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<section id="preconditioning-krylov-methods">
<h1><span class="section-number">7. </span>Preconditioning Krylov methods<a class="headerlink" href="#preconditioning-krylov-methods" title="Link to this heading">¶</a></h1>
<p>In this section we will discuss some preconditioners and how to
analyse them. The most important question is how quickly the
preconditioned GMRES algorithm converges for a given matrix and
preconditioner. We will focus on the link between stationary
iterative methods and preconditioners.</p>
<section id="stationary-iterative-methods">
<h2><span class="section-number">7.1. </span>Stationary iterative methods<a class="headerlink" href="#stationary-iterative-methods" title="Link to this heading">¶</a></h2>
<p>As we have already discussed, given a matrix equation <span class="math notranslate nohighlight">\(Ax=b\)</span>,
iterative methods provide a way of obtaining a (hopefully) better
approximate solution <span class="math notranslate nohighlight">\({x}^{k+1}\)</span> from a previous approximate
<span class="math notranslate nohighlight">\({x}^k\)</span>. Stationary iterative methods are defined from splittings
as follows.</p>
<div class="proof proof-type-definition" id="id1">
<div class="proof-title">
<span class="proof-type">Definition 7.1</span>
<span class="proof-title-name">(Stationary iterative methods)</span>
</div><div class="proof-content">
<p>A stationary iterative method is constructed from matrices <span class="math notranslate nohighlight">\(M\)</span> and
<span class="math notranslate nohighlight">\(N\)</span> with <span class="math notranslate nohighlight">\(A=M+N\)</span>. Then the iterative method is defined by</p>
<div class="math notranslate nohighlight">
\[M{x}^{k+1}=-N{x}^k+{b}.\]</div>
</div></div><p>The word “stationary” refers to the fact that exactly the same thing
is done at each iteration. This contrasts with Krylov methods such as
GMRES, where the sequence of operations depends on the previous
iterations (e.g. a different size least square system is solve in each
GMRES iteration).</p>
<p>In this section we will introduce/recall some classic stationary
methods.</p>
<div class="proof proof-type-definition" id="id2">
<div class="proof-title">
<span class="proof-type">Definition 7.2</span>
<span class="proof-title-name">(Richardson iteration)</span>
</div><div class="proof-content">
<p>For a chosen parameter <span class="math notranslate nohighlight">\(\omega>0\)</span>, take <span class="math notranslate nohighlight">\(M=I/\omega\)</span>. This
defines the iterative method given by</p>
<div class="math notranslate nohighlight">
\[{x}^{k+1} = {x}^k+\omega\left({b}-A{x}^k\right).\]</div>
</div></div><p>Richardson, L.F. (1910). <em>The approximate arithmetical solution
by finite differences of physical problems involving differential
equations, with an application to the stresses in a masonry
dam</em>. Philos. Trans. Roy. Soc. London Ser. A 210: 307-357.</p>
<p>This approach is convenient for parallel computing, because each entry in
<span class="math notranslate nohighlight">\(x^{k+1}\)</span> can be updated independently, once <span class="math notranslate nohighlight">\(Ax^k\)</span> has been evaluated.</p>
<div class="proof proof-type-definition" id="id3">
<div class="proof-title">
<span class="proof-type">Definition 7.3</span>
<span class="proof-title-name">(Jacobi’s method)</span>
</div><div class="proof-content">
<p>Split <span class="math notranslate nohighlight">\(A=L+D+U\)</span> with <span class="math notranslate nohighlight">\(L\)</span> strictly lower triangular, <span class="math notranslate nohighlight">\(D\)</span>
diagonal and <span class="math notranslate nohighlight">\(U\)</span> strictly upper triangular, i.e.</p>
<div class="math notranslate nohighlight">
\[L_{ij}=0, \, j\geq i, \quad D_{ij}=0, \, i\neq j, \quad U_{ij}, \, i\geq j.\]</div>
<p>Then, Jacobi’s method is</p>
<div class="math notranslate nohighlight">
\[D{x}^{k+1} = {b}-(L+U){x}^k.\]</div>
</div></div><p><span class="math notranslate nohighlight">\(D\)</span> is very cheap to invert because it is diagonal; entries in
<span class="math notranslate nohighlight">\(x^{k+1}\)</span> can be updated independently once <span class="math notranslate nohighlight">\((L+U)x^k\)</span> has been evaluated.</p>
<p>Jacobi, C.G.J. (1845). <em>Ueber eine neue Aufloesungsart der bei der
Methode der kleinsten Quadrate vorkommenden linearen Gleichungen</em>,
Astronomische Nachrichten, 22, 297-306.</p>
<div class="proof proof-type-definition" id="id4">
<div class="proof-title">
<span class="proof-type">Definition 7.4</span>
<span class="proof-title-name">(Gauss-Seidel Method)</span>
</div><div class="proof-content">
<p>Split <span class="math notranslate nohighlight">\(A=L+D+U\)</span> with <span class="math notranslate nohighlight">\(L\)</span> strictly lower triangular, <span class="math notranslate nohighlight">\(D\)</span>
diagonal and <span class="math notranslate nohighlight">\(U\)</span> strictly upper triangular.
The Gauss-Seidel method (forward or backwards) is</p>
<div class="math notranslate nohighlight">
\[(L+D){x}^{k+1} = {b}-U{x}^k, \quad\mbox{or},\quad
(U+D){x}^{k+1} = {b}-L{x}^k.\]</div>
</div></div><p>Each Gauss-Seidel iteration requires the solution of a triangular
system by forward/backward substitution.</p>
<div class="proof proof-type-exercise" id="id5">
<div class="proof-title">
<span class="proof-type">Exercise 7.5</span>
</div><div class="proof-content">
<p>Show that forward Gauss-Seidel is a modification Jacobi’s method
but using new values as soon as possible.</p>
</div></div><div class="proof proof-type-definition" id="id6">
<div class="proof-title">
<span class="proof-type">Definition 7.6</span>
<span class="proof-title-name">(Scaled Gauss-Seidel method)</span>
</div><div class="proof-content">
<p>We introduce a scaling/relaxation parameter <span class="math notranslate nohighlight">\(\omega>0\)</span> and
take <span class="math notranslate nohighlight">\(M=D/\omega+L\)</span>, so that</p>
<div class="math notranslate nohighlight">
\[\left(\frac{1}{\omega}D+L\right){x}^{k+1}
= {b}+\left(\left(\frac{1}{\omega}-1\right)D-U\right){x}^k.\]</div>
</div></div><p>For <span class="math notranslate nohighlight">\(\omega=1\)</span>, we recover Gauss-Seidel. For <span class="math notranslate nohighlight">\(1<\omega<2\)</span>, we often
obtain faster convergence. This is called Successive Over-Relaxation
(SOR). The optimal value of <span class="math notranslate nohighlight">\(\omega\)</span> is known for some problems.
This was state of the art for numerical solution of PDEs in the 50s
and 60s.</p>
<ul class="simple">
<li><p>Richardson and Jacobi are <em>simultaneous displacement
methods</em>: updates can be done simultaneously (e.g. on a GPU).
Changing variables by a permutation does not alter the algorithm.</p></li>
<li><p>Gauss-Seidel and SOR are <em>successive displacement methods</em>:
we can only overwrite the old vector with the new one element by element.
Successive displacement methods usually converge faster, and changing
variables by a permutation does alter the algorithm.</p></li>
</ul>
</section>
<section id="using-splitting-methods-as-preconditioners">
<h2><span class="section-number">7.2. </span>Using splitting methods as preconditioners<a class="headerlink" href="#using-splitting-methods-as-preconditioners" title="Link to this heading">¶</a></h2>
<p>A (non-symmetric) preconditioner <span class="math notranslate nohighlight">\(\hat{A}\)</span> can be built from a
splitting method by applying one iteration with initial guess
<span class="math notranslate nohighlight">\({x}^0={0}\)</span>.</p>
<p>A preconditioner is used to compute <span class="math notranslate nohighlight">\(v\)</span> by solving</p>
<div class="math notranslate nohighlight">
\[\hat{A}v = r,\]</div>
<p>such that <span class="math notranslate nohighlight">\(Av \approx r\)</span>, and the above equation is easy to solve.
