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\newcommand{\IR}{\mathbb{R}}
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Linear Algebra for Team-Based Inquiry Learning
2023 Edition
| Steven Clontz |
Drew Lewis |
| University of South Alabama |
|
|
|
August 24, 2023
Section 2.6: Subspace Basis and Dimension (EV6)
Observation 2.6.1
Recall from section Section 2.3 Subspaces (EV3) that a subspace of a vector space is the result of spanning a set of vectors from that space.
Recall also that a linearly dependent set contains “redundant” vectors. For example, only two of the three vectors in Figure 14 are needed to span the planar subspace.
Activity 2.6.1 (~10 min)
Consider the subspace of \(\IR^4\) given by \(W=\vspan\left\{
\left[\begin{array}{c}2\\3\\0\\1\end{array}\right],
\left[\begin{array}{c}2\\0\\1\\-1\end{array}\right],
\left[\begin{array}{c}2\\-3\\2\\-3\end{array}\right],
\left[\begin{array}{c}1\\5\\-1\\0\end{array}\right]
\right\}
\text{.}\)
Part 1.
Mark the column 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.
Activity 2.6.1 (~10 min)
Consider the subspace of \(\IR^4\) given by \(W=\vspan\left\{
\left[\begin{array}{c}2\\3\\0\\1\end{array}\right],
\left[\begin{array}{c}2\\0\\1\\-1\end{array}\right],
\left[\begin{array}{c}2\\-3\\2\\-3\end{array}\right],
\left[\begin{array}{c}1\\5\\-1\\0\end{array}\right]
\right\}
\text{.}\)
Part 2.
What would be the result of removing the vector that gave us this column?
- The set still spans \(W\text{,}\) and remains linearly dependent.
- The set still spans \(W\text{,}\) but is now also linearly independent.
- The set no longer spans \(W\text{,}\) and remains linearly dependent.
- The set no longer spans \(W\text{,}\) but is now linearly independent.
Definition 2.6.2
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 2.6.3
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 & 2 \\
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
\(W=\vspan\setList{
\left[\begin{array}{c}1\\2\\3\end{array}\right],
\left[\begin{array}{c}2\\4\\6\end{array}\right],
\left[\begin{array}{c}0\\-2\\-2\end{array}\right],
\left[\begin{array}{c}1\\2\\1\end{array}\right]
}\) 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 2.6.2 (~10 min)
Part 1.
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*}
Activity 2.6.2 (~10 min)
Part 2.
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 2.6.4
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 the basis for a subspace is not unique in general.
Fact 2.6.5
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 2.6.6
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.
Activity 2.6.3
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*}
Part 1.
Explain and demonstrate how to find a basis of \(W\text{.}\)
Activity 2.6.3
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*}
Part 2.
Explain and demonstrate how to find the dimension of \(W\text{.}\)