Skip to content
New issue

Have a question about this project? Sign up for a free GitHub account to open an issue and contact its maintainers and the community.

Sign up for GitHub

By clicking “Sign up for GitHub”, you agree to our terms of service and privacy statement. We’ll occasionally send you account related emails.

Already on GitHub? Sign in to your account

Jump to bottom

update ev5 sample #359

Merged
merged 1 commit into from
Oct 15, 2024
Merged

update ev5 sample #359

Changes from all commits
Commits
Show all changes
1 commit
Select commit
File filter

Filter by extension

Filter by extension
Conversations
Failed to load comments.
Loading
Jump to
Jump to file
Failed to load files.
Loading
Diff view
Diff view
214 changes: 79 additions & 135 deletions source/linear-algebra/source/02-EV/samples/05.ptx
View file
Open in desktop
Original file line number Diff line number Diff line change
@@ -1,138 +1,82 @@
<?xml version='1.0' encoding='UTF-8'?>
<example xml:id="sample-EV5"><title>EV5</title>
<statement>
<ol>
<li>
<p>
Write a statement involving spanning and independence properties
that's equivalent to each claim below.
</p>
<ul>
<li>
<p>
The set of vectors <m>\left\{ \left[\begin{array}{c}
1 \\
3 \\
4 \\
-4
\end{array}\right] , \left[\begin{array}{c}
0 \\
1 \\
3 \\
-3
\end{array}\right] , \left[\begin{array}{c}
3 \\
11 \\
18 \\
-18
\end{array}\right] , \left[\begin{array}{c}
-2 \\
-7 \\
-11 \\
11
\end{array}\right] \right\}</m> is a <em>basis</em>
of <m>\mathbb{R}^4</m>.
</p>
</li>
<li>
<p>
The set of vectors <m>\left\{ \left[\begin{array}{c}
1 \\
3 \\
4 \\
-4
\end{array}\right] , \left[\begin{array}{c}
0 \\
1 \\
3 \\
-3
\end{array}\right] , \left[\begin{array}{c}
3 \\
11 \\
18 \\
-18
\end{array}\right] , \left[\begin{array}{c}
-2 \\
-7 \\
-11 \\
11
\end{array}\right] \right\}</m> is <em>not</em> a basis
of <m>\mathbb{R}^4</m>.
</p>
</li>
</ul>
</li>
<li>
<p>
Explain how to determine which of these statements is true.
</p>
</li>
</ol>
</statement>
<solution>
<p>The set of vectors <m>\left\{ \left[\begin{array}{c}
1 \\
3 \\
4 \\
-4
\end{array}\right] , \left[\begin{array}{c}
0 \\
1 \\
3 \\
-3
\end{array}\right] , \left[\begin{array}{c}
3 \\
11 \\
18 \\
-18
\end{array}\right] , \left[\begin{array}{c}
-2 \\
-7 \\
-11 \\
11
\end{array}\right] \right\}</m> is a basis
of <m>\mathbb{R}^4</m> exactly when it is linearly independent <em>and</em> the set spans <m>\mathbb{R}^4</m>.
If it is either linearly dependent, <em>or</em> the set does not span <m>\mathbb{R}^4</m>, then the set is not a basis.
</p>
<p> To answer this, we compute
<me>
\mathrm{RREF}\, \left[\begin{array}{cccc}
1 &amp; 0 &amp; 3 &amp; -2 \\
3 &amp; 1 &amp; 11 &amp; -7 \\
4 &amp; 3 &amp; 18 &amp; -11 \\
-4 &amp; -3 &amp; -18 &amp; 11
\end{array}\right] = \left[\begin{array}{cccc}
1 &amp; 0 &amp; 3 &amp; -2 \\
0 &amp; 1 &amp; 2 &amp; -1 \\
0 &amp; 0 &amp; 0 &amp; 0 \\
0 &amp; 0 &amp; 0 &amp; 0
\end{array}\right]
</me>.
</p>
<p>We see that this set of vectors is linearly dependent, so therefore
the set of vectors <m>\left\{ \left[\begin{array}{c}
1 \\
3 \\
4 \\
-4
\end{array}\right] , \left[\begin{array}{c}
0 \\
1 \\
3 \\
-3
\end{array}\right] , \left[\begin{array}{c}
3 \\
11 \\
18 \\
-18
\end{array}\right] , \left[\begin{array}{c}
-2 \\
-7 \\
-11 \\
11
\end{array}\right] \right\}</m> is

