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add task to GT3 to associate an eigenvector with an eigenvalue (#948)
* add task to EV3 to associate an eigenvector with an eigenvalue * add sample solution for updated GT3 * add activities
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‎source/linear-algebra/exercises/outcomes/GT/GT3/generator.sage‎

Lines changed: 14 additions & 4 deletions
Original file line numberDiff line numberDiff line change
@@ -4,16 +4,26 @@ TBIL.config_matrix_typesetting()
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class Generator(BaseGenerator):
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def data(self):
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while True:
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ls = [choice([-1,1])*i for i in range(2,7)]
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l1,l2 = sample(ls,2)
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S=random_matrix(QQ, 2, 2, algorithm='echelonizable', rank=2, upper_bound=2)
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# eigenvalues will be two distinct small integers with
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# different absolute values
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l1,l2 = sample(range(2,6),2)
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l1,l2 = l1*choice([-1,1]),l2*choice([-1,1])
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S=random_matrix(QQ, 2, 2, algorithm='echelonizable', rank=2, upper_bound=6)
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A=S.inverse()*matrix([[l1,1],[0,l2]])*S
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if all(a!=0 for a in A.list()):
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# to get roughly consistent difficulty
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if all(abs(a)>5 for a in A.list()):
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break
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# Get an eigenvector
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eigenvector = column_matrix((A-matrix([[l1,0],[0,l1]])).right_kernel(basis='pivot').basis()[0])
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# Scale to get whole numbers
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eigenvector = eigenvector[0].denominator()*eigenvector[1].denominator()*eigenvector
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return {
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"matrix": A,
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"e1": l1,
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"e2": l2,
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"charpoly": A.charpoly('lambda_'),
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"eigenvector": eigenvector,
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"scaled_eigenvector": l1*eigenvector,
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}

‎source/linear-algebra/exercises/outcomes/GT/GT3/template.xml‎

Lines changed: 20 additions & 7 deletions
Original file line numberDiff line numberDiff line change
@@ -1,10 +1,23 @@
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<?xml version='1.0' encoding='UTF-8'?>
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<knowl mode="exercise" xmlns="https://spatext.clontz.org" version="0.2">
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<content>
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<p>Explain and demonstrate how to find the eigenvalues of the matrix <m>{{matrix}}</m>.</p>
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</content>
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<outtro>
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<p>The characteristic polynomial of <m>{{matrix}}</m> is <m>{{charpoly}}</m>.</p>
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<p>The eigenvalues of <m>{{matrix}}</m> are <m>{{e1}}</m> and <m>{{e2}}</m>.</p>
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</outtro>
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<knowl>
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<content>
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<p>Explain and demonstrate how to find the eigenvalues of the matrix <m>{{matrix}}</m>.</p>
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</content>
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<outtro>
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<p>The characteristic polynomial of <m>{{matrix}}</m> is <m>{{charpoly}}</m>.</p>
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<p>The eigenvalues of <m>{{matrix}}</m> are <m>{{e1}}</m> and <m>{{e2}}</m>.</p>
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</outtro>
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</knowl>
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<knowl>
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<content>
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<p>Explain and demonstrate which of these eigenvalues is associated to the eigenvector <m>{{eigenvector}}</m>.</p>
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</content>
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<outtro>
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<p>
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<m>{{eigenvector}}</m> is associated with the eigenvalue <m>{{e1}}</m> because
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<me>{{matrix}}{{eigenvector}}={{scaled_eigenvector}}={{e1}}{{eigenvector}}</me>
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</p>
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</outtro>
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</knowl>
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</knowl>