We can use a stationary method by setting <span class="math notranslate nohighlight">\(v=x^1\)</span> and <span class="math notranslate nohighlight">\(x^0=0\)</span>,</p>
<p>to get</p>
<div class="math notranslate nohighlight">
\[Mv:= M{x}^1 = -N\underbrace{x^0}_{=0} + r = r,\]</div>
<p>i.e. we are choosing <span class="math notranslate nohighlight">\(\hat{A}=M\)</span>. Later we shall see how to relate
convergence properties of splitting methods to the convergence of
preconditioned CG using <span class="math notranslate nohighlight">\(\hat{A}=M\)</span>.</p>
</section>
<section id="symmetric-iterative-methods">
<h2><span class="section-number">7.3. </span>Symmetric iterative methods<a class="headerlink" href="#symmetric-iterative-methods" title="Link to this heading">¶</a></h2>
<p>Consider a symmetric matrix <span class="math notranslate nohighlight">\(A=A^T\)</span>
If we can build iterative methods from the splitting
<span class="math notranslate nohighlight">\(A=M+N\)</span>, then we can also build iterative methods from the splitting
<span class="math notranslate nohighlight">\(A=A^T=M^T+N^T\)</span>. We can then combine them together.</p>
<div class="proof proof-type-definition" id="id7">
<div class="proof-title">
<span class="proof-type">Definition 7.7</span>
<span class="proof-title-name">(Symmetric iterative method)</span>
</div><div class="proof-content">
<p>Given a splitting <span class="math notranslate nohighlight">\(A=M+N\)</span>, a symmetric method performs one
stationary iteration using <span class="math notranslate nohighlight">\(M+N\)</span>, followed by one stationary
iteration using <span class="math notranslate nohighlight">\(M^T+N^T\)</span>, i.e.</p>
<div class="math notranslate nohighlight">
\[M{x}^{k+\frac{1}{2}}=-N{x}^k + {b}, \quad
M^T{x}^{k+1}=-N^T{x}^{k+\frac{1}{2}} + {b}.\]</div>
</div></div><div class="proof proof-type-example" id="id8">
<div class="proof-title">
<span class="proof-type">Example 7.8</span>
<span class="proof-title-name">(Symmetric Successive Over-Relaxation (SSOR).)</span>
</div><div class="proof-content">
<p>For a symmetric matrix <span class="math notranslate nohighlight">\(A=L+D+U\)</span>, <span class="math notranslate nohighlight">\(L=U^T\)</span>. The symmetric version
of SOR is then</p>
<div class="math notranslate nohighlight">
\[\begin{split}(L+\frac{1}{\omega}D){x}^{k+\frac{1}{2}}&=\left(\left(\frac{1}{\omega}-1\right)
D-U\right){x}^k + {b}, \\
(U+\frac{1}{\omega}D){x}^{k+1}&=\left(\left(\frac{1}{\omega}-1\right)
D-L\right){x}^{k+\frac{1}{2}} + {b}.\end{split}\]</div>
</div></div><p>Some Krylov methods, notably the Conjugate Gradient method, require
the preconditioner <span class="math notranslate nohighlight">\(\hat{A}\)</span> to be symmetric.
We can build symmetric preconditioners from symmetric splitting methods.
Write the symmetric iteration as a single step with
<span class="math notranslate nohighlight">\({x}^0={0}\)</span>.</p>
<div class="math notranslate nohighlight">
\[\begin{split}M^T{x}^{1}&=(M-A)^T{x}^{\frac{1}{2}} + {b}, \\
&= (M-A)^TM^{-1}{b} + {b}, \\
& = (M^T + M-A)M^{-1}{b},\end{split}\]</div>
<p>so that</p>
<div class="math notranslate nohighlight">
\[x^1 = M^{-T}(M^T + M - A)M^{-1}b,\]</div>
<p>i.e. <span class="math notranslate nohighlight">\(\hat{A}^{-1}=M^{-T}(M^T + M-A)M^{-1}\)</span>.</p>
<div class="proof proof-type-example" id="id9">
<div class="proof-title">
<span class="proof-type">Example 7.9</span>
<span class="proof-title-name">(Symmetric Gauss-Seidel preconditioner)</span>
</div><div class="proof-content">
<p><span class="math notranslate nohighlight">\(\hat{A}^{-1} = (L+D)^{-T}D(L+D)^{-1}\)</span>.</p>
</div></div></section>
<section id="convergence-criteria-for-stationary-methods">
<h2><span class="section-number">7.4. </span>Convergence criteria for stationary methods<a class="headerlink" href="#convergence-criteria-for-stationary-methods" title="Link to this heading">¶</a></h2>
<p>In this section we will look at the convergence of stationary
methods. This is relevant because it relates directly to the
convergence properties of the corresponding preconditioned Krylov
method when the stationary method is used as a preconditioner.</p>
<p>For a splitting <span class="math notranslate nohighlight">\(A=M+N\)</span>, recall that the iterative method is</p>
<div class="math notranslate nohighlight">
\[M{x}^{k+1} = -N{x}^k + {b}.\]</div>
<p>On the other hand, the solution <span class="math notranslate nohighlight">\({x}^*\)</span> of <span class="math notranslate nohighlight">\(A{x}={b}\)</span> satisfies</p>
<div class="math notranslate nohighlight">
\[M{x}^* = -N{x}^* + {b}.\]</div>
<p>Subtracting these two equations gives</p>
<div class="math notranslate nohighlight">
\[M{e}^{k+1} = -N{e}^k, \quad {e}^k = {x}^*-{x}^k,\]</div>
<p>so</p>
<div class="math notranslate nohighlight">
\[{e}^{k+1}=C{e}^k \implies {e}^k = C^k{e}^0, \quad
C:=-M^{-1}N = -M^{-1}(A-M) = I - M^{-1}A.\]</div>
<p><span class="math notranslate nohighlight">\(C\)</span> is called the <em>iteration matrix</em>.</p>
<p>For a symmetric iterative method,</p>
<div class="math notranslate nohighlight">
\[M{x}^{k+\frac{1}{2}}=-N{x}^k + {b}, \quad
M^T{x}^{k+1}=-N^T{x}^{k+\frac{1}{2}} + {b},\]</div>
<p>we subtract <span class="math notranslate nohighlight">\(Ax^*=b\)</span> from both equations to get</p>
<div class="math notranslate nohighlight">
\[M{e}^{k+\frac{1}{2}}=-N{e}^k, \quad
M^T{e}^{k+1}=-N^T{e}^{k+\frac{1}{2}}.\]</div>
<p>Then eliminating <span class="math notranslate nohighlight">\(e^{k+1/2}\)</span> gives</p>
<div class="math notranslate nohighlight">
\[M^Te^{k+1} = N^TM^{-1}Ne^k,\]</div>
<p>i.e. the iteration matrix is</p>
<div class="math notranslate nohighlight">
\[C = M^{-T}N^TM^{-1}N\]</div>
<div class="proof proof-type-exercise" id="id10">
<div class="proof-title">
<span class="proof-type">Exercise 7.10</span>
</div><div class="proof-content">
<p>Show that</p>
</div></div><div class="math notranslate nohighlight">
\[C = I-\left(M_s\right)^{-1}A,\]</div>
<p>where</p>
<div class="math notranslate nohighlight">
\[M_s = M(M+M^T-A)^{-1}M^T.\]</div>
<p>From the above exercise, note the relationship
between <span class="math notranslate nohighlight">\(M_s\)</span> and <span class="math notranslate nohighlight">\(\hat{A}\)</span> for symmetric methods.</p>
<div class="proof proof-type-definition" id="id11">
<div class="proof-title">
<span class="proof-type">Definition 7.11</span>
<span class="proof-title-name">(Convergence of stationary methods)</span>
</div><div class="proof-content">
<p>An iterative method based on the splitting <span class="math notranslate nohighlight">\(A=M+N\)</span> with iteration
matrix <span class="math notranslate nohighlight">\(C=-M^{-1}N\)</span> is called {convergent} if</p>
<div class="math notranslate nohighlight">
\[{y}^k = C^k{y}^0 \to {0}\]</div>
<p>for any initial vector <span class="math notranslate nohighlight">\({y}^0\)</span>.</p>
</div></div><div class="proof proof-type-exercise" id="id12">
<div class="proof-title">
<span class="proof-type">Exercise 7.12</span>
</div><div class="proof-content">
<p>Show that this implies that <span class="math notranslate nohighlight">\({e}^k={x}^*-{x}^k\to{0}\)</span> i.e.
