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Computationallinearalgebracourse.pdf

@@ -155,7 +155,7 @@ <h2><span class="section-number">2.2. </span>QR factorisation by classical Gram-
 <p>with <span class="math notranslate nohighlight">\((q_1,q_2,\ldots,q_n)\)</span> an orthonormal set. The non-diagonal
 entries of <span class="math notranslate nohighlight">\(R\)</span> are found by inner products, i.e.,</p>
 <div class="math notranslate nohighlight">
-\[r_{ij} = q_i^*a_j, \, i &gt; j,\]</div>
+\[r_{ij} = q_i^*a_j, \, i &lt; j,\]</div>
 <p>and the diagonal entries are chosen so that <span class="math notranslate nohighlight">\(\|q_i\|=1\)</span>, for
 <span class="math notranslate nohighlight">\(i=1,2,\ldots,n\)</span>, i.e.</p>
 <div class="math notranslate nohighlight">
@@ -347,7 +347,7 @@ <h2><span class="section-number">2.5. </span>Modified Gram-Schmidt as triangular
 \underbrace{
 \begin{pmatrix}
 \frac{1}{r_{11}} &amp; -\frac{r_{12}}{r_{11}} &amp; \ldots &amp;
-\ldots &amp; -\frac{r_{11}}{r_{11}} \\
+\ldots &amp; -\frac{r_{1n}}{r_{11}} \\
 0 &amp; 1 &amp; 0 &amp; \ldots &amp; 0 \\
 0 &amp; 0 &amp; 1 &amp; \ldots &amp; 0 \\
 \vdots &amp; \ddots &amp; \ddots &amp; \ldots &amp; \vdots \\
@@ -681,7 +681,7 @@ <h2><span class="section-number">2.6. </span>Householder triangulation<a class="
 done so, you will need to modified
 <a class="reference internal" href="cla_utils.html#cla_utils.exercises3.householder" title="cla_utils.exercises3.householder"><code class="xref py py-func docutils literal notranslate"><span class="pre">cla_utils.exercises3.householder()</span></code></a> to use the <code class="docutils literal notranslate"><span class="pre">kmax</span></code>
 argument. You may make use of the built-in triangular solve
-algorithm <a class="reference external" href="http://scipy.github.io/devdocs/reference/generated/scipy.linalg.solve_triangular.html#scipy.linalg.solve_triangular" title="(in SciPy v1.8.0.dev0+1849.a542b67)"><code class="xref py py-func docutils literal notranslate"><span class="pre">scipy.linalg.solve_triangular()</span></code></a> (we shall consider
+algorithm <a class="reference external" href="http://scipy.github.io/devdocs/reference/generated/scipy.linalg.solve_triangular.html#scipy.linalg.solve_triangular" title="(in SciPy v1.8.0.dev0+1869.838cfbe)"><code class="xref py py-func docutils literal notranslate"><span class="pre">scipy.linalg.solve_triangular()</span></code></a> (we shall consider
 triangular matrix algorithms briefly later). The test script
 <code class="docutils literal notranslate"><span class="pre">test_exercises3.py</span></code> in the <code class="docutils literal notranslate"><span class="pre">test</span></code> directory will also test this
 function.</p>
@@ -793,7 +793,7 @@ <h2><span class="section-number">2.7. </span>Application: Least squares problems
 appropriate augmented matrix <span class="math notranslate nohighlight">\(\hat{A}\)</span>, calling
 <a class="reference internal" href="cla_utils.html#cla_utils.exercises3.householder" title="cla_utils.exercises3.householder"><code class="xref py py-func docutils literal notranslate"><span class="pre">cla_utils.exercises3.householder()</span></code></a> and extracting appropriate
 subarrays using slice notation, before using
-<a class="reference external" href="http://scipy.github.io/devdocs/reference/generated/scipy.linalg.solve_triangular.html#scipy.linalg.solve_triangular" title="(in SciPy v1.8.0.dev0+1849.a542b67)"><code class="xref py py-func docutils literal notranslate"><span class="pre">scipy.linalg.solve_triangular()</span></code></a> to solve the resulting upper triangular
+<a class="reference external" href="http://scipy.github.io/devdocs/reference/generated/scipy.linalg.solve_triangular.html#scipy.linalg.solve_triangular" title="(in SciPy v1.8.0.dev0+1869.838cfbe)"><code class="xref py py-func docutils literal notranslate"><span class="pre">scipy.linalg.solve_triangular()</span></code></a> to solve the resulting upper triangular
 system, before returning the solution <span class="math notranslate nohighlight">\(x\)</span>. The test script
 <code class="docutils literal notranslate"><span class="pre">test_exercises3.py</span></code> in the <code class="docutils literal notranslate"><span class="pre">test</span></code> directory will also test this
 function.</p>

