Estimate the number of iterations












0














Construct $A =QLambda Q^T$. $Q$ is found by applying $QR$ factorization to $B=$randn($n$), and $Lambda$ is defined to be
begin{align*}
Lambda = mathrm{diag}(lambda_1,lambda_2,ldots,lambda_n),
end{align*}
where $(lambda_i)_{i=1}^n$ is a colloction of iid random variables and $lambda_1$ is uniform on $[-1,1]$. It's clear that $|A|<1$. Let $vec b=$rand($n,1$), which means its entries are uniformly distributed random variables on $[0,1]$.



When solving $(I - A)vec x = vec b$, I apply the Neumann series iteration as follows:
begin{align*}
vec x_0 &= vec 0,\
vec x_j &= Avec x_{j-1} + b, quad j = 1,2,3ldots.
end{align*}
There are two ways to define the number of iterations.



First, define the number of iterations $hat k_epsilon(A)$ as
begin{align*}
hat k_epsilon(A) = min left{ k : frac{|A|^k}{1 - |A|} sqrt{n}< epsilonright}.
end{align*}



Also, define the number of iterations $tilde k_epsilon(A,vec b, vec x_0)$ as
begin{align*}
tilde k_epsilon(A,vec b, vec x_0) &= min left{ k : |vec x - vec x_k| < epsilonright}.
end{align*}
When $vec b sim text{rand(n)}$, and $vec x_0 = 0$, I want to show that
begin{align*}
hat k_epsilon(A) geq tilde k_epsilon(A,vec b, vec x_0).
end{align*}
Besides, I want to find an expression for $hat k_epsilon(A)$ in terms of the eigenvalues of $A$? Since $hat k_epsilon(A)$ is affected only by $A$ and $epsilon$, can I just write $frac{|A|^k}{1 - |A|} sqrt{n}= epsilon$ and move everything except $k$ to the RHS and estimate the value of $k$? Is this the desired expression of $hat k_epsilon(A)$?






linear-algebra numerical-linear-algebra







share|cite|improve this question
























  • Any hint on how to solve this problem?
    – Jiexiong687691
    Nov 29 at 9:08
















0














Construct $A =QLambda Q^T$. $Q$ is found by applying $QR$ factorization to $B=$randn($n$), and $Lambda$ is defined to be
begin{align*}
Lambda = mathrm{diag}(lambda_1,lambda_2,ldots,lambda_n),
end{align*}
where $(lambda_i)_{i=1}^n$ is a colloction of iid random variables and $lambda_1$ is uniform on $[-1,1]$. It's clear that $|A|<1$. Let $vec b=$rand($n,1$), which means its entries are uniformly distributed random variables on $[0,1]$.



When solving $(I - A)vec x = vec b$, I apply the Neumann series iteration as follows:
begin{align*}
vec x_0 &= vec 0,\
vec x_j &= Avec x_{j-1} + b, quad j = 1,2,3ldots.
end{align*}
There are two ways to define the number of iterations.



First, define the number of iterations $hat k_epsilon(A)$ as
begin{align*}
hat k_epsilon(A) = min left{ k : frac{|A|^k}{1 - |A|} sqrt{n}< epsilonright}.
end{align*}



Also, define the number of iterations $tilde k_epsilon(A,vec b, vec x_0)$ as
begin{align*}
tilde k_epsilon(A,vec b, vec x_0) &= min left{ k : |vec x - vec x_k| < epsilonright}.
end{align*}
When $vec b sim text{rand(n)}$, and $vec x_0 = 0$, I want to show that
begin{align*}
hat k_epsilon(A) geq tilde k_epsilon(A,vec b, vec x_0).
end{align*}
Besides, I want to find an expression for $hat k_epsilon(A)$ in terms of the eigenvalues of $A$? Since $hat k_epsilon(A)$ is affected only by $A$ and $epsilon$, can I just write $frac{|A|^k}{1 - |A|} sqrt{n}= epsilon$ and move everything except $k$ to the RHS and estimate the value of $k$? Is this the desired expression of $hat k_epsilon(A)$?






linear-algebra numerical-linear-algebra







share|cite|improve this question
























  • Any hint on how to solve this problem?
    – Jiexiong687691
    Nov 29 at 9:08














0












0








0







Construct $A =QLambda Q^T$. $Q$ is found by applying $QR$ factorization to $B=$randn($n$), and $Lambda$ is defined to be
begin{align*}
Lambda = mathrm{diag}(lambda_1,lambda_2,ldots,lambda_n),
end{align*}
where $(lambda_i)_{i=1}^n$ is a colloction of iid random variables and $lambda_1$ is uniform on $[-1,1]$. It's clear that $|A|<1$. Let $vec b=$rand($n,1$), which means its entries are uniformly distributed random variables on $[0,1]$.