<span class="math notranslate nohighlight">\({x}^k\to x^*\)</span> as <span class="math notranslate nohighlight">\(k\to\infty\)</span>.</p>
</div></div><div class="proof proof-type-theorem" id="id13">
<div class="proof-title">
<span class="proof-type">Theorem 7.13</span>
<span class="proof-title-name">(A first convergence criterion)</span>
</div><div class="proof-content">
<dl class="simple">
<dt>If <span class="math notranslate nohighlight">\(\|C\|<1\)</span>, using the operator norm for some chosen vector norm,</dt><dd><p>then the iterative method converges.</p>
</dd>
</dl>
</div></div><div class="proof proof-type-proof">
<div class="proof-title">
<span class="proof-type">Proof </span>
</div><div class="proof-content">
<div class="math notranslate nohighlight">
\[\begin{split}\|{y}^k\| = & \|C^k{y}^0\| \\
\leq & \|C^k\|\|{y}^0\| \\
\leq & \ \left(\|C\|\right)^k\|{y}^0\|
\to 0\quad\mbox{as}\,k\to\infty.\end{split}\]</div>
</div></div><p>This is only a sufficient condition. There may be matrices <span class="math notranslate nohighlight">\(C\)</span> with
<span class="math notranslate nohighlight">\(\|C\|>1\)</span>, but the method is still
convergent.</p>
<p>To obtain a necessary condition, we need to use the spectral radius.</p>
<div class="proof proof-type-definition" id="id14">
<div class="proof-title">
<span class="proof-type">Definition 7.14</span>
</div><div class="proof-content">
<p>The spectral radius <span class="math notranslate nohighlight">\(\rho(C)\)</span> of a matrix <span class="math notranslate nohighlight">\(C\)</span> is
the maximum of the absolute values of all the eigenvalues <span class="math notranslate nohighlight">\(\lambda_i\)</span>
of <span class="math notranslate nohighlight">\(C\)</span>:</p>
<div class="math notranslate nohighlight">
\[\rho(C) = \max_{1\leq i\leq n}|\lambda_i|.\]</div>
</div></div><div class="proof proof-type-theorem" id="id15">
<div class="proof-title">
<span class="proof-type">Theorem 7.15</span>
</div><div class="proof-content">
<p>An iterative method converges <span class="math notranslate nohighlight">\(\iff \rho(C)<1\)</span>.</p>
</div></div><div class="proof proof-type-proof">
<div class="proof-title">
<span class="proof-type">Proof </span>
</div><div class="proof-content">
<p>[Proof that <span class="math notranslate nohighlight">\(\rho(C)\geq 1\implies\)</span> non-convergence]</p>
<p>If <span class="math notranslate nohighlight">\(\rho(C)\geq 1\)</span>, then <span class="math notranslate nohighlight">\(C\)</span> has an eigenvector <span class="math notranslate nohighlight">\({v}\)</span> with
<span class="math notranslate nohighlight">\(\|{v}\|_2=1\)</span> and eigenvalue <span class="math notranslate nohighlight">\(\lambda\)</span> with <span class="math notranslate nohighlight">\(|\lambda|>1\)</span>. Then</p>
<div class="math notranslate nohighlight">
\[\|C^k{v}\|_2 = \|\lambda^k{v}\|_2 = |\lambda|^k\|{v}\|_2\geq 1,\]</div>
<p>which does not converge to zero.</p>
<p>[Proof that <span class="math notranslate nohighlight">\(\rho(C)< 1\implies\)</span> convergence]</p>
<p>Assume a linearly independent eigenvalue expansion (not necessary
for the proof but it simplifies things a lot)
<span class="math notranslate nohighlight">\({z} = \sum_{i=1}^n\alpha_i{v}_i\)</span>. Then,</p>
<div class="math notranslate nohighlight">
\[C^k{z} = \sum_{i=1}^n\alpha_iC^k{v}_i
= \sum_{i=1}^n\alpha_i\lambda^k{v}_i\to 0.\]</div>
</div></div><ul class="simple">
<li><p>For symmetric matrices <span class="math notranslate nohighlight">\(B\)</span>, <span class="math notranslate nohighlight">\(\rho(B)=\|B\|_2\)</span>, so</p></li>
</ul>
<p>the two convergence theorems are related.</p>
<ul>
<li><p>If <span class="math notranslate nohighlight">\(\|C\|=c<1\)</span>, then</p>
<div class="math notranslate nohighlight">
\[\|{e}^{k+1}\| = \|C{e}^k\| \leq \|C\|\|{e}^k\|
= c\|{e}^k\|.\]</div>
<p>This guarantees that the error will be reduced by a factor of at least
<span class="math notranslate nohighlight">\(c\)</span> in each iteration. If we only have <span class="math notranslate nohighlight">\(\rho(C)<1\)</span>, not <span class="math notranslate nohighlight">\(\|C\|<1\)</span>
then the error may not converge monotonically.</p>
</li>
</ul>
<div class="proof proof-type-example" id="id16">
<div class="proof-title">
<span class="proof-type">Example 7.16</span>
<span class="proof-title-name">(Range of SOR parameter)</span>
</div><div class="proof-content">
<blockquote>
<div><p>We can use this to analyse the SOR parameter <span class="math notranslate nohighlight">\(\omega\)</span>.</p>
<div class="math notranslate nohighlight">
\[\left(\frac{1}{\omega}D+L\right){x}^{k+1} =
{b}+\left(\left(\frac{1}{\omega}-1\right)D-U\right){x}^k\]</div>
<p>What values of <span class="math notranslate nohighlight">\(\omega\)</span>? For SOR,iteration matrix <span class="math notranslate nohighlight">\(C\)</span> is</p>
<div class="math notranslate nohighlight">
\[C = \left(\frac{1}{\omega}D+L\right)^{-1}
\left(\frac{1-\omega}{\omega}D-U\right) = (D+\omega L)^{-1}
((1-\omega)D-\omega U).\]</div>
<p>so</p>
</div></blockquote>
<div class="math notranslate nohighlight">
\[\begin{split}\det(C) & =
\det\left((D+\omega L)^{-1}
((1-\omega)D-\omega U)\right) \\
& = \det\left((D+\omega L)^{-1}\right)\det\left(
(1-\omega)D - \omega U\right) \\
& = \det\left(D^{-1}\right)\det(D)\det\left((I-\omega I) -
\omega D^{-1}U\right)\\
& = \det\left((1-\omega)I\right) = (1-\omega)^n.\end{split}\]</div>
