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Update EV6 class activities to clarify spanning R^n and spanning subspace (#933)
* Update EV6 class activities to clarify spanning R^n and spanning subspace * Apply suggestion from @StevenClontz * tweaks, rewordings, remove stale comments * add example of subspace dimension * Apply suggestion from @StevenClontz * combine fact+definition for dimension * Apply suggestion from @jkostiuk Co-authored-by: jkostiuk <kostiuk.jordan@gmail.com> --------- Co-authored-by: jkostiuk <kostiuk.jordan@gmail.com>
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‎source/linear-algebra/source/02-EV/06.ptx‎

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</subsection>
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<subsection><title>Class Activities</title>
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<observation>
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<p>
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Recall from section <xref ref="EV3"></xref> that a <term>subspace</term> of a vector space is
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the result of spanning a set of vectors from that vector space.
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In <xref ref="EV3-planar-subspace-vs-r2" /> we saw an example of
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two linearly independent vectors spanning a planar subspace of <m>\IR^3</m>.
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</p>
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<p>
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Recall also that a linearly dependent set contains <q>redundant</q> vectors. For example,
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only two of the three vectors in <xref ref="EV4-figure-linearly-dependent"/> are needed to span
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the planar subspace.
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Because these independent vectors fail to span <m>\IR^3</m>, they are not
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a basis for <m>\IR^3</m>. However, they still span a <em>subspace</em> of
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<m>\IR^3</m>...
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</p>
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</observation>
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<activity estimated-time='10'>
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<activity>
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<introduction>
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<p>
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Consider the subspace of <m>\IR^4</m> given by <m>W=\vspan\left\{
@@ -68,33 +69,61 @@ the planar subspace.
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</p>
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</introduction>
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<task>
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<p>
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Mark the column of <m>\RREF\left[\begin{array}{cccc}
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2&amp;2&amp;2&amp;1\\
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3&amp;0&amp;-3&amp;5\\
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0&amp;1&amp;2&amp;-1\\
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1&amp;-1&amp;-3&amp;0
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\end{array}\right]</m> that shows that <m>W</m>'s spanning set
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is linearly dependent.
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</p>
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<statement>
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<p>
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Which feature of <md>\RREF\left[\begin{array}{cccc}
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2&amp;2&amp;2&amp;1\\
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3&amp;0&amp;-3&amp;5\\
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0&amp;1&amp;2&amp;-1\\
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1&amp;-1&amp;-3&amp;0
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\end{array}\right]=
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\left[\begin{array}{cccc}
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1&amp;0&amp;-1&amp;0\\
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0&amp;1&amp; 2&amp;0\\
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0&amp;0&amp; 0&amp;1\\
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0&amp;0&amp; 0&amp;0
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\end{array}\right]</md> shows that <m>W</m>'s spanning set
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is linearly dependent?
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<ol marker="A." cols="2">
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<li>The third column.</li>
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<li>The fourth column.</li>
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<li>The third row.</li>
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<li>The fourth row.</li>
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</ol>
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</p>
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</statement>
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<answer>
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<p>A.</p>
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<p>
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The third columns lacks a pivot, introducing a free variable that
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prevents uniqueness of linear combinations.
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</p>
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</answer>
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</task>
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<task>
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<p>
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What would be the result of removing the vector that gave us this column?
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<ol marker="A.">
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<li>The set still spans <m>W</m>, and remains linearly dependent.</li>
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<li>The set still spans <m>W</m>, but is now also linearly independent.</li>
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<li>The set no longer spans <m>W</m>, and remains linearly dependent.</li>
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<li>The set no longer spans <m>W</m>, but is now linearly independent.</li>
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</ol>
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</p>
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</task>
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<statement>
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<p>
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If we removed the vector that causes this issue,
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what could we say about that set of three vectors?
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<ol marker="A.">
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<li>The set spans the vector space <m>\IR^4</m>, but remains linearly dependent.</li>
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<li>The set spans the subspace <m>W</m> of <m>\IR^4</m>, but remains linearly dependent.</li>
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<li>The set spans the subspace <m>W</m> of <m>\IR^4</m>, and is now linearly independent.</li>
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<li>The set no longer spans the subspace <m>W</m> of <m>\IR^4</m>, but is now linearly independent.</li>
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</ol>
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</p>
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</statement>
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<answer>
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<p>C.</p>
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<p>
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Because the removed vector was already a linear combination of the others,
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we still span <m>W</m>. Now that all vectors yield pivot columns, the set
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is now independent.
