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Linear Algebra for Team-Based Inquiry Learning
2022 Edition
Steven Clontz
Drew Lewis
University of South Alabama
University of South Alabama
August 2, 2022
Section 2.7: Subspace Basis and Dimension (VS7) Observation 2.7.1
Recall from section Section 2.4 Subspaces (VS4) 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.
Figure 1. A linearly dependent set of three vectors Activity 2.7.2 (~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 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.
Activity 2.7.2 (~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.
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 (~10 min)
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 (~10 min)
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 (~10 min)
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 (~10 min)
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*}
Part 1.
Find a basis for \(\vspan S\text{.}\)
Activity 2.7.7 (~10 min)
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*}
Part 2.
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 (~10 min)
Find the dimension of each subspace of \(\IR^4\) by finding \(\RREF\) for each corresponding matrix.
Part 1.
\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*}
Activity 2.7.11 (~10 min)
Find the dimension of each subspace of \(\IR^4\) by finding \(\RREF\) for each corresponding matrix.
Part 2.
\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*}
Activity 2.7.11 (~10 min)
Find the dimension of each subspace of \(\IR^4\) by finding \(\RREF\) for each corresponding matrix.
Part 3.
\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*}
Activity 2.7.11 (~10 min)
Find the dimension of each subspace of \(\IR^4\) by finding \(\RREF\) for each corresponding matrix.
Part 4.
\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*}
