Question about matrices and eigenvectors












3












$begingroup$


For two $ntimes n$ square matrices $A$ and $B$ satisfying $AB=A+B,;$
suppose $A$ has an eigenvector $u.$

How to show that $u$ is also and eigenvector of $B?$






linear-algebra







share|cite|improve this question











$endgroup$












  • $begingroup$
    Show us what you tried please.
    $endgroup$
    – user376343
    Dec 13 '18 at 12:14






  • 4




    $begingroup$
    I tried from definition $Au=lambda u$ and multiply both side by u i.e. ABu=Au+Bu, then Bu=ABu-Au=(AB-A)u. But I don’t know how to continue onward.
    $endgroup$
    – Real_Galois
    Dec 13 '18 at 12:19








  • 2




    $begingroup$
    But AB doesnt have to equal BA
    $endgroup$
    – Real_Galois
    Dec 13 '18 at 12:23






  • 1




    $begingroup$
    Also, it seems that proof by contradiction is not working (or I don’t know the way)
    $endgroup$
    – Real_Galois
    Dec 13 '18 at 12:27










  • $begingroup$
    Well! Could you please enclose your comments above into your question?
    $endgroup$
    – user376343
    Dec 13 '18 at 13:29
















3












$begingroup$


For two $ntimes n$ square matrices $A$ and $B$ satisfying $AB=A+B,;$
suppose $A$ has an eigenvector $u.$

How to show that $u$ is also and eigenvector of $B?$






linear-algebra







share|cite|improve this question











$endgroup$












  • $begingroup$
    Show us what you tried please.
    $endgroup$
    – user376343
    Dec 13 '18 at 12:14






  • 4




    $begingroup$
    I tried from definition $Au=lambda u$ and multiply both side by u i.e. ABu=Au+Bu, then Bu=ABu-Au=(AB-A)u. But I don’t know how to continue onward.
    $endgroup$
    – Real_Galois
    Dec 13 '18 at 12:19








  • 2




    $begingroup$
    But AB doesnt have to equal BA
    $endgroup$
    – Real_Galois
    Dec 13 '18 at 12:23






  • 1




    $begingroup$
    Also, it seems that proof by contradiction is not working (or I don’t know the way)
    $endgroup$
    – Real_Galois
    Dec 13 '18 at 12:27










  • $begingroup$
    Well! Could you please enclose your comments above into your question?
    $endgroup$
    – user376343
    Dec 13 '18 at 13:29














3












3








3


1



$begingroup$


For two $ntimes n$ square matrices $A$ and $B$ satisfying $AB=A+B,;$
suppose $A$ has an eigenvector $u.$

How to show that $u$ is also and eigenvector of $B?$






linear-algebra







share|cite|improve this question











$endgroup$




For two $ntimes n$ square matrices $A$ and $B$ satisfying $AB=A+B,;$
suppose $A$ has an eigenvector $u.$

How to show that $u$ is also and eigenvector of $B?$






linear-algebra




linear-algebra






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 13 '18 at 14:44









user376343

3,7383827




3,7383827










asked Dec 13 '18 at 12:12









Real_GaloisReal_Galois

255




255












  • $begingroup$
    Show us what you tried please.
    $endgroup$
    – user376343
    Dec 13 '18 at 12:14






  • 4




    $begingroup$
    I tried from definition $Au=lambda u$ and multiply both side by u i.e. ABu=Au+Bu, then Bu=ABu-Au=(AB-A)u. But I don’t know how to continue onward.
    $endgroup$
    – Real_Galois
    Dec 13 '18 at 12:19








  • 2




    $begingroup$
    But AB doesnt have to equal BA
    $endgroup$
    – Real_Galois
    Dec 13 '18 at 12:23






  • 1




    $begingroup$
    Also, it seems that proof by contradiction is not working (or I don’t know the way)
    $endgroup$
    – Real_Galois
    Dec 13 '18 at 12:27










  • $begingroup$
    Well! Could you please enclose your comments above into your question?
    $endgroup$
    – user376343
    Dec 13 '18 at 13:29


















  • $begingroup$
    Show us what you tried please.
    $endgroup$
    – user376343
    Dec 13 '18 at 12:14






  • 4




    $begingroup$
    I tried from definition $Au=lambda u$ and multiply both side by u i.e. ABu=Au+Bu, then Bu=ABu-Au=(AB-A)u. But I don’t know how to continue onward.
    $endgroup$
    – Real_Galois
    Dec 13 '18 at 12:19