<em>not</em> a basis.
</p>
</solution>
<task>
<introduction>
<p> Consider the set of vectors <me>\left\{ \left[\begin{array}{c} 1 \\ -2 \\ 0 \\ 1 \end{array}\right] , \left[\begin{array}{c} 2 \\ -4 \\ 0 \\ 2 \end{array}\right] , \left[\begin{array}{c} 3 \\ -6 \\ 0 \\ 3 \end{array}\right] , \left[\begin{array}{c} 5 \\ -5 \\ -2 \\ 4 \end{array}\right] \right\}</me> </p>
</introduction>
<task>
<statement>
<p> Write a statement involving the solutions of a vector equation that's equivalent to this claim: <q>The set of vectors is a basis for</q> </p>
</statement>
<solution>
<p> <q>The vector equation
<me>x_1 \left[\begin{array}{c} 1 \\ -2 \\ 0 \\ 1 \end{array}\right] +x_2 \left[\begin{array}{c} 2 \\ -4 \\ 0 \\ 2 \end{array}\right] +x_3 \left[\begin{array}{c} 3 \\ -6 \\ 0 \\ 3 \end{array}\right] +x_4 \left[\begin{array}{c} 5 \\ -5 \\ -2 \\ 4 \end{array}\right] =\vec w</me>
has exactly one solution for every <m>\vec w\in\mathbb R^4</m>.</q> </p>
</solution>
</task>
<task>
<statement>
<p> Explain and demonstrate how to determine whether or not this statement is true. </p>
</statement>
<solution>
<p> Since <m>\mathrm{RREF}\, \left[\begin{array}{cccc} 1 &amp; 2 &amp; 3 &amp; 5 \\ -2 &amp; -4 &amp; -6 &amp; -5 \\ 0 &amp; 0 &amp; 0 &amp; -2 \\ 1 &amp; 2 &amp; 3 &amp; 4 \end{array}\right] = \left[\begin{array}{cccc} 1 &amp; 2 &amp; 3 &amp; 0 \\ 0 &amp; 0 &amp; 0 &amp; 1 \\ 0 &amp; 0 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 0 &amp; 0 \end{array}\right]</m>,
we see from the zero row that there are some vectors <m>\vec w</m> for which the equation is not true, so the set fails to span and therefore fails to be a basis.</p>
</solution>
<solution>
<p> Since <m>\mathrm{RREF}\, \left[\begin{array}{cccc} 1 &amp; 2 &amp; 3 &amp; 5 \\ -2 &amp; -4 &amp; -6 &amp; -5 \\ 0 &amp; 0 &amp; 0 &amp; -2 \\ 1 &amp; 2 &amp; 3 &amp; 4 \end{array}\right] = \left[\begin{array}{cccc} 1 &amp; 2 &amp; 3 &amp; 0 \\ 0 &amp; 0 &amp; 0 &amp; 1 \\ 0 &amp; 0 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 0 &amp; 0 \end{array}\right]</m>,
we see from the non-pivot column that there are some vectors <m>\vec w</m> for which the equation has infinitely-many solutions, so the set is linearly dependent and therefore fails to be a basis.</p>
</solution>
</task>
</task>
<task>
<introduction>
<p> Consider the set of vectors <me>\left\{ \left[\begin{array}{c} 1 \\ 3 \\ 4 \\ -4 \end{array}\right] , \left[\begin{array}{c} -1 \\ -3 \\ -4 \\ 4 \end{array}\right] , \left[\begin{array}{c} 0 \\ 1 \\ 3 \\ -3 \end{array}\right] \right\}</me> </p>
</introduction>
<task>
<statement>
<p> Write a statement involving the solutions of a vector equation that's equivalent to this claim: <q>The set of vectors is a basis for</q> </p>
</statement>
<solution>
<p> <q>The vector equation