<p>The determinant is the product of the eigenvalues, hence <span class="math notranslate nohighlight">\(\rho(C)<1\)</span>
requires <span class="math notranslate nohighlight">\(|1-\omega|<1\)</span>.</p>
</div></div></section>
<section id="splitting-methods-as-preconditioners">
<h2><span class="section-number">7.5. </span>Splitting methods as preconditioners<a class="headerlink" href="#splitting-methods-as-preconditioners" title="Link to this heading">¶</a></h2>
<p>Recall that preconditioned GMRES converges well if the eigenvalues
of <span class="math notranslate nohighlight">\(\hat{A}^{-1}A\)</span> are clustered together.</p>
<div class="proof proof-type-theorem" id="id17">
<div class="proof-title">
<span class="proof-type">Theorem 7.17</span>
</div><div class="proof-content">
<p>Let <span class="math notranslate nohighlight">\(A\)</span> be a matrix with splitting <span class="math notranslate nohighlight">\(M+N\)</span>, such that <span class="math notranslate nohighlight">\(\rho(C) < c <
1\)</span>. Then, the eigenvalues of the left preconditioned matrix
<span class="math notranslate nohighlight">\(\hat{A}^{-1}A\)</span> with <span class="math notranslate nohighlight">\(\hat{A}=M\)</span> are located in a disk of radius
<span class="math notranslate nohighlight">\(c\)</span> around <span class="math notranslate nohighlight">\(1\)</span> in the complex plane.</p>
</div></div><div class="proof proof-type-proof">
<div class="proof-title">
<span class="proof-type">Proof </span>
</div><div class="proof-content">
<div class="math notranslate nohighlight">
\[C=-M^{-1}N = M^{-1}(M-A) = I-M^{-1}A.\]</div>
<p>Then,</p>
<div class="math notranslate nohighlight">
\[1>c>\rho(C)=\rho(I-M^{-1}A),\]</div>
<p>and the result follows since <span class="math notranslate nohighlight">\(I\)</span> and <span class="math notranslate nohighlight">\(M^{-1}A\)</span> have a simultaneous
eigendecomposition.</p>
</div></div><p>We deduce that good convergence of the GMRES algorithm occurs when <span class="math notranslate nohighlight">\(c\)</span>
is small.</p>
<p>For symmetric splittings, we have already observed that the iteration
matrix is</p>
<div class="math notranslate nohighlight">
\[C = I-\left(M_s\right)^{-1}A,\]</div>
<p>where</p>
<div class="math notranslate nohighlight">
\[M_s = M(M+M^T-A)^{-1}M^T.\]</div>
<p>For symmetric splittings we can say a little more about the preconditioner.</p>
<div class="proof proof-type-theorem" id="id18">
<div class="proof-title">
<span class="proof-type">Theorem 7.18</span>
</div><div class="proof-content">
<p>Let <span class="math notranslate nohighlight">\(A\)</span> be a matrix with splitting
<span class="math notranslate nohighlight">\(M+N\)</span>, such that the symmetric splitting has iteration matrix</p>
<div class="math notranslate nohighlight">
\[\rho(C) = c < 1,\]</div>
<p>and assume further that <span class="math notranslate nohighlight">\(M_s\)</span> is positive definite.</p>
<p>Then, the eigenvalues of the symmetric preconditioned matrix
<span class="math notranslate nohighlight">\(\hat{A}^{-1}A\)</span> are contained in the interval <span class="math notranslate nohighlight">\([1-c,1+c]\)</span>.</p>
</div></div><div class="proof proof-type-proof">
<div class="proof-title">
<span class="proof-type">Proof </span>
</div><div class="proof-content">
<p>We have</p>
<div class="math notranslate nohighlight">
\[\begin{split}C & = I-\left(M_s\right)^{-1}A, \\
& = I - \hat{A}^{-1}A,\end{split}\]</div>
<p>so <span class="math notranslate nohighlight">\(\rho(I- \hat{A}^{-1}A) = \rho(C) = c\)</span>. Further, <span class="math notranslate nohighlight">\(M_s\)</span> is
symmetric and positive definite, so there exists a unique symmetric
positive definite matrix square root <span class="math notranslate nohighlight">\(S\)</span> such that <span class="math notranslate nohighlight">\(SS =
M_s\)</span>. Then,</p>
<div class="math notranslate nohighlight">
\[M_s^{-1}A = SSA = S(SAS)S^{-1}.\]</div>
<p>Thus, <span class="math notranslate nohighlight">\(M_s^{-1}A\)</span> is similar to (and therefore has the same eigenvalues as)
<span class="math notranslate nohighlight">\(SAS\)</span>, which is symmetric, and therefore has real eigenvalues,
and the result follows.</p>
</div></div></section>
<section id="convergence-analysis-for-richardson">
<h2><span class="section-number">7.6. </span>Convergence analysis for Richardson<a class="headerlink" href="#convergence-analysis-for-richardson" title="Link to this heading">¶</a></h2>
<p>First we examine Richardson iteration. In the unscaled case,</p>
<div class="math notranslate nohighlight">
\[{x}^{k+1} = {x}^k - \left(A{x}^k - {b}\right), \quad
M = I, \, N = A-I, \, \implies C = I-A.\]</div>
<p>Let <span class="math notranslate nohighlight">\({e}\)</span> be an eigenvector of <span class="math notranslate nohighlight">\(A\)</span> with eigenvalue <span class="math notranslate nohighlight">\(\lambda\)</span>, so
<span class="math notranslate nohighlight">\(A{e}=\lambda{e}\)</span>. Then <span class="math notranslate nohighlight">\((I-A){e}={e}-\lambda{e}=(1-\lambda){e}\)</span>.
So, <span class="math notranslate nohighlight">\({e}\)</span> is an eigenvector of <span class="math notranslate nohighlight">\(I-A\)</span> with eigenvalue <span class="math notranslate nohighlight">\(1-\lambda\)</span>.
Richardson’s method will converge if <span class="math notranslate nohighlight">\(\rho(C)<1\)</span> i.e.
