Deployed docs from commit acef4f9

docs/Contents/eigenvalues_and_eigenvectors.html

@@ -649,6 +649,25 @@ <h3><span class="math notranslate nohighlight">\(Ax = 0\)</span> has a non-trivi
 \end{bmatrix}\end{split}\]</div>
 <p>Since the columns of <span class="math notranslate nohighlight">\(A\)</span> are linearly independent, <span class="math notranslate nohighlight">\(\mathbf{AB}_i = \mathbf{e}_i\)</span> has a unique solution. Thus we can find each column <span class="math notranslate nohighlight">\(\mathbf{B}_i\)</span> thus <span class="math notranslate nohighlight">\(\mathbf{A}\)</span> is invertible.</p>
 </aside>
+<aside class="topic">
+<p class="topic-title">Theorem</p>
+<p>A symmetric matrix <span class="math notranslate nohighlight">\(\mathbf{A}\)</span> with distinct eigenvalues has orthogonal eigenvectors.</p>
+<p><strong>Proof</strong></p>
+<p>We only prove for the <span class="math notranslate nohighlight">\(2 \times 2\)</span> case where <span class="math notranslate nohighlight">\(\mathbf{A}\)</span> has two eigenvectors.</p>
+<div class="math notranslate nohighlight">
+\[\begin{split}\mathbf{A} \mathbf{x}_1 &amp; = \lambda_1 \mathbf{x}_1 \\
+\mathbf{x}_2^T \mathbf{A} \mathbf{x}_1 &amp; = \lambda_1 \mathbf{x}_2^T \mathbf{x}_1 \\\end{split}\]</div>
+<p>And,</p>
+<div class="math notranslate nohighlight">
+\[\begin{split}\mathbf{A x}_2 &amp; = \lambda_2 \mathbf{x}_2 \\
+\mathbf{x}_1^T \mathbf{A x}_2 &amp; = \lambda_2 \mathbf{x}_1^T \mathbf{x}_2 \\
+\mathbf{x}_2^T \mathbf{A x}_1 &amp; = \lambda_2 \mathbf{x}_2^T \mathbf{x}_1 \\\end{split}\]</div>
+<p>Taking these away from each other, we have,</p>
+<div class="math notranslate nohighlight">
+\[\begin{split}\mathbf{x}_2^T \mathbf{A x}_1 - \mathbf{x}_2^T \mathbf{A x}_1 &amp; = (\lambda_1 - \lambda_2) \mathbf{x}_2^T \mathbf{x}_1 \\
+0 &amp; = (\lambda_1 - \lambda_2) \mathbf{x}_2^T \mathbf{x}_1\end{split}\]</div>
+<p>Since <span class="math notranslate nohighlight">\(\lambda_1\)</span> and <span class="math notranslate nohighlight">\(\lambda_2\)</span> are distinct, their difference is non-zero. Therefore, <span class="math notranslate nohighlight">\(\mathbf{x}_2^T \mathbf{x}_1\)</span> must be <span class="math notranslate nohighlight">\(0\)</span>.</p>
+</aside>
 </section>
 </section>
 </section>

docs/_sources/Contents/eigenvalues_and_eigenvectors.rst.txt

@@ -172,3 +172,33 @@ Since the columns of :math:`A` are linearly independent, some column :math:`v_i`
       \end{bmatrix}
 
    Since the columns of :math:`A` are linearly independent, :math:`\mathbf{AB}_i = \mathbf{e}_i` has a unique solution. Thus we can find each column :math:`\mathbf{B}_i` thus :math:`\mathbf{A}` is invertible.
+
+.. topic:: Theorem
+
+   A symmetric matrix :math:`\mathbf{A}` with distinct eigenvalues has orthogonal eigenvectors.
+
+   **Proof**
+
+   We only prove for the :math:`2 \times 2` case where :math:`\mathbf{A}` has two eigenvectors.
+
+   .. math::
+
+      \mathbf{A} \mathbf{x}_1 & = \lambda_1 \mathbf{x}_1 \\
+      \mathbf{x}_2^T \mathbf{A} \mathbf{x}_1 & = \lambda_1 \mathbf{x}_2^T \mathbf{x}_1 \\
+
+   And,
+
+   .. math::
+
+      \mathbf{A x}_2 & = \lambda_2 \mathbf{x}_2 \\
+      \mathbf{x}_1^T \mathbf{A x}_2 & = \lambda_2 \mathbf{x}_1^T \mathbf{x}_2 \\
+      \mathbf{x}_2^T \mathbf{A x}_1 & = \lambda_2 \mathbf{x}_2^T \mathbf{x}_1 \\
+
+   Taking these away from each other, we have,
+
+   .. math::
+
+      \mathbf{x}_2^T \mathbf{A x}_1 - \mathbf{x}_2^T \mathbf{A x}_1 & = (\lambda_1 - \lambda_2) \mathbf{x}_2^T \mathbf{x}_1 \\
+      0 & = (\lambda_1 - \lambda_2) \mathbf{x}_2^T \mathbf{x}_1
+
+   Since :math:`\lambda_1` and :math:`\lambda_2` are distinct, their difference is non-zero. Therefore, :math:`\mathbf{x}_2^T \mathbf{x}_1` must be :math:`0`.