  • 2




    $begingroup$
    But AB doesnt have to equal BA
    $endgroup$
    – Real_Galois
    Dec 13 '18 at 12:23






  • 1




    $begingroup$
    Also, it seems that proof by contradiction is not working (or I don’t know the way)
    $endgroup$
    – Real_Galois
    Dec 13 '18 at 12:27










  • $begingroup$
    Well! Could you please enclose your comments above into your question?
    $endgroup$
    – user376343
    Dec 13 '18 at 13:29
















$begingroup$
Show us what you tried please.
$endgroup$
– user376343
Dec 13 '18 at 12:14




$begingroup$
Show us what you tried please.
$endgroup$
– user376343
Dec 13 '18 at 12:14




4




4




$begingroup$
I tried from definition $Au=lambda u$ and multiply both side by u i.e. ABu=Au+Bu, then Bu=ABu-Au=(AB-A)u. But I don’t know how to continue onward.
$endgroup$
– Real_Galois
Dec 13 '18 at 12:19






$begingroup$
I tried from definition $Au=lambda u$ and multiply both side by u i.e. ABu=Au+Bu, then Bu=ABu-Au=(AB-A)u. But I don’t know how to continue onward.
$endgroup$
– Real_Galois
Dec 13 '18 at 12:19






2




2




$begingroup$
But AB doesnt have to equal BA
$endgroup$
– Real_Galois
Dec 13 '18 at 12:23




$begingroup$
But AB doesnt have to equal BA
$endgroup$
– Real_Galois
Dec 13 '18 at 12:23




1




1




$begingroup$
Also, it seems that proof by contradiction is not working (or I don’t know the way)
$endgroup$
– Real_Galois
Dec 13 '18 at 12:27




$begingroup$
Also, it seems that proof by contradiction is not working (or I don’t know the way)
$endgroup$
– Real_Galois
Dec 13 '18 at 12:27












$begingroup$
Well! Could you please enclose your comments above into your question?
$endgroup$
– user376343
Dec 13 '18 at 13:29




$begingroup$
Well! Could you please enclose your comments above into your question?
$endgroup$
– user376343
Dec 13 '18 at 13:29










2 Answers
2






active

oldest

votes


















4












$begingroup$

Hint: from the given condition show that $A-I$ and $B-I$ are inverse of each of each other. Then from the given equation solve for $B$.



It is easy to check that $(A-I)(B-I)=I$. Now $$(A-I)B=A. $$ Multiply bothsides by $B-I$ to get $$B= (B-I)A=BA-A.$$



You should be able to continue from here since now $AB= BA$.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Show $B=(A-I)^{-1}A=(B-I) A$.
    $endgroup$
    – user9077
    Dec 13 '18 at 13:49








  • 2




    $begingroup$
    If you have show that $(A−I)(B−I)=I$ it's enough. $A−I$ has the same eigenvectors as $A$ , $B−I$ as inverse the same as $A−I$ and $B$ the same as $B−I$
    $endgroup$
    – Widawensen
    Dec 13 '18 at 16:31






  • 1




    $begingroup$
    Nice observations, thanks!
    $endgroup$
    – user9077
    Dec 13 '18 at 16:48



















1












$begingroup$

With the hint of @user9077, and without need of commutativity of $A,B,$ assume $(lambda,v);$ is an eigenpair of $A.$
Then $$Bv=(B-I)Av=(B-I)lambda v=lambda B v- lambda v$$ hence $$(1-lambda)Bv=-lambda v.$$ This means that $(frac{lambda}{lambda-1},v)$ is an eigenpair of $B.$






share|cite|improve this answer









$endgroup$













    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%2f3037923%2fquestion-about-matrices-and-eigenvectors%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4












    $begingroup$

    Hint: from the given condition show that $A-I$ and $B-I$ are inverse of each of each other. Then from the given equation solve for $B$.



    It is easy to check that $(A-I)(B-I)=I$. Now $$(A-I)B=A. $$ Multiply bothsides by $B-I$ to get $$B= (B-I)A=BA-A.$$



    You should be able to continue from here since now $AB= BA$.






