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‎Computationallinearalgebracourse.pdf‎

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‎L2_QR_factorisation.html‎

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@@ -660,7 +660,30 @@ <h2><span class="section-number">2.6. </span>Householder triangulation<a class="
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<summary>
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Supplementary video</summary><div class="video_wrapper" style="">
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<iframe allowfullscreen="true" src="https://player.vimeo.com/video/450202242" style="border: 0; height: 345px; width: 560px">
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</iframe></div></details><div class="proof proof-type-exercise" id="id9">
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</iframe></div></details><p>In this section we will frequently encounter systems of the form</p>
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<div class="math notranslate nohighlight">
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\[\hat{R}x = y.\]</div>
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<p>This is an upper triangular system that can be solved efficiently
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using back-substitution.</p>
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<p>Written in components, this equation is</p>
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<blockquote>
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<div><div class="math notranslate nohighlight">
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\[ \begin{align}\begin{aligned}R_{11}x_1 + R_{12}x_2 + \ldots + R_{1(m-1)}x_{m-1} + R_{1m}x_m = y_1,\\0x_1 + R_{22}x_2 + \ldots + R_{2(m-1)}x_{m-1} + R_{2m}x_m = y_2,\\\vdots\\0x_1 + 0x_2 + \ldots + R_{(m-1)(m-1)}x_{m-1} + R_{(m-1)m}x_m = y_{m-1},\\ 0x_1 + 0x_2 + \ldots + 0x_{m-1} + R_{mm}x_m = y_{m}.\end{aligned}\end{align} \]</div>
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</div></blockquote>
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<p>The last equation yields <span class="math notranslate nohighlight">\(x_m\)</span> directly by dividing by <span class="math notranslate nohighlight">\(R_{mm}\)</span>, then
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we can use this value to directly compute <span class="math notranslate nohighlight">\(x_{m-1}\)</span>. This is repeated
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for all of the entries of <span class="math notranslate nohighlight">\(x\)</span> from <span class="math notranslate nohighlight">\(m\)</span> down to 1. This procedure is
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called back substitution, which we summarise in the following
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pseudo-code.</p>
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<ul class="simple">
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<li><p><span class="math notranslate nohighlight">\(x_m \gets y_m/R_{mm}\)</span></p></li>
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<li><p>FOR <span class="math notranslate nohighlight">\(i= m-1\)</span> TO 1 (BACKWARDS)</p>
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<ul>
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<li><p><span class="math notranslate nohighlight">\(x_i \gets (y_i - \sum_{k=i+1}^mR_{ik}x_k)/R_{ii}\)</span></p></li>
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</ul>
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</li>
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</ul>
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<div class="proof proof-type-exercise" id="id9">
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<div class="proof-title">
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<span class="proof-type">Exercise 2.9</span>
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\[\hat{R}x = \hat{Q}^*b.\]</div>
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</div></blockquote>
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<p>This is an upper triangular system that can be solved efficiently
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using back-substitution. Written in components, this equation is</p>
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<blockquote>
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<div><div class="math notranslate nohighlight">
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\[ \begin{align}\begin{aligned}R_{11}x_1 + R_{12}x_2 + \ldots + R_{1(m-1)}x_{m-1} + R_{1m}x_m = y_1,\\0x_1 + R_{22}x_2 + \ldots + R_{2(m-1)}x_{m-1} + R_{2m}x_m = y_2,\\\vdots\\0x_1 + 0x_2 + \ldots + R_{(m-1)(m-1)}x_{m-1} + R_{(m-1)m}x_m = y_{m-1},\\ 0x_1 + 0x_2 + \ldots + 0x_{m-1} + R_{mm}x_m = y_{m}.\end{aligned}\end{align} \]</div>
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</div></blockquote>
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<p>The last equation yields <span class="math notranslate nohighlight">\(x_m\)</span> directly by dividing by <span class="math notranslate nohighlight">\(R_{mm}\)</span>, then
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we can use this value to directly compute <span class="math notranslate nohighlight">\(x_{m-1}\)</span>. This is repeated
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for all of the entries of <span class="math notranslate nohighlight">\(x\)</span> from <span class="math notranslate nohighlight">\(m\)</span> down to 1. This procedure is
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called back substitution, which we summarise in the following
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pseudo-code.</p>
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<ul class="simple">
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<li><p><span class="math notranslate nohighlight">\(x_m \gets y_m/R_{mm}\)</span></p></li>
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<li><p>FOR <span class="math notranslate nohighlight">\(i= m-1\)</span> TO 1 (BACKWARDS)</p>
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<ul>
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<li><p><span class="math notranslate nohighlight">\(x_i \gets (y_i - \sum_{k=i+1}^mR_{ik}x_k)/R_{ii}\)</span></p></li>
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</ul>
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</li>
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</ul>
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using back-substitution.</p>
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<div class="proof proof-type-exercise" id="id12">
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<div class="proof-title">
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by forming an appropriate augmented matrix <span class="math notranslate nohighlight">\(\hat{A}\)</span>, calling
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<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
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subarrays using slice notation, before using
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<a class="reference external" href="http://scipy.github.io/devdocs/reference/generated/scipy.linalg.solve_triangular.html#scipy.linalg.solve_triangular" title="(in SciPy v1.10.0.dev0+2059.a082eb4)"><code class="xref py py-func docutils literal notranslate"><span class="pre">scipy.linalg.solve_triangular()</span></code></a> to solve the resulting upper
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<a class="reference internal" href="cla_utils.html#cla_utils.exercises3.solve_U" title="cla_utils.exercises3.solve_U"><code class="xref py py-func docutils literal notranslate"><span class="pre">cla_utils.exercises3.solve_U()</span></code></a> to solve the resulting upper
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triangular system, before returning the solution <span class="math notranslate nohighlight">\(x\)</span>. The test
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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
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test this function.</p>