<span class="math notranslate nohighlight">\(|1-\lambda|<1\)</span> for all eigenvalues <span class="math notranslate nohighlight">\(\lambda\)</span> of <span class="math notranslate nohighlight">\(A\)</span>.</p>
<p>This is restrictive, which motivates the scaled Richardson iteration,</p>
<div class="math notranslate nohighlight">
\[{x}^{k+1} = {x}^k - \omega\left(A{x}^k - {b}\right), \quad
M = \frac{I}{\omega}, \, N = A-\frac{I}{\omega}, \, \implies C =
I-\omega A.\]</div>
<p>If <span class="math notranslate nohighlight">\(A\)</span> has eigenvalues <span class="math notranslate nohighlight">\(\lambda_1,\lambda_2,\ldots,\lambda_n\)</span> then the
iterative matrix <span class="math notranslate nohighlight">\(C\)</span> has eigenvalues
<span class="math notranslate nohighlight">\(1-\omega\lambda_1,1-\omega\lambda_2,\ldots,1-\omega\lambda_n\)</span>. This
requires <span class="math notranslate nohighlight">\(|1-\omega\lambda_i|<1\)</span>, <span class="math notranslate nohighlight">\(i=1,\ldots,n\)</span>, for convergence.</p>
<p>If, further, <span class="math notranslate nohighlight">\(A\)</span> is symmetric positive definite, then all eigenvalues
are real and positive. Then, all of the eigenvalues of <span class="math notranslate nohighlight">\(C\)</span> lie between
<span class="math notranslate nohighlight">\(1-\omega\lambda_{\min}\)</span> and <span class="math notranslate nohighlight">\(1-\omega\lambda_{\max}\)</span>. We can
minimise <span class="math notranslate nohighlight">\(\rho(C)\)</span> by choosing
<span class="math notranslate nohighlight">\(\omega=2/(\lambda_{\min}+\lambda_{\max})\)</span>. The resulting iteration
matrix has spectral radius</p>
<div class="math notranslate nohighlight">
\[\rho(C) = 1-2\frac{\lambda_{\min}}{\lambda_{\min}+\lambda_{\max}}
= \frac{\lambda_{\max}-\lambda_{\min}}{\lambda_{\min}+\lambda_{\max}}.\]</div>
</section>
<section id="convergence-analysis-for-symmetric-matrices">
<h2><span class="section-number">7.7. </span>Convergence analysis for symmetric matrices<a class="headerlink" href="#convergence-analysis-for-symmetric-matrices" title="Link to this heading">¶</a></h2>
<p>For a symmetric positive definite matrix <span class="math notranslate nohighlight">\(A\)</span>, recall the
Rayleigh Quotient formula,</p>
<div class="math notranslate nohighlight">
\[\lambda_{\max}=\max_{{x}\ne
0}\frac{{x}^TA{x}}{{x}^T{x}}\equiv \|A\|_2^2, \quad
\lambda_{\min}=\min_{{x}\ne
0}\frac{{x}^TA{x}}{{x}^T{x}},\]</div>
<p>implying that</p>
<div class="math notranslate nohighlight">
\[\lambda_{\min}\|{y}\|_2^2\leq {y}^TA{y}
\leq\lambda_{\max}\|{y}\|_2^2\]</div>
<p>for any non-zero vector <span class="math notranslate nohighlight">\({y}\)</span>.</p>
<div class="proof proof-type-definition" id="id19">
<div class="proof-title">
<span class="proof-type">Definition 7.19</span>
<span class="proof-title-name">(<span class="math notranslate nohighlight">\(A\)</span>-weighted norm)</span>
</div><div class="proof-content">
<p>For symmetric positive definite <span class="math notranslate nohighlight">\(A\)</span>, we can define the weighted
vector norm</p>
<div class="math notranslate nohighlight">
\[\|{x}\|_A = \sqrt{{x}^TA{x}},\]</div>
<p>and the corresponding matrix (operator) norm</p>
<div class="math notranslate nohighlight">
\[\|B\|_A = \|A^{1/2}BA^{-1/2}\|_2.\]</div>
</div></div><p>These norms are useful for studying convergence of iterative methods
for <span class="math notranslate nohighlight">\(A{x}={b}\)</span> in the symmetric positive definite case.</p>
<div class="proof proof-type-theorem" id="id20">
<div class="proof-title">
<span class="proof-type">Theorem 7.20</span>
</div><div class="proof-content">
<p>For a splitting <span class="math notranslate nohighlight">\(A=M+N\)</span>, if the (symmetric) matrix <span class="math notranslate nohighlight">\(M+M^T-A\)</span> is
positive definite then</p>
<div class="math notranslate nohighlight">
\[\|I-M^{-1}A\|_A<1.\]</div>
</div></div><div class="proof proof-type-proof">
<div class="proof-title">
<span class="proof-type">Proof </span>
</div><div class="proof-content">
<p>If <span class="math notranslate nohighlight">\({y}=(I-M^{-1}A){x}\)</span>, <span class="math notranslate nohighlight">\({w}=M^{-1}A{x}\)</span>, then</p>
<div class="math notranslate nohighlight">
\[\begin{split}\|{y}\|_A^2 & = ({x}-{w})^TA({x}-{w})
= {x}^TA{x}-2{w}^TM{w} + {w}^TA{w} \\
&= {x}^TA{x}-{w}^T(M+M^T){w} + {w}^TA{w} \\
&= {x}^TA{x}-{w}^T(M+M^T-A){w} \\
&\leq \|{x}\|_A^2 - \mu_{\min}\|{w}\|^2_2,\end{split}\]</div>
<p>where <span class="math notranslate nohighlight">\(\mu_{\min}\)</span> is the (positive) minimum eigenvalue of <span class="math notranslate nohighlight">\(M^T+M-A\)</span>.</p>
<p>Further,</p>
<div class="math notranslate nohighlight">
\[\begin{split}\|{w}\|_2^2 & = {x}^TA\left(M^{-1}\right)^TM^{-1}A{x} \\
&= \left(A^{1/2}{x}\right)^TA^{1/2}
\left(M^{-1}\right)^TM^{-1}A^{1/2}\left(A^{1/2}{x}\right) \\
&\geq \hat{\mu}_{\min}\|A^{{1/2}}{x}\|_2^2 =
\hat{\mu}_{\min}\|{x}\|^2_A,\end{split}\]</div>
<p>where <span class="math notranslate nohighlight">\(\hat{\mu}_{\min}\)</span> is the minimum eigenvalue of
<span class="math notranslate nohighlight">\(A^{1/2}\left(M^{-1}\right)^TM^{-1}A^{1/2}\)</span> i.e. the square
of the minimum eigenvalue of <span class="math notranslate nohighlight">\(M^{-1}A^{1/2}\)</span>, which is invertible so
<span class="math notranslate nohighlight">\(\hat{\mu}_{\min}>0\)</span>. If <span class="math notranslate nohighlight">\({y}=(I-M^{-1}A){x}\)</span>,
<span class="math notranslate nohighlight">\({w}=M^{-1}A{x}\)</span>, then</p>
<div class="math notranslate nohighlight">
\[\|{y}\|^2_A
\leq \left(1-\mu_{\min}\hat{\mu}_{\min}\right)\|{x}\|_A^2<\|{x}\|_A^2.\]</div>
</div></div><p>This enables us to show the following useful result for symmetric
positive definite matrices.</p>
<div class="proof proof-type-theorem" id="id21">
<div class="proof-title">
<span class="proof-type">Theorem 7.21</span>
</div><div class="proof-content">
<p>Let <span class="math notranslate nohighlight">\(A\)</span> be a symmetric positive definite matrix with
splitting <span class="math notranslate nohighlight">\(A=M+N\)</span>, if <span class="math notranslate nohighlight">\(M\)</span> is also symmetric positive definite, then</p>
<div class="math notranslate nohighlight">
\[\rho(I-M^{-1}A) = \|I-M^{-1}A\|_A=\|I-M^{-1}A\|_M.\]</div>
</div></div><div class="proof proof-type-proof">
<div class="proof-title">
<span class="proof-type">Proof </span>
</div><div class="proof-content">
<div class="math notranslate nohighlight">
\[\begin{split}I-A^{1/2}M^{-1}A^{1/2} &= A^{1/2}(I-M^{-1}A)A^{-1/2}, \\
I-M^{-1/2}AM^{-1/2} &= M^{1/2}(I-M^{-1}A)M^{-1/2}, \\\end{split}\]</div>
<p>so <span class="math notranslate nohighlight">\(I-M^{-1}A\)</span>, <span class="math notranslate nohighlight">\(I-A^{1/2}M^{-1}A^{1/2}\)</span>, and <span class="math notranslate nohighlight">\(I-M^{-1/2}AM^{-1/2}\)</span>
all have the same eigenvalues, since they are similar matrices.
Hence,</p>
<div class="math notranslate nohighlight">
\[\begin{split}\rho(I-M^{-1}A) & = \rho(I-A^{1/2}M^{-1}A^{1/2}) \\
& = \|I-A^{1/2}M^{-1}A^{1/2}\|_2 \\
& = \|I-M^{-1}A\|_A,\end{split}\]</div>
<p>and similarly for <span class="math notranslate nohighlight">\(I-M^{-1/2}AM^{-1/2}\)</span>.</p>
</div></div><p>The consequence of this is that if <span class="math notranslate nohighlight">\(M+M^T-A\)</span> is symmetric
positive definite then there is a guaranteed reduction in the <span class="math notranslate nohighlight">\(A\)</span>-norm
of the error in each iteration. If <span class="math notranslate nohighlight">\(M\)</span> is also symmetric
positive definite the there is guaranteed reduction in the <span class="math notranslate nohighlight">\(M\)</span>-norm of
the error in each iteration.</p>
<p>Now we apply this to the convergence of Jacobi iteration. In this
case <span class="math notranslate nohighlight">\(M=D\)</span>, so <span class="math notranslate nohighlight">\(M^T+M-A=2D-A\)</span> which may not be positive definite.