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</p>
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</answer>
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</task>
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</activity>
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<sage language="octave">
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<input>rref([2,2,2,1; 3,0,-3,5; 0,1,2,-1; 1,-1,-3,0])</input>
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</sage>
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<definition>
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<statement>
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<p>
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</p>
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</observation>
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<!-- <activity estimated-time='10'>
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<statement>
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<p>
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Let <m>W</m> be the subspace of <m>\IR^4</m> given by
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<me>W = \vspan \left\{
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\left[\begin{array}{c} 1 \\ 3 \\ 1 \\ -1 \end{array}\right],
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\left[\begin{array}{c} 2 \\ -1 \\ 1 \\ 2 \end{array}\right],
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\left[\begin{array}{c} 4 \\ 5 \\ 3 \\ 0 \end{array}\right],
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\left[\begin{array}{c} 3 \\ 2 \\ 2 \\ 1 \end{array}\right]
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\right\} </me>.
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Find a basis for <m>W</m>.
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</p>
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</statement>
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</activity>
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<sage language="octave">
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</sage> -->
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<!-- <activity estimated-time='10'>
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<statement>
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<p>
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Let <m>W</m> be the subspace of <m>\P_3</m> given by
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<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>
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Find a basis for <m>W</m>.
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</p>
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</statement>
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</activity> -->
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<!-- <sage language="octave">
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</sage> -->
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<!-- <activity estimated-time='10'>
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<statement>
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<p>
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Let <m>W</m> be the subspace of <m>M_{2,2}</m> given by
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<me>W = \vspan \left\{
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\left[\begin{array}{cc} 1 &amp; 3 \\ 1 &amp; -1 \end{array}\right],
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\left[\begin{array}{cc} 2 &amp; -1 \\ 1 &amp; 2 \end{array}\right],
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\left[\begin{array}{cc} 4 &amp; 5 \\ 3 &amp; 0 \end{array}\right],
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\left[\begin{array}{cc} 3 &amp; 2 \\ 2 &amp; 1 \end{array}\right]
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\right\}. </me>
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Find a basis for <m>W</m>.
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</p>
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</statement>
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</activity>
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<sage language="octave">
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</sage> -->
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<activity estimated-time='10'>
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<task>
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<p>
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</me>.
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</p>
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<p>
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Thus the basis for a subspace is not unique in general.
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Thus a given basis for a subspace need not be unique.
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</p>
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</observation>
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<fact>
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<definition>
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<statement>
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<p>
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Any non-trivial real vector space has infinitely-many different bases, but all
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the bases for a given vector space are exactly the same size.
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So we say the <term>dimension</term> of a vector space or subspace is equal to the
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size of any basis for the vector space.
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</p>
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<p>
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For example,
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As you'd expect, <m>\IR^n</m> has dimension <m>n</m>.
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For example, <m>\IR^3</m> has dimension <m>3</m> because any basis for <m>\IR^3</m>
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such as
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<me>
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\setList{\vec e_1,\vec e_2,\vec e_3}
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\text{ and }
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\left[\begin{array}{c}3\\-2\\5\end{array}\right]
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}
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</me>
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are all valid bases for <m>\IR^3</m>, and they all contain three vectors.
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</p>
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</statement>
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</fact>
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<definition>
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<statement>
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<p>
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The <term>dimension</term> of a vector space or subspace is equal to the size
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of any basis for the vector space.
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contains exactly three vectors.
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</p>
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<p>
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As you'd expect, <m>\IR^n</m> has dimension <m>n</m>.
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For example, <m>\IR^3</m> has dimension <m>3</m> because any basis for <m>\IR^3</m>
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such as
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Likewise, the planar subspace with the following two bases
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<me>
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\setList{\vec e_1,\vec e_2,\vec e_3}
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\text{ and }
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\setList{
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\left[\begin{array}{c}1\\0\\0\end{array}\right],
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\left[\begin{array}{c}0\\1\\0\end{array}\right],
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\left[\begin{array}{c}1\\1\\1\end{array}\right]
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\left[\begin{array}{c}1\\2\\3\end{array}\right],
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\left[\begin{array}{c}-2\\0\\5\end{array}\right]
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}
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\text{ and }
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\setList{
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\left[\begin{array}{c}1\\0\\-3\end{array}\right],
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\left[\begin{array}{c}2\\-2\\1\end{array}\right],
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\left[\begin{array}{c}3\\-2\\5\end{array}\right]
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\left[\begin{array}{c}-1\\2\\8\end{array}\right],
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\left[\begin{array}{c}0\\4\\11\end{array}\right]
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}
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</me>
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contains exactly three vectors.
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has dimension <m>2</m> because any basis for the subspace
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will have exactly two vectors.
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</p>
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</statement>
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</definition>

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