Learning Outcomes
- Compute a basis for the subspace spanned by a given set of Euclidean vectors, and determine the dimension of the subspace.
Section 2.7 Subspace Basis and Dimension (VS7)
Subsection 2.7.1 Class Activities
Observation 2.7.1.
Recall from section Section 2.4 that a subspace of a vector space is a subset that is itself a vector space.
One easy way to construct a subspace is to take the span of set, but a linearly dependent set contains “redundant” vectors. For example, only two of the three vectors in the following image are needed to span the planar subspace.
Activity 2.7.2.
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)
Mark the part of \(\RREF\left[\begin{array}{cccc}
2&2&2&1\\
3&0&-3&5\\
0&1&2&-1\\
1&-1&-3&0
\end{array}\right]\) that shows that \(W\)'s spanning set is linearly dependent.
(b)
Find a basis for \(W\) by removing a vector from its spanning set to make it linearly independent.
Fact 2.7.3.
Let \(S=\{\vec v_1,\dots,\vec v_m\}\text{.}\) The easiest basis describing \(\vspan S\) is the set of vectors in \(S\) given by the pivot columns of \(\RREF[\vec v_1\,\dots\,\vec v_m]\text{.}\)
Put another way, 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}{ccc}
1 & 2 & 3 \\
0 & -2 & -2 \\
-3 & 1 & -2
\end{array}\right]
=
\left[\begin{array}{ccc}
\circledNumber{1} & 0 & 1 \\
0 & \circledNumber{1} & 1 \\
0 & 0 & 0
\end{array}\right]
\end{equation*}
the subspace \(W=\vspan\setList{
\left[\begin{array}{c}1\\0\\-3\end{array}\right],
\left[\begin{array}{c}2\\-2\\1\end{array}\right],
\left[\begin{array}{c}3\\-2\\-2\end{array}\right]
}\) has \(\setList{
\left[\begin{array}{c}1\\0\\-3\end{array}\right],
\left[\begin{array}{c}2\\-2\\1\end{array}\right]
}\) as a basis.
Activity 2.7.4.
Let \(W\) be the subspace of \(\IR^4\) given by
\begin{equation*}
W = \vspan \left\{
\left[\begin{array}{c} 1 \\ 3 \\ 1 \\ -1 \end{array}\right],
\left[\begin{array}{c} 2 \\ -1 \\ 1 \\ 2 \end{array}\right],
\left[\begin{array}{c} 4 \\ 5 \\ 3 \\ 0 \end{array}\right],
\left[\begin{array}{c} 3 \\ 2 \\ 2 \\ 1 \end{array}\right]
\right\} \text{.}
\end{equation*}
Find a basis for \(W\text{.}\)
Activity 2.7.5.
Let \(W\) be the subspace of \(\P_3\) given by
\begin{equation*}
W = \vspan \left\{x^3+3x^2+x-1, 2x^3-x^2+x+2, 4x^3+5x^2+3x, 3x^3+2x^2+2x+1 \right\}
\end{equation*}
Find a basis for \(W\text{.}\)
Activity 2.7.6.
Let \(W\) be the subspace of \(M_{2,2}\) given by
\begin{equation*}
W = \vspan \left\{
\left[\begin{array}{cc} 1 & 3 \\ 1 & -1 \end{array}\right],
\left[\begin{array}{cc} 2 & -1 \\ 1 & 2 \end{array}\right],
\left[\begin{array}{cc} 4 & 5 \\ 3 & 0 \end{array}\right],
\left[\begin{array}{cc} 3 & 2 \\ 2 & 1 \end{array}\right]
\right\}.
\end{equation*}
Find a basis for \(W\text{.}\)
Activity 2.7.7.
Let
\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\}
\end{equation*}
and
\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*}
(a)
Find a basis for \(\vspan S\text{.}\)
(b)
Find a basis for \(\vspan T\text{.}\)
Observation 2.7.8.
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*}
Fact 2.7.9.
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.
For example,
\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*}
are all valid bases for \(\IR^3\text{,}\) and they all contain three vectors.
Definition 2.7.10.
The dimension of a vector space 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.
Activity 2.7.11.
Find the dimension of each subspace of \(\IR^4\) by finding \(\RREF\) for each corresponding matrix.
(a)
\begin{equation*}
\vspan\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}4\\3\\0\\2\end{array}\right],
\left[\begin{array}{c}-3\\0\\1\\3\end{array}\right]
\right\}
\end{equation*}
(b)
\begin{equation*}
\vspan\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\\2\end{array}\right]
\right\}
\end{equation*}
(c)
\begin{equation*}
\vspan\left\{
\left[\begin{array}{c}2\\3\\0\\-1\end{array}\right],
\left[\begin{array}{c}4\\3\\0\\2\end{array}\right],
\left[\begin{array}{c}-3\\0\\1\\3\end{array}\right],
\left[\begin{array}{c}3\\6\\1\\5\end{array}\right]
\right\}
\end{equation*}
(d)
\begin{equation*}
\vspan\left\{
\left[\begin{array}{c}5\\3\\0\\-1\end{array}\right],
\left[\begin{array}{c}-2\\1\\0\\3\end{array}\right],
\left[\begin{array}{c}4\\5\\1\\3\end{array}\right]
\right\}
\end{equation*}
Subsection 2.7.2 Videos
Subsection 2.7.3 Slideshow
Slideshow of activities available at
https://teambasedinquirylearning.github.io/linear-algebra/2023e/VS7.slides.html.Exercises 2.7.4 Exercises
Exercises available at
https://teambasedinquirylearning.github.io/linear-algebra/2023e/exercises/#/bank/VS7/.Subsection 2.7.5 Mathematical Writing Explorations
Exploration 2.7.12.
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 2.7.13.
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 2.7.14.
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 2.7.6 Sample Problem and Solution
Sample problem Example B.1.11.