We generalise to scaled Jacobi iteration with <span class="math notranslate nohighlight">\(M=D/\omega\)</span>.</p>
<div class="proof proof-type-proposition" id="id22">
<div class="proof-title">
<span class="proof-type">Proposition 7.22</span>
</div><div class="proof-content">
<p>Let <span class="math notranslate nohighlight">\(A\)</span> be a symmetric positive definite matrix. Let <span class="math notranslate nohighlight">\(\lambda\)</span> be
the (real) maximum eigenvalue of <span class="math notranslate nohighlight">\(D^{-1/2}AD^{-1/2}\)</span>. If <span class="math notranslate nohighlight">\(\omega <
2/\lambda\)</span> then scaled Jacobi iteration converges.</p>
</div></div><div class="proof proof-type-proof">
<div class="proof-title">
<span class="proof-type">Proof </span>
</div><div class="proof-content">
<p>For scaled Jacobi iteration with <span class="math notranslate nohighlight">\(M=D/\omega\)</span>, we have
<span class="math notranslate nohighlight">\(M^T+M-A=2D/\omega-A\)</span>. To check positive definiteness we need to
show that</p>
<div class="math notranslate nohighlight">
\[x^T\left(\frac{2}{\omega}D - A\right)x > 0,\]</div>
<p>for all <span class="math notranslate nohighlight">\(x\neq 0\)</span>.</p>
<p>To show this, we write <span class="math notranslate nohighlight">\(x = D^{-1/2}y\)</span>, so that</p>
<div class="math notranslate nohighlight">
\[\begin{split}x^T\left(\frac{2}{\omega}D - A\right)x & =
y^T\left(\frac{2}{\omega}I - D^{-1/2}AD^{-1/2}\right)y \\
& \geq \mu \|y\|^2 > 0,\end{split}\]</div>
<p>provided that the minimum eigenvalue <span class="math notranslate nohighlight">\(\mu\)</span> of
<span class="math notranslate nohighlight">\(F=\frac{2}{\omega}I - D^{-1/2}AD^{-1/2}\)</span> is positive (it is real
since <span class="math notranslate nohighlight">\(F\)</span> is symmetric). We have <span class="math notranslate nohighlight">\(\mu=2/\omega - \lambda\)</span>
Hence, <span class="math notranslate nohighlight">\(2D/\omega-A\)</span> is positive
definite (so scaled Jacobi converges) if <span class="math notranslate nohighlight">\(2/\omega-\lambda>0\)</span>
i.e. <span class="math notranslate nohighlight">\(\omega<2/\lambda\)</span>.</p>
</div></div><div class="proof proof-type-proposition" id="id23">
<div class="proof-title">
<span class="proof-type">Proposition 7.23</span>
</div><div class="proof-content">
<p>Let <span class="math notranslate nohighlight">\(A\)</span> be a symmetric positive definite matrix. Then Gauss-Seidel
iteration always converges.</p>
</div></div><div class="proof proof-type-proof">
<div class="proof-title">
<span class="proof-type">Proof </span>
</div><div class="proof-content">
<p>For Gauss-Seidel,</p>
<div class="math notranslate nohighlight">
\[M^T+M-A = (D+L)^T + D+L - A = D+U+D+L-A = D,\]</div>
<p>which is symmetric-positive definite, so Gauss-Seidel always converges.</p>
</div></div><div class="proof proof-type-proposition" id="id24">
<div class="proof-title">
<span class="proof-type">Proposition 7.24</span>
</div><div class="proof-content">
<p>Let <span class="math notranslate nohighlight">\(A\)</span> be a symmetric positive definite matrix. Then SOR converges
provided that <span class="math notranslate nohighlight">\(0<\omega 2\)</span>.</p>
</div></div><div class="proof proof-type-proof">
<div class="proof-title">
<span class="proof-type">Proof </span>
</div><div class="proof-content">
<p>For SOR,</p>
<div class="math notranslate nohighlight">
\[\begin{split}M^T+M-A =& \left(\frac{1}{\omega}D+L\right)^T +
\frac{1}{\omega}D+L - A \\
&= \frac{2}{\omega}D
+U+L-(L+D+U)=\left(\frac{2}{\omega}-1\right)D,\end{split}\]</div>
<p>which is symmetric positive definite provided that
<span class="math notranslate nohighlight">\(0<\omega<2\)</span>.</p>
</div></div></section>
<section id="an-example-matrix-non-examinable-in-2024-25">
<h2><span class="section-number">7.8. </span>An example matrix (Non-examinable in 2024/25)<a class="headerlink" href="#an-example-matrix-non-examinable-in-2024-25" title="Link to this heading">¶</a></h2>
<p>We consider stationary methods for an example arising from the
finite difference discretisation of the two point boundary value
problem</p>
<div class="math notranslate nohighlight">
\[-\frac{d^2u}{dx^2} = f, \quad u(0) = u(1) = 0.\]</div>
<p>Here, <span class="math notranslate nohighlight">\(f\)</span> is assumed known and we have to find <span class="math notranslate nohighlight">\(u\)</span>. We approximate
this problem by writing <span class="math notranslate nohighlight">\(u_k = u(k/(n+1))\)</span> for <span class="math notranslate nohighlight">\(k=0,1,2,\ldots,n+1\)</span>.
From the boundary conditions we have <span class="math notranslate nohighlight">\(u_0=u_{n+1}=0\)</span>, meaning we
just have to find <span class="math notranslate nohighlight">\(u_k\)</span> with <span class="math notranslate nohighlight">\(1\leq k \leq n\)</span>, that solve the
finite difference approximation</p>
<div class="math notranslate nohighlight">
\[-u_{k-1} + 2u_k - u_{k+1} = f_k, \quad 1\leq k \leq n,\]</div>
<p>where <span class="math notranslate nohighlight">\(f_k=f(k/n)/n^2\)</span>, <span class="math notranslate nohighlight">\(1\leq k\leq n\)</span>. Taking into account the
boundary conditions <span class="math notranslate nohighlight">\(u_0=u_{n+1}=0\)</span>, we can write this as a matrix
system <span class="math notranslate nohighlight">\(Ax=b\)</span> with</p>
<div class="math notranslate nohighlight">
\[\begin{split}A = \begin{pmatrix}
2 & -1 & \cdots & \cdots & 0 \\
-1 & 2 & -1 & \cdots & 0 \\
0 & -1 & 2 & \cdots &
\vdots \\
\vdots & & & & \vdots \\
\vdots & 0 & -1 & 2 & -1 \\
0 & 0 & \cdots & -1 & 2
\end{pmatrix},
\quad
x = \begin{pmatrix}
u_1 \\
u_2 \\
\vdots \\
u_n
\end{pmatrix},
\quad
b =
\begin{pmatrix}
f_1 \\
f_2 \\
\vdots \\
f_n
\end{pmatrix}.\end{split}\]</div>
<p>However, it is possible to evaluate <span class="math notranslate nohighlight">\(Ax\)</span> and to implement our classic
stationary iterative methods without ever forming <span class="math notranslate nohighlight">\(A\)</span>. This is critically
important for efficient implementations (especially when extending
to 2D and 3D problems).</p>
<p>We introduce this example matrix because it is possible to compute
spectral radii for all of the matrices arising in the analysis
of classic stationary methods. In the next example we consider
Jacobi.</p>
<div class="proof proof-type-example" id="id25">
<div class="proof-title">
<span class="proof-type">Example 7.25</span>
<span class="proof-title-name">(Jacobi iteration for the example matrix)</span>
</div><div class="proof-content">
<p>In this case, <span class="math notranslate nohighlight">\(D=2I\)</span>. Thus in fact, scaled Jacobi and scaled Richardson
are equivalent. We have to find the maximum eigenvalue of <span class="math notranslate nohighlight">\(K=D^{-1}A\)</span>.