When solving $(I - A)vec x = vec b$, I apply the Neumann series iteration as follows:
begin{align*}
vec x_0 &= vec 0,\
vec x_j &= Avec x_{j-1} + b, quad j = 1,2,3ldots.
end{align*}
There are two ways to define the number of iterations.



First, define the number of iterations $hat k_epsilon(A)$ as
begin{align*}
hat k_epsilon(A) = min left{ k : frac{|A|^k}{1 - |A|} sqrt{n}< epsilonright}.
end{align*}



Also, define the number of iterations $tilde k_epsilon(A,vec b, vec x_0)$ as
begin{align*}
tilde k_epsilon(A,vec b, vec x_0) &= min left{ k : |vec x - vec x_k| < epsilonright}.
end{align*}
When $vec b sim text{rand(n)}$, and $vec x_0 = 0$, I want to show that
begin{align*}
hat k_epsilon(A) geq tilde k_epsilon(A,vec b, vec x_0).
end{align*}
Besides, I want to find an expression for $hat k_epsilon(A)$ in terms of the eigenvalues of $A$? Since $hat k_epsilon(A)$ is affected only by $A$ and $epsilon$, can I just write $frac{|A|^k}{1 - |A|} sqrt{n}= epsilon$ and move everything except $k$ to the RHS and estimate the value of $k$? Is this the desired expression of $hat k_epsilon(A)$?






linear-algebra numerical-linear-algebra







share|cite|improve this question















Construct $A =QLambda Q^T$. $Q$ is found by applying $QR$ factorization to $B=$randn($n$), and $Lambda$ is defined to be
begin{align*}
Lambda = mathrm{diag}(lambda_1,lambda_2,ldots,lambda_n),
end{align*}
where $(lambda_i)_{i=1}^n$ is a colloction of iid random variables and $lambda_1$ is uniform on $[-1,1]$. It's clear that $|A|<1$. Let $vec b=$rand($n,1$), which means its entries are uniformly distributed random variables on $[0,1]$.



When solving $(I - A)vec x = vec b$, I apply the Neumann series iteration as follows:
begin{align*}
vec x_0 &= vec 0,\
vec x_j &= Avec x_{j-1} + b, quad j = 1,2,3ldots.
end{align*}
There are two ways to define the number of iterations.



First, define the number of iterations $hat k_epsilon(A)$ as
begin{align*}
hat k_epsilon(A) = min left{ k : frac{|A|^k}{1 - |A|} sqrt{n}< epsilonright}.
end{align*}



Also, define the number of iterations $tilde k_epsilon(A,vec b, vec x_0)$ as
begin{align*}
tilde k_epsilon(A,vec b, vec x_0) &= min left{ k : |vec x - vec x_k| < epsilonright}.
end{align*}
When $vec b sim text{rand(n)}$, and $vec x_0 = 0$, I want to show that
begin{align*}
hat k_epsilon(A) geq tilde k_epsilon(A,vec b, vec x_0).
end{align*}
Besides, I want to find an expression for $hat k_epsilon(A)$ in terms of the eigenvalues of $A$? Since $hat k_epsilon(A)$ is affected only by $A$ and $epsilon$, can I just write $frac{|A|^k}{1 - |A|} sqrt{n}= epsilon$ and move everything except $k$ to the RHS and estimate the value of $k$? Is this the desired expression of $hat k_epsilon(A)$?






linear-algebra numerical-linear-algebra




linear-algebra numerical-linear-algebra






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 3 at 8:11

























asked Nov 29 at 8:09









Jiexiong687691

705




705












  • Any hint on how to solve this problem?
    – Jiexiong687691
    Nov 29 at 9:08


















  • Any hint on how to solve this problem?
    – Jiexiong687691
    Nov 29 at 9:08
















Any hint on how to solve this problem?
– Jiexiong687691
Nov 29 at 9:08




Any hint on how to solve this problem?
– Jiexiong687691
Nov 29 at 9:08










1 Answer
1






active

oldest

votes


















1














Let $rho=sup_i(|lambda_i|)$. Then $||A||_2=rho$. If $rho<1$, then the sequence $(x_i)$ converges and you easily can calculate $k$.



Beware, if, for example $n$ is great and $rhoapprox 1-1/n$, then the convergence is very slow and the cost of the calculation of its limit becomes greater than the cost of the calculation of $(I-A)^{-1}$.



If you uniformly choose the $(lambda_i)$ in $[-1,1]$, then you will be very often in the above case!!






share|cite|improve this answer





















    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3018343%2festimate-the-number-of-iterations%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1














    Let $rho=sup_i(|lambda_i|)$. Then $||A||_2=rho$. If $rho<1$, then the sequence $(x_i)$ converges and you easily can calculate $k$.