L2_QR_factorisation.html

@@ -129,7 +129,7 @@ <h1>Source code for cla_utils.exercises2</h1><div class="highlight"><pre>
 
     <span class="k">raise</span> <span class="ne">NotImplementedError</span>
 
-    <span class="k">return</span> <span class="n">Q</span><span class="p">,</span> <span class="n">R</span></div>
+    <span class="k">return</span> <span class="n">R</span></div>
 
 <div class="viewcode-block" id="GS_modified"><a class="viewcode-back" href="../../cla_utils.html#cla_utils.exercises2.GS_modified">[docs]</a><span class="k">def</span> <span class="nf">GS_modified</span><span class="p">(</span><span class="n">A</span><span class="p">):</span>
     <span class="sd">&quot;&quot;&quot;</span>

_modules/cla_utils/exercises2.html

@@ -125,7 +125,7 @@ entries of `R` are found by inner products, i.e.,
 
 .. math::
 
-   r_{ij} = q_i^*a_j, \, i > j,
+   r_{ij} = q_i^*a_j, \, i < j,
 
 and the diagonal entries are chosen so that `\|q_i\|=1`, for
 `i=1,2,\ldots,n`, i.e.
@@ -350,7 +350,7 @@ example, at the first iteration,
       \underbrace{
       \begin{pmatrix}
       \frac{1}{r_{11}} & -\frac{r_{12}}{r_{11}} & \ldots &
-      \ldots & -\frac{r_{11}}{r_{11}} \\
+      \ldots & -\frac{r_{1n}}{r_{11}} \\
       0 & 1 & 0 & \ldots & 0 \\    
       0 & 0 & 1 & \ldots & 0 \\
       \vdots & \ddots & \ddots & \ldots & \vdots \\

_sources/L2_QR_factorisation.rst.txt

@@ -12,3 +12,11 @@ repositories for Autumn 2021. An up to date version is in the
    correct indices. Now reads: "Return the rank2 matrix A = u_1v_1^* +
    u_2*v_2^*." See the code on the course website for the up-to-date
    version.
+
+2. The skeleton code for exercises2.GS_classical was amended as the
+   function should just return R, not Q and R.
+
+3. Updated the sign of the inequality in the formula for $r$ in Section 2.2.
+
+4. Replaced $r_{11}/r_{11}$ by $r_{1n}/r_{11}$ in the final column of the
+   right-multiplied $R$ matrix in Section 2.5.

_sources/errata.rst.txt

@@ -57,6 +57,11 @@ <h1>Errata for 2021/22<a class="headerlink" href="#errata-for-2021-22" title="Pe
 correct indices. Now reads: “Return the rank2 matrix A = u_1v_1^* +
 u_2*v_2^*.” See the code on the course website for the up-to-date
 version.</p></li>
+<li><p>The skeleton code for exercises2.GS_classical was amended as the
+function should just return R, not Q and R.</p></li>
+<li><p>Updated the sign of the inequality in the formula for $r$ in Section 2.2.</p></li>
+<li><p>Replaced $r_{11}/r_{11}$ by $r_{1n}/r_{11}$ in the final column of the
+right-multiplied $R$ matrix in Section 2.5.</p></li>
 </ol>
 </section>