add task to EV3 to associate an eigenvector with an eigenvalue

source/linear-algebra/exercises/outcomes/GT/GT3/generator.sage

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

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

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

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

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

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

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