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jkostiuk merged 1 commit into from
Feb 25, 2026
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Align EV7 activities with EV6 #936

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129 changes: 74 additions & 55 deletions source/linear-algebra/source/02-EV/07.ptx
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Original file line number Diff line number Diff line change
Expand Up @@ -125,49 +125,75 @@ Find its solution set (a subspace of <m>\IR^4</m>).
</task>
<task>
<p>
Rewrite this solution space in the form <me>\setBuilder{ a \left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown\end{array}\right] + b \left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown \end{array}\right] }{a,b \in \IR}.</me>
Rewrite this solution space in the following forms
<md>\setBuilder{ a \left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown\end{array}\right] + b \left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown \end{array}\right] }{a,b \in \IR}</md><md>=
\vspan \left\{\left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown\end{array}\right], \left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown \end{array}\right]\right\}</md>.
</p>
</task>
<task>
<statement>
<p>
Which of these choices best describes the set of two vectors
<m>\left\{\left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown\end{array}\right], \left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown \end{array}\right]\right\}</m>
used in this solution space?
So, how can we use this to find a basis for the solution space?
<ol marker="A.">
<li>
<p>
The set is linearly dependent.
Take RREF of an appropriate matrix and remove the vectors that caused
a non-pivot row.
</p>
</li>
</li>
<li>
<p>
The set is linearly independent.
Take RREF of an appropriate matrix and remove the vectors that caused
a non-pivot column.
</p>
</li>
<li>
<p>
The set spans the solution space.
</p>
</li>
<li>
</li>
<li>
<p>
The set is a basis of the solution space.
Take RREF of an appropriate matrix and remove the vectors that caused
a non-pivot row or column.
</p>
</li>
</li>
<li>
<p>
This set cannot be a basis for the solution space because it will always
have in a zero row in the RREF.
</p>
</li>
</ol>
</p>
</statement>
<answer>
<p>
D.
B.
</p>
<p>
This is exactly the technique used in <xref ref="EV6"/> to find
a basis from a set of spanning vectors.
</p>
</answer>
</task>


</activity>

<observation>
<p>
To find a basis for the subspace
<md>
\vspan \left\{\left[\begin{array}{c} -2 \\ 1 \\ 0 \\ 0\end{array}\right], \left[\begin{array}{c} -1 \\ 0 \\ -4 \\ 1 \end{array}\right]\right\}
</md>
we may compute
<md>
\left[\begin{array}{cc} -2 &amp; -1 \\ 1 &amp; 0 \\ 0 &amp; -4 \\ 0 &amp; 1\end{array}\right] \sim
\left[\begin{array}{cc} 1 &amp; 0 \\ 0 &amp; 1 \\ 0 &amp; 0 \\ 0 &amp; 0\end{array}\right]
</md>.
Because all columns are pivot columns, the set
<md>
\left\{\left[\begin{array}{c} -2 \\ 1 \\ 0 \\ 0\end{array}\right], \left[\begin{array}{c} -1 \\ 0 \\ -4 \\ 1 \end{array}\right]\right\}
</md>
is already linearly independent and therefore already a basis for the subspace.
</p>
</observation>



<sage language="octave">
</sage>
Expand Down Expand Up @@ -195,44 +221,18 @@ Find its solution set (a subspace of <m>\IR^7</m>).
</task>
<task>
<p>
Rewrite this solution space in the form <me>\setBuilder{ a \left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown\\ \unknown\\ \unknown \\ \unknown\end{array}\right] + b \left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown\\ \unknown\\ \unknown \\ \unknown\end{array}\right]+c \left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown\\ \unknown\\ \unknown \\ \unknown\end{array}\right]+d \left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown\\ \unknown\\ \unknown \\ \unknown\end{array}\right] }{a,b,c,d \in \IR}.</me>
Rewrite this solution space in the following forms: <md>\setBuilder{ a \left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown\\ \unknown\\ \unknown \\ \unknown\end{array}\right] + b \left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown\\ \unknown\\ \unknown \\ \unknown\end{array}\right]+c \left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown\\ \unknown\\ \unknown \\ \unknown\end{array}\right]+d \left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown\\ \unknown\\ \unknown \\ \unknown\end{array}\right] }{a,b,c,d \in \IR}.</md>
<md>
=\vspan\left\{\left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown\\ \unknown\\ \unknown \\ \unknown\end{array}\right], \left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown\\ \unknown\\ \unknown \\ \unknown\end{array}\right],\left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown\\ \unknown\\ \unknown \\ \unknown\end{array}\right],\left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown\\ \unknown\\ \unknown \\ \unknown\end{array}\right]\right\}
</md>.
</p>
</task>
<task>
<statement>
<p>
Which of these choices best describes the set of vectors
<m>\left\{\left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown\\ \unknown\\ \unknown \\ \unknown\end{array}\right], \left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown\\ \unknown\\ \unknown \\ \unknown\end{array}\right],\left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown\\ \unknown\\ \unknown \\ \unknown\end{array}\right],\left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown\\ \unknown\\ \unknown \\ \unknown\end{array}\right]\right\}</m>
used in this solution space?
<ol marker="A.">
<li>
<p>
The set is linearly dependent.
</p>
</li>
<li>
<p>
The set is linearly independent.
</p>
</li>
<li>
<p>
The set spans the solution space.
</p>
</li>
<li>
<p>
The set is a basis for the solution space.
</p>
</li>
</ol>
Find a basis for this solution space.
</p>
</statement>
<answer>
<p>
D.
</p>
</answer>
</task>
</activity>

Expand Down Expand Up @@ -318,6 +318,10 @@ solution space?
<p>
A.
</p>
<p>
In <xref ref="EV4"/> we established that sets that contain the zero
vector cannot be independent.
</p>
</answer>
</task>
</activity>
Expand All @@ -333,31 +337,46 @@ To create a computer-animated film, an animator first models a scene
as a subset of <m>\mathbb R^3</m>. Then to transform this three-dimensional
visual data for display on a two-dimensional movie screen or television set,
the computer could apply a linear transformation that maps visual information
at the point <m>(x,y,z)\in\mathbb R^3</m> onto the pixel located at
<m>(x+y,y-z)\in\mathbb R^2</m>.
at the vector <m>\left[\begin{array}{c}x\\y\\z\end{array}\right]\in\mathbb R^3</m> onto the pixel located at
<m>\left[\begin{array}{c}x+y\\y-z\end{array}\right]\in\mathbb R^2</m>.
</p>
</introduction>
<task>
<statement>
<p>
What homogeneous linear system describes the positions <m>(x,y,z)</m>
What homogeneous linear system describes the positions
<m>\left[\begin{array}{c}x\\y\\z\end{array}\right]</m>
within the original scene that would be aligned with the
pixel <m>(0,0)</m> on the screen?
pixel <m>\left[\begin{array}{c}0\\0\end{array}\right]</m> on the screen?
</p>
</statement>
<answer>
<p><md>x+y=0</md><md>y-z=0</md></p>
</answer>
</task>
<task>
<statement>
<p>
Solve this system to describe these locations.
Find a basis for the solution set of this system. What best describes the
shape of the data that gets projected onto this single point of the screen?
<ol marker="A.">
<li><p>A point</p></li>
<li><p>A line</p></li>
<li><p>A plane</p></li>
</ol>
</p>
</statement>
<answer>
<p>B.</p>
<p>The basis <m>\left\{\left[\begin{array}{c}-1\\1\\1\end{array}\right]\right\}</m> describes a one-dimensional line.</p>
</answer>
</task>
</activity>
<sage language="octave">
</sage>



</subsection>

<subsection>
Expand Down