    Beware, if, for example $n$ is great and $rhoapprox 1-1/n$, then the convergence is very slow and the cost of the calculation of its limit becomes greater than the cost of the calculation of $(I-A)^{-1}$.



    If you uniformly choose the $(lambda_i)$ in $[-1,1]$, then you will be very often in the above case!!






    share|cite|improve this answer


























      1














      Let $rho=sup_i(|lambda_i|)$. Then $||A||_2=rho$. If $rho<1$, then the sequence $(x_i)$ converges and you easily can calculate $k$.



      Beware, if, for example $n$ is great and $rhoapprox 1-1/n$, then the convergence is very slow and the cost of the calculation of its limit becomes greater than the cost of the calculation of $(I-A)^{-1}$.



      If you uniformly choose the $(lambda_i)$ in $[-1,1]$, then you will be very often in the above case!!






      share|cite|improve this answer
























        1












        1








        1






        Let $rho=sup_i(|lambda_i|)$. Then $||A||_2=rho$. If $rho<1$, then the sequence $(x_i)$ converges and you easily can calculate $k$.



        Beware, if, for example $n$ is great and $rhoapprox 1-1/n$, then the convergence is very slow and the cost of the calculation of its limit becomes greater than the cost of the calculation of $(I-A)^{-1}$.



        If you uniformly choose the $(lambda_i)$ in $[-1,1]$, then you will be very often in the above case!!






        share|cite|improve this answer












        Let $rho=sup_i(|lambda_i|)$. Then $||A||_2=rho$. If $rho<1$, then the sequence $(x_i)$ converges and you easily can calculate $k$.



        Beware, if, for example $n$ is great and $rhoapprox 1-1/n$, then the convergence is very slow and the cost of the calculation of its limit becomes greater than the cost of the calculation of $(I-A)^{-1}$.



        If you uniformly choose the $(lambda_i)$ in $[-1,1]$, then you will be very often in the above case!!







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 30 at 11:24









        loup blanc

        22.4k21750




        22.4k21750






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.





            Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


            Please pay close attention to the following guidance:


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3018343%2festimate-the-number-of-iterations%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Wiesbaden

            Image
            Wappen Deutschlandkarte 50.0825 8.24 117 Koordinaten: 50° 5′  N , 8° 14′  O Basisdaten Bundesland: Hessen Regierungsbezirk: Darmstadt Höhe: 117 m ü. NHN Fläche: 203,93 km 2 Einwohner: 278.654 (31. Dez. 2017) [1] Bevölkerungsdichte: 1366 Einwohner je km 2 Postleitzahlen: 65183–65207, 55246, 55252 Vorlage:Infobox Gemeinde in Deutschland/Wartung/PLZ enthält Text Vorwahlen: 0611, 06122, 06127, 06134 Kfz-Kennzeichen: WI Gemeindeschlüssel: 06 4 14 000 LOCODE: DE WIB NUTS: DE714 Stadtgliederung: 26 Ortsbezirke Adresse der Stadtverwaltung: Schlossplatz 6 65183 Wiesbaden Webpräsenz: www.wiesbaden.de Oberbürgermeister: Sven Gerich (SPD) Lage der Landeshauptstadt Wiesbaden in Hessen und im Regierungsbezirk Darmstadt Offizielles Logo der Landeshauptstadt Wiesbaden Flagge der Landeshauptstadt Wiesbaden Wiesbaden is
            Read more

            Marschland

            Image
            Kilometerweite Marsch in den Vier- und Marschlanden am Elbdeich Die Wedeler Marschlandschaft in Schleswig-Holstein Der Hadelner Kanal zur Entwässerung der Marsch Die tief gelegene Marsch und ein Entwässerungskanal Das Schöpfwerk Otterndorf im Land Hadeln zur Entwässerung der Medem verfügte damals bereits über die größte Kreiselpumpe Europas. Das Schöpfwerk in Neuhaus für die Aue und des Neuhaus-Bülkauer Kanals Als Marsch(land) (v. niederdt., altsächs. mersc ) – auch Masch , Mersch oder Schwemmland genannt – bezeichnet man eine nacheiszeitlich entstandene geomorphologische Landform im Gebiet der nordwestdeutschen Küsten und Flüsse sowie vergleichbare Landformen weltweit. Inhaltsverzeichnis 1 Landschaft 2 Entstehung 3 Marschböden 4 Nutzung 5 Marschgebiete in Deutschland 6 Siehe auch 7 Literatur 8 Weblinks 9 Einzelnachweise Landschaft | Marschen sind generell flache Landstriche ohne natürliche Erhebungen. Si
            Read more

            Dieringhausen

            Image
            Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
            Read more