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jkostiuk merged 7 commits into from
Feb 18, 2026
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Update EV6 class activities to clarify spanning R^n and spanning subspace #933

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Update EV6 class activities to clarify spanning R^n and spanning subs…
StevenClontz Feb 18, 2026
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tweaks, rewordings, remove stale comments
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add example of subspace dimension
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177 changes: 73 additions & 104 deletions source/linear-algebra/source/02-EV/06.ptx
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Expand Up @@ -43,19 +43,20 @@
</subsection>

<subsection><title>Class Activities</title>

<observation>
<p>
Recall from section <xref ref="EV3"></xref> that a <term>subspace</term> of a vector space is
the result of spanning a set of vectors from that vector space.
In <xref ref="EV3-planar-subspace-vs-r2" /> we saw an example of
two linearly independent vectors spanning a planar subspace of <m>\IR^3</m>.
</p>
<p>
Recall also that a linearly dependent set contains <q>redundant</q> vectors. For example,
only two of the three vectors in <xref ref="EV4-figure-linearly-dependent"/> are needed to span
the planar subspace.
Because these independent vectors fail to span <m>\IR^3</m>, they are not
a basis for <m>\IR^3</m>. However, they still span a <em>subspace</em> of
<m>\IR^3</m>...
</p>
</observation>

<activity estimated-time='10'>
<activity>
<introduction>
<p>
Consider the subspace of <m>\IR^4</m> given by <m>W=\vspan\left\{
Expand All @@ -68,33 +69,61 @@ the planar subspace.
</p>
</introduction>
<task>
<p>
Mark the column of <m>\RREF\left[\begin{array}{cccc}
2&amp;2&amp;2&amp;1\\
3&amp;0&amp;-3&amp;5\\
0&amp;1&amp;2&amp;-1\\
1&amp;-1&amp;-3&amp;0
\end{array}\right]</m> that shows that <m>W</m>'s spanning set
is linearly dependent.
</p>
<statement>
<p>
Which feature of <md>\RREF\left[\begin{array}{cccc}
2&amp;2&amp;2&amp;1\\
3&amp;0&amp;-3&amp;5\\
0&amp;1&amp;2&amp;-1\\
1&amp;-1&amp;-3&amp;0
\end{array}\right]=
\left[\begin{array}{cccc}
1&amp;0&amp;-1&amp;0\\
0&amp;1&amp; 2&amp;0\\
0&amp;0&amp; 0&amp;1\\
0&amp;0&amp; 0&amp;0
\end{array}\right]</md> shows that <m>W</m>'s spanning set
is linearly dependent?
<ol marker="A." cols="2">
<li>The third column.</li>
<li>The fourth column.</li>
<li>The third row.</li>
<li>The fourth row.</li>
</ol>
</p>
</statement>
<answer>
<p>A.</p>
<p>
The third columns lacks a pivot, introducing a free variable that
prevents uniqueness of linear combinations.
</p>
</answer>
</task>
<task>
<p>
What would be the result of removing the vector that gave us this column?
<ol marker="A.">
<li>The set still spans <m>W</m>, and remains linearly dependent.</li>
<li>The set still spans <m>W</m>, but is now also linearly independent.</li>
<li>The set no longer spans <m>W</m>, and remains linearly dependent.</li>
<li>The set no longer spans <m>W</m>, but is now linearly independent.</li>
</ol>
</p>
</task>
<statement>
<p>
If we removed the vector that causes this issue,
what could we say about that set of three vectors?
<ol marker="A.">
<li>The set spans the vector space <m>\IR^4</m>, but remains linearly dependent.</li>
<li>The set spans the subspace <m>W</m> of <m>\IR^4</m>, but remains linearly dependent.</li>
<li>The set spans the subspace <m>W</m> of <m>\IR^4</m>, and is now linearly independent.</li>
<li>The set no longer spans the subspace <m>W</m> of <m>\IR^4</m>, but is now linearly independent.</li>
</ol>
</p>
</statement>
<answer>
<p>C.</p>
<p>
Because the removed vector was already a linear combination of the others,
we still span <m>W</m>. Now that all vectors yield pivot columns, the set
is now independent.
</p>
</answer>
</task>
</activity>

<sage language="octave">
<input>rref([2,2,2,1; 3,0,-3,5; 0,1,2,-1; 1,-1,-3,0])</input>
</sage>

<definition>
<statement>
<p>
Expand Down Expand Up @@ -145,55 +174,6 @@ What would be the result of removing the vector that gave us this column?
</p>
</observation>

<!-- <activity estimated-time='10'>
<statement>
<p>
Let <m>W</m> be the subspace of <m>\IR^4</m> given by
<me>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\} </me>.
Find a basis for <m>W</m>.
</p>
</statement>
</activity>

<sage language="octave">
</sage> -->

<!-- <activity estimated-time='10'>
<statement>
<p>
Let <m>W</m> be the subspace of <m>\P_3</m> given by
<me>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\} </me>
Find a basis for <m>W</m>.
</p>
</statement>
</activity> -->

<!-- <sage language="octave">
</sage> -->

<!-- <activity estimated-time='10'>
<statement>
<p>
Let <m>W</m> be the subspace of <m>M_{2,2}</m> given by
<me>W = \vspan \left\{
\left[\begin{array}{cc} 1 &amp; 3 \\ 1 &amp; -1 \end{array}\right],
\left[\begin{array}{cc} 2 &amp; -1 \\ 1 &amp; 2 \end{array}\right],
\left[\begin{array}{cc} 4 &amp; 5 \\ 3 &amp; 0 \end{array}\right],
\left[\begin{array}{cc} 3 &amp; 2 \\ 2 &amp; 1 \end{array}\right]
\right\}. </me>
Find a basis for <m>W</m>.
</p>
</statement>
</activity>

<sage language="octave">
</sage> -->

<activity estimated-time='10'>
<task>
<p>
Expand Down Expand Up @@ -248,19 +228,22 @@ T=\left\{
</me>.
</p>
<p>
Thus the basis for a subspace is not unique in general.
Thus a given basis for a subspace need not be unique.
</p>
</observation>


<fact>
<definition>
<statement>
<p>
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 <term>dimension</term> of a vector space or subspace is equal to the
size of any basis for the vector space.
</p>
<p>
For example,
As you'd expect, <m>\IR^n</m> has dimension <m>n</m>.
For example, <m>\IR^3</m> has dimension <m>3</m> because any basis for <m>\IR^3</m>
such as
<me>
\setList{\vec e_1,\vec e_2,\vec e_3}
\text{ and }
Expand All @@ -276,37 +259,23 @@ Thus the basis for a subspace is not unique in general.
\left[\begin{array}{c}3\\-2\\5\end{array}\right]
}
</me>
are all valid bases for <m>\IR^3</m>, and they all contain three vectors.
</p>
</statement>
</fact>

<definition>
<statement>
<p>
The <term>dimension</term> of a vector space or subspace is equal to the size
of any basis for the vector space.
contains exactly three vectors.
</p>
<p>
As you'd expect, <m>\IR^n</m> has dimension <m>n</m>.
For example, <m>\IR^3</m> has dimension <m>3</m> because any basis for <m>\IR^3</m>
such as
Likewise, the planar subspace with the following two bases
<me>
\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]
\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\\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]
\left[\begin{array}{c}-1\\2\\8\end{array}\right],
\left[\begin{array}{c}0\\4\\11\end{array}\right]
}
</me>
contains exactly three vectors.
has dimension <m>2</m> because any basis for the subspace
will have exactly two vectors.
</p>
</statement>
</definition>
Expand Down