Skip to main content
Logo image

Linear Algebra for Team-Based Inquiry Learning PREVIEW Edition

Steven Clontz, Drew Lewis

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)
What do the spaces \(\vspan{S}\) and \(\IR^2\) have in common? In what ways do they differ?
🔗
🔗
🔗
🔗

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?
  1. The third column.
    🔗
  2. The fourth column.
    🔗
  3. The third row.
    🔗
  4. The fourth row.
    🔗
🔗
Answer.
A.
🔗
The third columns lacks a pivot, introducing a free variable that prevents uniqueness of linear combinations.
🔗
🔗
🔗
(b)
If we removed the vector that causes this issue, what could we say about that set of three vectors?
  1. The set spans the vector space \(\IR^4\text{,}\) but remains linearly dependent.
    🔗
  2. The set spans subspace \(W\subset \IR^4\text{,}\) but remains linearly dependent.
    🔗
  3. The set spans subspace \(W\subset \IR^4\text{,}\) and is now linearly independent.
    🔗
  4. The set not longer spans the subspace \(W\subset \IR^4\text{,}\) but is now linearly independent.
    🔗
🔗
Answer.
C.
🔗
Because the removed vector was already a linear combination of the others, we still span \(W\text{.}\) Now that all vectors yield pivot columns, the set is now 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 the basis for a subspace is not unique in general.
🔗
🔗

Definition 2.6.9.

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 2.6.10.

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 2.6.11.

The dimension of a subspace may be found by doing what with an appropriate RREF matrix?
  1. Count the rows.
    🔗
  2. Count the non-pivot columns.
    🔗
  3. Count the pivots.
    🔗
  4. Add the number of pivot rows and pivot columns.
    🔗
🔗
🔗
🔗

Subsection 2.6.3 Individual Practice

Activity 2.6.12.

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

Figure 22. Video: Finding a basis of a subspace and computing the dimension of a subspace
🔗
🔗

Subsection 2.6.5 Exercises

🔗

Subsection 2.6.6 Mathematical Writing Explorations

Exploration 2.6.13.

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 2.6.14.

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 2.6.15.

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.
🔗
🔗
🔗
PrevTopNext