<me>x_1 \left[\begin{array}{c} 1 \\ 3 \\ 4 \\ -4 \end{array}\right] +x_2 \left[\begin{array}{c} -1 \\ -3 \\ -4 \\ 4 \end{array}\right] +x_3 \left[\begin{array}{c} 0 \\ 1 \\ 3 \\ -3 \end{array}\right] =\vec w</me>
has exactly one solution for every <m>\vec w\in\mathbb R^4</m>.</q> </p>
</solution>
</task>
<task>
<statement>
<p> Explain and demonstrate how to determine whether or not this statement is true. </p>
</statement>
<solution>
<p> Since <m>\mathrm{RREF}\, \left[\begin{array}{ccc} 1 &amp; -1 &amp; 0 \\ 3 &amp; -3 &amp; 1 \\ 4 &amp; -4 &amp; 3 \\ -4 &amp; 4 &amp; -3 \end{array}\right] = \left[\begin{array}{ccc} 1 &amp; -1 &amp; 0 \\ 0 &amp; 0 &amp; 1 \\ 0 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 0 \end{array}\right]</m>
we see from the zero row that there are some vectors <m>\vec w</m> for which the equation is not true, so the set fails to span and therefore fails to be a basis.</p>
</solution>
<solution>
<p>The set has only three vectors, so the set cannot span and there must be vectors for which the equation has no solutions. Therefore the set is not a basis.</p>
</solution>
</task>
</task>
<task>
<introduction>
<p> Consider the set of vectors <me>\left\{ \left[\begin{array}{c} 3 \\ 2 \\ -1 \\ 0 \end{array}\right] , \left[\begin{array}{c} -2 \\ -1 \\ 0 \\ -1 \end{array}\right] , \left[\begin{array}{c} -2 \\ -1 \\ 1 \\ 0 \end{array}\right] , \left[\begin{array}{c} -4 \\ -1 \\ 0 \\ -2 \end{array}\right] \right\}</me> </p>
</introduction>
<task>
<statement>
<p> Write a statement involving the solutions of a vector equation that's equivalent to this claim: <q>The set of vectors is a basis for</q> </p>
</statement>
<solution>
<p> <q>The vector equation
<me>x_1 \left[\begin{array}{c} 3 \\ 2 \\ -1 \\ 0 \end{array}\right] +x_2 \left[\begin{array}{c} -2 \\ -1 \\ 0 \\ -1 \end{array}\right] +x_3 \left[\begin{array}{c} -2 \\ -1 \\ 1 \\ 0 \end{array}\right] + x_4\left[\begin{array}{c} -4 \\ -1 \\ 0 \\ -2 \end{array}\right] =\vec w</me>
has exactly one solution for every <m>\vec w\in\mathbb R^4</m>.</q> </p>
</solution>
</task>
<task>
<statement>
<p> Explain and demonstrate how to determine whether or not this statement is true. </p>
</statement>
<solution>
<p>Since <m>\mathrm{RREF}\, \left[\begin{array}{cccc} 3 &amp; -2 &amp; -2 &amp; -4 \\ 2 &amp; -1 &amp; -1 &amp; -1 \\ -1 &amp; 0 &amp; 1 &amp; 0 \\ 0 &amp; -1 &amp; 0 &amp; -2 \end{array}\right] = \left[\begin{array}{cccc} 1 &amp; 0 &amp; 0 &amp; 0 \\ 0 &amp; 1 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 1 &amp; 0 \\ 0 &amp; 0 &amp; 0 &amp; 1 \end{array}\right]</m>
we see the equation always has exactly one solution (each row and column has a pivot). Therefore the set is spanning and linearly independent, and therefore the set is a basis.</p>
</solution>
</task>
</task>
</example>
Loading