We can compute this by knowing that the eigenvectors <span class="math notranslate nohighlight">\(v\)</span> of <span class="math notranslate nohighlight">\(K\)</span>
are all of the form</p>
<div class="math notranslate nohighlight">
\[\begin{split}v =
\left(
\begin{pmatrix}
\sin(l\pi/(n+1)) \\
\sin(2l\pi/(n+1)) \\
\sin(nl\pi/(n+1)) \\
\end{pmatrix}
\right),\end{split}\]</div>
<p>with one eigenvector for each value of <span class="math notranslate nohighlight">\(0< l <n+1\)</span>. This can be proved
by considering symmetries of the matrix, but here we just assume this
form and establish that we have eigenvectors after substituting into the
definition of an eigenvector <span class="math notranslate nohighlight">\(Av=\lambda v\)</span>. This is a general approach
that can be tried for any matrices arising in the analysis of convergence
of classic stationary methods for this example matrix.</p>
<div class="math notranslate nohighlight">
\[(D^{-1}A u)_k =
-u_{k-1}/2 + u_k - u_{k+1}/2 =
\lambda u_k\]</div>
<p>which becomes</p>
<div class="math notranslate nohighlight">
\[\lambda\sin(kl\pi/(n+1)) = -\sin((l-1)k\pi/(n+1))/2 + \sin(kl\pi/(n+1)) - \sin((l+1)k\pi/(n+1))/2,\]</div>
<p>and you can use trigonometric formulae, or write</p>
<div class="math notranslate nohighlight">
\[\begin{split}\lambda\sin(kl\pi/(n+1) & =
-\sin((l-1)k\pi/(n+1))/2 + \sin(lk\pi/(n+1)) - \sin((l+1)k\pi/(n+1))/2\\
& = \sin(kl\pi/(n+1)) - \Im\left(\exp(ik(l-1)\pi/(n+1)) + \exp(ik(l+1)\pi/(n+1))\right)/2 \\
& = \sin(kl\pi/(n+1)) - \Im\left(\left(
\exp(-ik\pi/(n+1)) + \exp(ik\pi/(n+1))\right)\exp(ikl\pi/(n+1))\right)/2 \\
& = \sin(kl\pi/(n+1)) - \Im\left(\sin(k\pi/(n+1))\exp(ikl\pi/(n+1))\right) \\
& = \sin(kl\pi/(n+1))(1 - \sin(k\pi/(n+1)))\end{split}\]</div>
</div></div><p>and we conclude that <span class="math notranslate nohighlight">\(\lambda=1-\sin(k\pi/(n+1))\)</span> are the eigenvalues
with <span class="math notranslate nohighlight">\(0<k<n+1\)</span>. The maximum eigenvalue corresponds to <span class="math notranslate nohighlight">\(k=1\)</span> and <span class="math notranslate nohighlight">\(k=n\)</span>,
with <span class="math notranslate nohighlight">\(\lambda=1-\sin(\pi/(n+1))\)</span>.</p>
<p>The condition <span class="math notranslate nohighlight">\(\omega<2/\lambda\)</span> thus requires that</p>
<div class="math notranslate nohighlight">
\[\omega < \frac{2}{1-\sin(\pi/(n+1))}.\]</div>
<div class="proof proof-type-exercise" id="id26">
<div class="proof-title">
<span class="proof-type">Exercise 7.26</span>
</div><div class="proof-content">
<p>Find the value of <span class="math notranslate nohighlight">\(\omega\)</span>
for scaled Jacobi such that the convergence rate is maximised,
i.e. so that <span class="math notranslate nohighlight">\(\rho(C)\)</span> is minimised. What happens to this rate
as <span class="math notranslate nohighlight">\(n\to \infty\)</span>?</p>
</div></div></section>
<section id="chebyshev-acceleration-examinable-in-2024-25">
<h2><span class="section-number">7.9. </span>Chebyshev acceleration (examinable in 2024/25)<a class="headerlink" href="#chebyshev-acceleration-examinable-in-2024-25" title="Link to this heading">¶</a></h2>
<p>Say we have computed iterates <span class="math notranslate nohighlight">\({x}^0,{x}^1,\ldots,{x}^k\)</span> using</p>
<div class="math notranslate nohighlight">
\[M{x}^{k+1} = -N{x}^k + {b}.\]</div>
<p>If the method is convergent, then these iterates are homing in on the
solution. Can we use extrapolation through these iterates to obtain a
better guess for the solution?</p>
<div class="math notranslate nohighlight">
\[\mbox{Find}\,c_{jk},\,j=1,\ldots,k,\,\mbox{with}\,
{y}^k = \sum_{j=0}^kc_{jk}{x}^j,\]</div>
<p>with <span class="math notranslate nohighlight">\({y}^k\)</span> the best possible approximation to <span class="math notranslate nohighlight">\({x}^*\)</span>.</p>
<p>The usual iterative method has <span class="math notranslate nohighlight">\(c_{kk}=1\)</span>, and <span class="math notranslate nohighlight">\(c_{jk}=0\)</span> for <span class="math notranslate nohighlight">\(j<k\)</span>.
If <span class="math notranslate nohighlight">\({x}^i={x}^*\)</span>, <span class="math notranslate nohighlight">\(i=0,1,\ldots,k\)</span> then</p>
<div class="math notranslate nohighlight">
\[{y}^k =
\sum_{j=0}^kc_{jk}{x}^*={x}^*\sum_{j=0}^kc_{jk},\]</div>
<dl class="simple">
<dt>so we need <span class="math notranslate nohighlight">\(\sum_{j=0}^kc_{jk}=1\)</span>. Subject to this constraint, we</dt><dd><p>seek to minimise <span class="math notranslate nohighlight">\({y}^k-{x}^* =
\sum_{j=0}^kc_{jk}({x}^j-{x}^*)\)</span>.</p>
</dd>
</dl>
<p>We can interpret this in terms of matrix polynomials
by writing</p>
<div class="math notranslate nohighlight">
\[\begin{split}{x}^*-{y}^k &= \sum_{j=0}^kc_{jk}({x}^*-{x}^j), \\
& = \sum_{j=0}^kc_{jk}\left(-M^{-1}N\right)^j{e}^0, \\
& = p_k\left(-M^{-1}N\right){e}^0,\end{split}\]</div>
<p>where</p>
<div class="math notranslate nohighlight">
\[p_k(X) = c_{0k} + c_{1k}X + c_{2k}X^2 + \ldots + c_{kk}X^k,\]</div>
<p>with <span class="math notranslate nohighlight">\(p_k(1)=1\)</span> (from our condition <span class="math notranslate nohighlight">\(\sum_{j=0}^kc_{jk}=1)\)</span>.</p>
<p>We want to try to minimise <span class="math notranslate nohighlight">\({y}^k-{x}^*\)</span> by choosing <span class="math notranslate nohighlight">\(c_{0k}\)</span>,
<span class="math notranslate nohighlight">\(c_{1k}\)</span>, <span class="math notranslate nohighlight">\(\ldots\)</span>, <span class="math notranslate nohighlight">\(c_{kk}\)</span> so that the eigenvalues of <span class="math notranslate nohighlight">\(p_k\)</span> are as
small as possible. If <span class="math notranslate nohighlight">\(\lambda\)</span> is an eigenvalue of <span class="math notranslate nohighlight">\(C=-M^{-1}N\)</span>,
then <span class="math notranslate nohighlight">\(p_k(\lambda)\)</span> is an eigenvalue of <span class="math notranslate nohighlight">\(p_k(C)\)</span>. It is not practical
to know all the eigenvalues of a large matrix, so we will develop
methods that work if we know that all eigenvalues of <span class="math notranslate nohighlight">\(C\)</span> are real, and
satisfy <span class="math notranslate nohighlight">\(-1<\alpha<\lambda<\beta<1\)</span>, for some constants <span class="math notranslate nohighlight">\(\alpha\)</span>
and <span class="math notranslate nohighlight">\(\beta\)</span> (we know that <span class="math notranslate nohighlight">\(|\lambda|<1\)</span> otherwise the basic method is
not convergent.</p>
<p>If all eigenvalues of <span class="math notranslate nohighlight">\(C\)</span> are real, and satisfy
<span class="math notranslate nohighlight">\(-1<\alpha<\lambda<\beta<1\)</span>,
then we try to make <span class="math notranslate nohighlight">\(\rho_{\max} = \max_{\alpha\leq t\leq\beta}|p_k(t)|\)</span>
as small as possible.
