ and }
\setList{
\left[\begin{array}{c}1\\0\\0\end{array}\right],
\left[\begin{array}{c}0\\1\\0\end{array}\right],
\left[\begin{array}{c}1\\1\\1\end{array}\right]
}
\text{ and }
\setList{
\left[\begin{array}{c}1\\0\\-3\end{array}\right],
\left[\begin{array}{c}2\\-2\\1\end{array}\right],
\left[\begin{array}{c}3\\-2\\5\end{array}\right]
}
\end{equation*}
contains exactly three vectors.
Likewise, the planar subspace with the following two bases
\begin{equation*}
\setList{
\left[\begin{array}{c}1\\2\\3\end{array}\right],
\left[\begin{array}{c}-2\\0\\5\end{array}\right]
}
\text{ and }
\setList{
\left[\begin{array}{c}-1\\2\\8\end{array}\right],
\left[\begin{array}{c}0\\4\\11\end{array}\right]
}
\end{equation*}
has dimension \(2\) because any basis for the subspace will have exactly two vectors.
Activity 2.6.9.
Consider the following subspace \(W\) of \(\mathbb R^4\text{:}\)
\begin{equation*}
W=\mathrm{span}\,\left\{ \left[\begin{array}{c} 1 \\ 0 \\ 0 \\ -1 \end{array}\right] , \left[\begin{array}{c} -2 \\ 0 \\ 0 \\ 2 \end{array}\right] , \left[\begin{array}{c} -3 \\ 1 \\ -5 \\ 5 \end{array}\right] , \left[\begin{array}{c} 12 \\ -3 \\ 15 \\ -18 \end{array}\right] \right\}.
\end{equation*}
(a)
Explain and demonstrate how to find a basis of \(W\text{.}\)
(b)
Explain and demonstrate how to find the dimension of \(W\text{.}\)
Activity 2.6.10.
The dimension of a subspace may be found by doing what with an appropriate RREF matrix?
-
Count the rows.
-
Count the non-pivot columns.
-
Count the pivots.
-
Add the number of pivot rows and pivot columns.
Subsection 2.6.3 Individual Practice
Activity 2.6.11.
In Observation 2.6.5, we found a basis for the subspace
\begin{equation*}
W=\vspan\setList{
\left[\begin{array}{c}1\\2\\3\end{array}\right],
\left[\begin{array}{c}2\\4\\6\end{array}\right],
\left[\begin{array}{c}0\\-2\\-2\end{array}\right],
\left[\begin{array}{c}1\\0\\1\end{array}\right]
}.
\end{equation*}
To do so, we use the results of the calculation:
\begin{equation*}
\RREF
\left[\begin{array}{cccc}
1 & 2 & 0 & 1 \\
2 & 4 & -2 & 0 \\
3 & 6 & -2 & 1 \\
\end{array}\right]
=
\left[\begin{array}{cccc}
\markedPivot{1} & 2 & 0 & 1 \\
0 & 0 & \markedPivot{1} & 1 \\
0 & 0 & 0 & 0
\end{array}\right]
\end{equation*}
to conclude that the set \(\setList{
\left[\begin{array}{c}1\\2\\3\end{array}\right],
\left[\begin{array}{c}0\\-2\\-2\end{array}\right]
}\text{,}\) the set of vectors corresponding to the pivot columns of the RREF, is a basis for \(W\text{.}\)
(a)
Explain why neither of the vectors \(\left[\begin{array}{c}1\\0\\0\end{array}\right],
\left[\begin{array}{c}0\\1\\0\end{array}\right]\) are elements of \(W\text{.}\)
(b)
Explain why this shows that, in general, when we calculate a basis for \(W=\vspan\{\vec{v}_1,\dots, \vec{v}_n\}\text{,}\) the pivot columns of \(\RREF[\vec{v}_1\dots \vec{v}_n]\) themselves do not form a basis for \(W\text{.}\)
Subsection 2.6.4 Videos
Subsection 2.6.5 Exercises
Exercises available at
https://tbil.org/preview/linear-algebra/exercises/#/bank/EV6/
Subsection 2.6.6 Mathematical Writing Explorations
Exploration 2.6.12.
Prove each of the following statements is true.
-
If \(\{\vec{b}_1, \vec{b}_2,\ldots, \vec{b}_m\}\) and \(\{\vec{c}_1,\vec{c}_2,\ldots,\vec{c}_n\}\) are each a basis for a vector space \(V\text{,}\) then \(m=n.\)
-
If \(\{\vec{v}_1,\vec{v}_2\ldots, \vec{v}_n\}\) is linearly independent, then so is \(\{\vec{v}_1,\vec{v}_1 + \vec{v}_2, \ldots, \vec{v}_1 + \vec{v}_2 + \cdots + \vec{v}_n\}\text{.}\)
-
Let \(V\) be a vector space of dimension \(n\text{,}\) and \(\vec{v} \in V\text{.}\) Then there exists a basis for \(V\) which contains \(\vec{v}\text{.}\)
Exploration 2.6.13.
Suppose we have the set of all function \(f:S \rightarrow \mathbb{R}\text{.}\) We claim that this is a vector space under the usual operation of function addition and scalar multiplication. What is the dimension of this space for each choice of \(S\) below:
-
\(\displaystyle S = \{1\}\)
-
\(\displaystyle S = \{1,2\}\)
-
\(\displaystyle S = \{1,2,\ldots ,n\}\)
-
\(\displaystyle S = \mathbb{R}\)
Exploration 2.6.14.
Suppose you have the vector space \(V = \left\{\left(\begin{array}{c}x\\y\\z\end{array}\right)\in \mathbb{R}^3: x + y + z = 1\right\}\) with the operations \(\left(\begin{array}{c}x_1\\y_1\\z_1\end{array}\right) \oplus \left(\begin{array}{c}x_2\\y_2\\z_2\end{array}\right) = \left(\begin{array}{c}x_1 + x_2 - 1\\y_1 + y_2\\z_1+z_2\end{array}\right) \mbox{ and } \alpha\odot\left(\begin{array}{c}x_1\\y_1\\z_1\end{array}\right) = \left(\begin{array}{c}\alpha x_1 - \alpha +1\\\alpha y_1\\\alpha z_1\end{array}\right).\) Find a basis for \(V\) and determine it’s dimension.
Subsection 2.6.7 Sample Problem and Solution
Sample problem Example B.1.10.