‎_sources/L2_QR_factorisation.rst.txt‎

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.. vimeo:: 450202242
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In this section we will frequently encounter systems of the form
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.. math::
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\hat{R}x = y.
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This is an upper triangular system that can be solved efficiently
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using back-substitution.
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Written in components, this equation is
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.. math::
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R_{11}x_1 + R_{12}x_2 + \ldots + R_{1(m-1)}x_{m-1} + R_{1m}x_m = y_1,
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0x_1 + R_{22}x_2 + \ldots + R_{2(m-1)}x_{m-1} + R_{2m}x_m = y_2,
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\vdots
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0x_1 + 0x_2 + \ldots + R_{(m-1)(m-1)}x_{m-1} + R_{(m-1)m}x_m = y_{m-1},
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0x_1 + 0x_2 + \ldots + 0x_{m-1} + R_{mm}x_m = y_{m}.
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The last equation yields `x_m` directly by dividing by `R_{mm}`, then
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we can use this value to directly compute `x_{m-1}`. This is repeated
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for all of the entries of `x` from `m` down to 1. This procedure is
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called back substitution, which we summarise in the following
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pseudo-code.
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* `x_m \gets y_m/R_{mm}`
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* FOR `i= m-1` TO 1 (BACKWARDS)
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* `x_i \gets (y_i - \sum_{k=i+1}^mR_{ik}x_k)/R_{ii}`
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.. proof:exercise::
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`(\ddagger)` The function :func:`cla_utils.exercises3.solve_U` has
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\hat{R}x = \hat{Q}^*b.
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This is an upper triangular system that can be solved efficiently
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using back-substitution. Written in components, this equation is
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.. math::
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R_{11}x_1 + R_{12}x_2 + \ldots + R_{1(m-1)}x_{m-1} + R_{1m}x_m = y_1,
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0x_1 + R_{22}x_2 + \ldots + R_{2(m-1)}x_{m-1} + R_{2m}x_m = y_2,
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\vdots
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0x_1 + 0x_2 + \ldots + R_{(m-1)(m-1)}x_{m-1} + R_{(m-1)m}x_m = y_{m-1},
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0x_1 + 0x_2 + \ldots + 0x_{m-1} + R_{mm}x_m = y_{m}.
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The last equation yields `x_m` directly by dividing by `R_{mm}`, then
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we can use this value to directly compute `x_{m-1}`. This is repeated
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for all of the entries of `x` from `m` down to 1. This procedure is
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called back substitution, which we summarise in the following
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pseudo-code.
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* `x_m \gets y_m/R_{mm}`
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* FOR `i= m-1` TO 1 (BACKWARDS)
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* `x_i \gets (y_i - \sum_{k=i+1}^mR_{ik}x_k)/R_{ii}`
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using back-substitution.
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.. proof:exercise::
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by forming an appropriate augmented matrix `\hat{A}`, calling
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:func:`cla_utils.exercises3.householder` and extracting appropriate
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subarrays using slice notation, before using
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:func:`scipy.linalg.solve_triangular` to solve the resulting upper
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:func:`cla_utils.exercises3.solve_U` to solve the resulting upper
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triangular system, before returning the solution `x`. The test
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script ``test_exercises3.py`` in the ``test`` directory will also
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test this function.

‎_sources/errata.rst.txt‎

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* Exercises 2.4 and 2.5: the tests expect you to change the array
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provided as input and not make a copy.
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* Exercise 2.12 was updated as we don't need to use scipy anymore (because
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solve\_U was moved earlier.
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* solveQ should be renamed to solve\_Q in test\_exercises1.test\_solveQ.

‎errata.html‎

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<ul class="simple">
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<li><p>Exercises 2.4 and 2.5: the tests expect you to change the array
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provided as input and not make a copy.</p></li>
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<li><p>Exercise 2.12 was updated as we don’t need to use scipy anymore (because
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solve_U was moved earlier.</p></li>
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<li><p>solveQ should be renamed to solve_Q in test_exercises1.test_solveQ.</p></li>
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</ul>
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</section>
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‎searchindex.js‎

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