Then, if <span class="math notranslate nohighlight">\(\lambda\)</span> is an eigenvalue of <span class="math notranslate nohighlight">\(C\)</span>, then the corresponding
eigenvalue of <span class="math notranslate nohighlight">\(p_k(C)\)</span> will satisfy
<span class="math notranslate nohighlight">\(|\lambda_{p_k}| = |p_k(\lambda)| \leq \rho_{\max}\)</span>.
We have reduced the problem to trying to find polynomials <span class="math notranslate nohighlight">\(p(t)\)</span> that have the
smallest absolute value in a given range, subject to <span class="math notranslate nohighlight">\(p(1)=1\)</span>.
The solution to this problem is known: Chebyshev polynomials.</p>
<div class="proof proof-type-definition" id="id27">
<div class="proof-title">
<span class="proof-type">Definition 7.27</span>
<span class="proof-title-name">(The Chebyshev polynomial of degree <span class="math notranslate nohighlight">\(k\)</span>, <span class="math notranslate nohighlight">\(T_k(t)\)</span> is defined by
the recurrence)</span>
</div><div class="proof-content">
<div class="math notranslate nohighlight">
\[T_0(t) = 1, \, T_1(t)=t, \, T_k(t)=2tT_{k-1}(t)-T_{k-2}(t).\]</div>
</div></div><p>For example: <span class="math notranslate nohighlight">\(T_2(t) = 2tT_1(t)-T_0(t) = 2t^2-1\)</span>.</p>
<p>If we search for the <span class="math notranslate nohighlight">\(k\)</span>-th degree polynomial <span class="math notranslate nohighlight">\(p_k(t)\)</span> that
minimises</p>
<div class="math notranslate nohighlight">
\[\max_{-1\leq t\leq 1}|p_k(t)|\]</div>
<p>subject to the constraint that the coefficient of <span class="math notranslate nohighlight">\(t^k\)</span> is <span class="math notranslate nohighlight">\(2^{k-1}\)</span>
then we get the <span class="math notranslate nohighlight">\(k\)</span>-th order Chebyshev polynomial <span class="math notranslate nohighlight">\(T_k(t)\)</span>. The
maximum value is <span class="math notranslate nohighlight">\(1\)</span>.</p>
<p>This is not quite what we want, so we change variables, to get</p>
<div class="math notranslate nohighlight">
\[T_k\left(\frac{2t-\beta-\alpha}{\beta-\alpha}\right)\quad\mbox{minimises}
\quad \max_{\alpha\leq t\leq \beta}|p_k(t)|\]</div>
<p>subject to the constraint that the coefficient of <span class="math notranslate nohighlight">\(t^k\)</span> is
<span class="math notranslate nohighlight">\(2^{2k-1}/(\beta-\alpha)\)</span>.
The maximum value is <span class="math notranslate nohighlight">\(1\)</span>.</p>
<p>Then we scale the polynomial to reach the condition <span class="math notranslate nohighlight">\(p_k(1)=1\)</span>.</p>
<div class="math notranslate nohighlight">
\[p_k=\frac{T_k\left(\frac{2t-\beta-\alpha}{\beta-\alpha}\right)}
{T_k\left(\frac{2-\beta-\alpha}{\beta-\alpha}\right)}
\quad\mbox{minimises}
\quad \max_{\alpha\leq t\leq \beta}|p_k(t)|\]</div>
<p>subject to the constraint that <span class="math notranslate nohighlight">\(p_k(1)=1\)</span>.
The maximum value is</p>
<div class="math notranslate nohighlight">
\[\frac{1}{T_k\left(\frac{2-\beta-\alpha}{\beta-\alpha}\right)}.\]</div>
<p>Say we have computed iterates <span class="math notranslate nohighlight">\({x}^0,{x}^1,\ldots,{x}^k\)</span> using</p>
<div class="math notranslate nohighlight">
\[M{x}^{k+1} = -N{x}^k + {b}.\]</div>
<p>Write</p>
<div class="math notranslate nohighlight">
\[p_k=\frac{T_k\left(\frac{2t-\beta-\alpha}{\beta-\alpha}\right)}
{T_k\left(\frac{2-\beta-\alpha}{\beta-\alpha}\right)}\]</div>
<p>in the form</p>
<div class="math notranslate nohighlight">
\[p_k(t) = c_{0k} + c_{1k}t + c_{2k}t^2 + \ldots + c_{kk}t^k,\]</div>
<p>then</p>
<div class="math notranslate nohighlight">
\[{y}^k = \sum_{j=0}^kc_{jk}{x}^k.\]</div>
<p>There appears to be a practical problem: we need to store <span class="math notranslate nohighlight">\({x}^0\)</span>,
<span class="math notranslate nohighlight">\({x}^1\)</span>, <span class="math notranslate nohighlight">\(\ldots\)</span>, <span class="math notranslate nohighlight">\({x}^k\)</span> in order to calculate <span class="math notranslate nohighlight">\({y}^k\)</span>. However,
we can get a formula for <span class="math notranslate nohighlight">\({y}^k\)</span> in terms of <span class="math notranslate nohighlight">\({y}^{k-1}\)</span> and
<span class="math notranslate nohighlight">\({y}^{k-2}\)</span> by using</p>
<div class="math notranslate nohighlight">
\[T_k(t) = 2tT_{k-1}(t)-T_{k-2}(t).\]</div>
<p>We get</p>
<div class="math notranslate nohighlight">
\[p_k(t) = 2\frac{2t-\beta-\alpha}{\beta-\alpha}
\frac{T_{k-1}(s)}{T_k(s)}p_{k-1}(t) -
\frac{T_{k-2}(s)}{T_k(s)}p_{k-2}(t),\]</div>
<p>where <span class="math notranslate nohighlight">\(s=\frac{2-\beta-\alpha}{\beta-\alpha}\)</span>.</p>
<p>After some manipulations we obtain</p>
<div class="math notranslate nohighlight">
\[{y}^k = \omega_k\left({y}^{k-1}-{y}^{k-2}+\gamma{z}^{k-1}
\right)+{y}^{k-2},\]</div>
<p>where</p>
<div class="math notranslate nohighlight">
\[\gamma=\frac{2}{2-\beta-\alpha}, \quad M{z}^{k-1}={b}-A{y}^{k-1}.\]</div>
<p>with starting formulas</p>
<div class="math notranslate nohighlight">
\[\begin{split}{y}^0 & = {x}^0 \\
{y}^1 & = {x}^0 + \gamma M^{-1}({b}-A{x}^0).\end{split}\]</div>
<p>Also,</p>
<div class="math notranslate nohighlight">
\[\omega_k = \frac{1}{1-\omega_{k-1}/(4s^2)}, \, \omega_1=2.\]</div>
<p>(See Golub and Van Loan for details).</p>
<p>Chebyshev can dramatically accelerate preconditioners provided that
the preconditioned operator is positive definite and upper
and lower bounds on the eigenvalues are known.</p>
</section>
</section>
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