‎source/linear-algebra/source/05-GT/03.ptx‎

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Original file line numberDiff line numberDiff line change
@@ -294,6 +294,19 @@ Thus the characteristic polynomial of <m>A</m> is
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</me>
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and its eigenvalues are the solutions <m>-1,6</m> to <m>\lambda^2-5\lambda-6=0</m>.
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</p>
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<p>
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In particular, we can see by
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<md>
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\left[\begin{array}{cc}1 &amp; 2 \\ 5 &amp; 4\end{array}\right]
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\left[\begin{array}{c}1 \\-1\end{array}\right]
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=
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\left[\begin{array}{c}-1 \\1\end{array}\right]
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=
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-1\left[\begin{array}{c}1 \\-1\end{array}\right]
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</md>
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that <m>\left[\begin{array}{c}1 \\-1\end{array}\right]</m> is an eigenvector
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associated with the eigenvalue <m>-1</m>.
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</p>
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</statement>
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</definition>
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@@ -304,14 +317,44 @@ Let <m>A = \left[\begin{array}{cc} 5 &amp; 2 \\ -3 &amp; -2 \end{array}\right]</
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</p>
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</introduction>
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<task>
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<statement>
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<p>
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Compute <m>\det (A-\lambda I)</m> to determine the characteristic polynomial of <m>A</m>.
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</p>
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</statement>
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<answer>
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<p>
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<m>\lambda^2-3\lambda-4</m>
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</p>
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</answer>
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</task>
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<task>
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<statement>
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<p>
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Set this characteristic polynomial equal to zero and factor to determine the eigenvalues of <m>A</m>.
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</p>
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</statement>
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<answer>
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<p>
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Solve <m>\lambda^2-3\lambda-4=(\lambda-4)(\lambda+1)=0</m>
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to find <m>\lambda=4,-1</m>.
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</p>
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</answer>
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</task>
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<task>
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<statement>
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<p>
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Use technology to calculate <m>\left[\begin{array}{cc} 5 &amp; 2 \\ -3 &amp; -2 \end{array}\right]\left[\begin{array}{c} 2 \\ -1 \end{array}\right]</m> to determine
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which of these eigenvalues is associated to the eigenvector <m>\left[\begin{array}{c} 2 \\ -1 \end{array}\right]</m>.
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</p>
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</statement>
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<answer>
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<p>
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Since
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<md>\left[\begin{array}{cc} 5 &amp; 2 \\ -3 &amp; -2 \end{array}\right]\left[\begin{array}{c} 2 \\ -1 \end{array}\right]=\left[\begin{array}{c} 8 \\ -4 \end{array}\right]=4\left[\begin{array}{c} 2 \\ -1 \end{array}\right]</md>
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the associated eigenvalue is <m>\lambda=4</m>.
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</p>
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</answer>
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</task>
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</activity>
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‎source/linear-algebra/source/05-GT/samples/03.ptx‎

Lines changed: 45 additions & 14 deletions
Original file line numberDiff line numberDiff line change
@@ -1,20 +1,51 @@
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<?xml version='1.0' encoding='UTF-8'?>
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<example xml:id="sample-GT3">
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<title>GT3</title>
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<task>
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<statement>
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<p>
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Explain and demonstrate how to find the eigenvalues of the matrix <m>\left[\begin{array}{cc} -2 &amp; -2 \\ 10 &amp; 7 \end{array}\right] </m>.
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</p>
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</statement>
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<solution>
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<p>
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Compute the characteristic polynomial:
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<me>
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\det(A-\lambda I) = \det \left[\begin{array}{cc} -2 - \lambda &amp; -2 \\ 10 &amp; 7-\lambda \end{array}\right]
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</me>
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<me>
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= (-2-\lambda)(7-\lambda)+20 = \lambda ^2 -5\lambda +6 = (\lambda -2)(\lambda -3)
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</me>
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The eigenvalues are the roots of the characteristic polynomial, namely <m>2</m> and <m>3</m>.
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</p>
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</solution>
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</task>
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<example xml:id="sample-GT3"><title>GT3</title>
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<statement>
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<p>
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Explain how to find the eigenvalues of the matrix <m>\left[\begin{array}{cc} -2 &amp; -2 \\ 10 &amp; 7 \end{array}\right] </m>.
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</p>
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</statement>
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<solution>
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<p>
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Compute the characteristic polynomial:
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<me>\det(A-\lambda I) = \det \left[\begin{array}{cc} -2 - \lambda &amp; -2 \\ 10 &amp; 7-\lambda \end{array}\right]
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</me><me>= (-2-\lambda)(7-\lambda)+20 = \lambda ^2 -5\lambda +6 = (\lambda -2)(\lambda -3)</me>
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The eigenvalues are the roots of the characteristic polynomial, namely <m>2</m> and <m>3</m>.
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</p>
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</solution>
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<task>
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<statement>
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<p>
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Explain and demonstrate which of these eigenvalues is associated to the eigenvector <m>\left[\begin{array}{cc} -1 \\ 2 \end{array}\right]</m>.
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</p>
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</statement>
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<solution>
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<p>
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We can compute
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<md>
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\left[\begin{array}{cc} -2 &amp; -2 \\ 10 &amp; 7 \end{array}\right]
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\left[\begin{array}{cc} -1 \\ 2 \end{array}\right] =
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\left[\begin{array}{cc} -2 \\ 4 \end{array}\right]
42+
</md> and <md>
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2\left[\begin{array}{cc} -1 \\ 2 \end{array}\right] =
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\left[\begin{array}{cc} -2 \\ 4 \end{array}\right]
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</md>
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which shows that <m>\left[\begin{array}{cc} -1 \\ 2 \end{array}\right]</m> is an
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eigenvector associated with the eigenvalue <m>2</m>.
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</p>
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</solution>
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</task>
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</example>

‎source/linear-algebra/source/meta/sample-exercises.ptx‎

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@@ -28,7 +28,6 @@ for a complete solution.
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<xi:include href="../03-AT/samples/05.ptx"/>
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<xi:include href="../03-AT/samples/06.ptx"/>
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<xi:include href="../04-MX/samples/01.ptx"/>
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<xi:include href="../04-MX/samples/02.ptx"/>
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<xi:include href="../04-MX/samples/03.ptx"/>

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