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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">
-\[r_{ij} = q_i^*a_j, \, i &lt; j,\]</div>
+\[r_{ij} = q_i^*a_j, \, i &gt; j,\]</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">
@@ -171,6 +171,11 @@ <h2><span class="section-number">2.2. </span>QR factorisation by classical Gram-
 <li><p>FOR <span class="math notranslate nohighlight">\(i = 1\)</span> TO <span class="math notranslate nohighlight">\(j-1\)</span></p>
 <ul>
 <li><p><span class="math notranslate nohighlight">\(r_{ij} \gets q_i^*a_j\)</span></p></li>
+</ul>
+</li>
+<li><p>END FOR</p></li>
+<li><p>FOR <span class="math notranslate nohighlight">\(i = 1\)</span> TO <span class="math notranslate nohighlight">\(j-1\)</span></p>
+<ul>
 <li><p><span class="math notranslate nohighlight">\(v_j \gets v_j - r_{ij}q_i\)</span></p></li>
 </ul>
 </li>
@@ -193,11 +198,11 @@ <h2><span class="section-number">2.2. </span>QR factorisation by classical Gram-
 <p>The <a class="reference internal" href="cla_utils.html#cla_utils.exercises2.GS_classical" title="cla_utils.exercises2.GS_classical"><code class="xref py py-func docutils literal notranslate"><span class="pre">cla_utils.exercises2.GS_classical()</span></code></a> function has been
 left unimplemented. It should implement the classical Gram-Schmidt
 algorithm above, using Numpy slice notation so that only one Python
-for loop is used. The function should work “in place” by changing
-the values in <span class="math notranslate nohighlight">\(A\)</span>, without introducing additional intermediate
-arrays (you will need to create a new array to store <span class="math notranslate nohighlight">\(R\)</span>). The test
-script <code class="docutils literal notranslate"><span class="pre">test_exercises2.py</span></code> in the <code class="docutils literal notranslate"><span class="pre">test</span></code> directory will test
-this function.</p>
+for loop is used. The function should work “in place” by making a
+copy of <span class="math notranslate nohighlight">\(A\)</span> and then changing the values in the copy, without
+introducing additional intermediate arrays (you will need to create
+a new array to store <span class="math notranslate nohighlight">\(R\)</span>). The test script <code class="docutils literal notranslate"><span class="pre">test_exercises2.py</span></code>
+in the <code class="docutils literal notranslate"><span class="pre">test</span></code> directory will test this function.</p>
 </div></div></section>
 <section id="projector-interpretation-of-gram-schmidt">
 <h2><span class="section-number">2.3. </span>Projector interpretation of Gram-Schmidt<a class="headerlink" href="#projector-interpretation-of-gram-schmidt" title="Permalink to this headline">¶</a></h2>
@@ -307,8 +312,9 @@ <h2><span class="section-number">2.4. </span>Modified Gram-Schmidt<a class="head
 algorithm above, using Numpy slice notation where possible.
 What is the minimal number of Python
 for loops possible?</p>
-<p>The function should work “in place” by changing the values of <span class="math notranslate nohighlight">\(A\)</span>,
-without introducing additional intermediate arrays. The test script
+<p>The function should work “in place” by making a
+copy of <span class="math notranslate nohighlight">\(A\)</span> and then changing those values, without introducing
+additional intermediate arrays. The test script
 <code class="docutils literal notranslate"><span class="pre">test_exercises2.py</span></code> in the <code class="docutils literal notranslate"><span class="pre">test</span></code> directory will test this
 function.</p>
 </div></div><div class="proof proof-type-exercise" id="id6">
@@ -681,7 +687,7 @@ <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+1859.14341ea)"><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+1987.d7fa27f)"><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 +799,7 @@ <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+1859.14341ea)"><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+1987.d7fa27f)"><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

@@ -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.
@@ -146,6 +146,9 @@ We now present the classical Gram-Schmidt algorithm as pseudo-code.
   * FOR `i = 1` TO `j-1`
 
     * `r_{ij} \gets q_i^*a_j`
+  * END FOR
+  * FOR `i = 1` TO `j-1`
+    
     * `v_j \gets v_j - r_{ij}q_i`
   * END FOR
   * `r_{jj} \gets \|v_j\|_2`
@@ -161,11 +164,11 @@ for code blocks in pseudo-code in these notes.)
    The :func:`cla_utils.exercises2.GS_classical` function has been
    left unimplemented. It should implement the classical Gram-Schmidt
    algorithm above, using Numpy slice notation so that only one Python
-   for loop is used. The function should work "in place" by changing
-   the values in `A`, without introducing additional intermediate
-   arrays (you will need to create a new array to store `R`). The test
-   script ``test_exercises2.py`` in the ``test`` directory will test
-   this function.
+   for loop is used. The function should work "in place" by making a
+   copy of `A` and then changing the values in the copy, without
+   introducing additional intermediate arrays (you will need to create
+   a new array to store `R`). The test script ``test_exercises2.py``
+   in the ``test`` directory will test this function.
 
 Projector interpretation of Gram-Schmidt
 ----------------------------------------
@@ -289,7 +292,7 @@ We now present this modified Gram-Schmidt algorithm as pseudo-code.
 
   * `r_{ii} \gets \|v_i\|_2`
   * `q_i = v_i/r_{ii}`
-    
+
   * FOR `j = i+1` TO `n`
 
     * `r_{ij} \gets q_i^*v_j`
@@ -308,8 +311,9 @@ in `A` with the `v` s and eventually the `q` s.
    What is the minimal number of Python
    for loops possible?
 
-   The function should work "in place" by changing the values of `A`,
-   without introducing additional intermediate arrays. The test script
+   The function should work "in place" by making a
+   copy of `A` and then changing those values, without introducing
+   additional intermediate arrays. The test script
    ``test_exercises2.py`` in the ``test`` directory will test this
    function.
 

_sources/L2_QR_factorisation.rst.txt

@@ -17,3 +17,6 @@ repositories for Autumn 2021. An up to date version is in 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

@@ -60,6 +60,8 @@ <h1>Errata for 2021/22<a class="headerlink" href="#errata-for-2021-22" title="Pe
 <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>