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      Show $B=(A-I)^{-1}A=(B-I) A$.
      $endgroup$
      – user9077
      Dec 13 '18 at 13:49








    • 2




      $begingroup$
      If you have show that $(A−I)(B−I)=I$ it's enough. $A−I$ has the same eigenvectors as $A$ , $B−I$ as inverse the same as $A−I$ and $B$ the same as $B−I$
      $endgroup$
      – Widawensen
      Dec 13 '18 at 16:31






    • 1




      $begingroup$
      Nice observations, thanks!
      $endgroup$
      – user9077
      Dec 13 '18 at 16:48
















    4












    $begingroup$

    Hint: from the given condition show that $A-I$ and $B-I$ are inverse of each of each other. Then from the given equation solve for $B$.



    It is easy to check that $(A-I)(B-I)=I$. Now $$(A-I)B=A. $$ Multiply bothsides by $B-I$ to get $$B= (B-I)A=BA-A.$$



    You should be able to continue from here since now $AB= BA$.






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      Show $B=(A-I)^{-1}A=(B-I) A$.
      $endgroup$
      – user9077
      Dec 13 '18 at 13:49








    • 2




      $begingroup$
      If you have show that $(A−I)(B−I)=I$ it's enough. $A−I$ has the same eigenvectors as $A$ , $B−I$ as inverse the same as $A−I$ and $B$ the same as $B−I$
      $endgroup$
      – Widawensen
      Dec 13 '18 at 16:31






    • 1




      $begingroup$
      Nice observations, thanks!
      $endgroup$
      – user9077
      Dec 13 '18 at 16:48














    4












    4








    4





    $begingroup$

    Hint: from the given condition show that $A-I$ and $B-I$ are inverse of each of each other. Then from the given equation solve for $B$.



    It is easy to check that $(A-I)(B-I)=I$. Now $$(A-I)B=A. $$ Multiply bothsides by $B-I$ to get $$B= (B-I)A=BA-A.$$



    You should be able to continue from here since now $AB= BA$.






    share|cite|improve this answer











    $endgroup$



    Hint: from the given condition show that $A-I$ and $B-I$ are inverse of each of each other. Then from the given equation solve for $B$.



    It is easy to check that $(A-I)(B-I)=I$. Now $$(A-I)B=A. $$ Multiply bothsides by $B-I$ to get $$B= (B-I)A=BA-A.$$



    You should be able to continue from here since now $AB= BA$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 13 '18 at 13:58

























    answered Dec 13 '18 at 13:25









    user9077user9077

    1,239612




    1,239612








    • 1




      $begingroup$
      Show $B=(A-I)^{-1}A=(B-I) A$.
      $endgroup$
      – user9077
      Dec 13 '18 at 13:49








    • 2




      $begingroup$
      If you have show that $(A−I)(B−I)=I$ it's enough. $A−I$ has the same eigenvectors as $A$ , $B−I$ as inverse the same as $A−I$ and $B$ the same as $B−I$
      $endgroup$
      – Widawensen
      Dec 13 '18 at 16:31






    • 1




      $begingroup$
      Nice observations, thanks!
      $endgroup$
      – user9077
      Dec 13 '18 at 16:48














    • 1




      $begingroup$
      Show $B=(A-I)^{-1}A=(B-I) A$.
      $endgroup$
      – user9077
      Dec 13 '18 at 13:49








    • 2




      $begingroup$
      If you have show that $(A−I)(B−I)=I$ it's enough. $A−I$ has the same eigenvectors as $A$ , $B−I$ as inverse the same as $A−I$ and $B$ the same as $B−I$
      $endgroup$
      – Widawensen
      Dec 13 '18 at 16:31






    • 1




      $begingroup$
      Nice observations, thanks!
      $endgroup$
      – user9077
      Dec 13 '18 at 16:48








    1




    1




    $begingroup$
    Show $B=(A-I)^{-1}A=(B-I) A$.
    $endgroup$
    – user9077
    Dec 13 '18 at 13:49






    $begingroup$
    Show $B=(A-I)^{-1}A=(B-I) A$.
    $endgroup$
    – user9077
    Dec 13 '18 at 13:49






    2




    2




    $begingroup$
    If you have show that $(A−I)(B−I)=I$ it's enough. $A−I$ has the same eigenvectors as $A$ , $B−I$ as inverse the same as $A−I$ and $B$ the same as $B−I$
    $endgroup$
    – Widawensen
    Dec 13 '18 at 16:31




    $begingroup$
    If you have show that $(A−I)(B−I)=I$ it's enough. $A−I$ has the same eigenvectors as $A$ , $B−I$ as inverse the same as $A−I$ and $B$ the same as $B−I$
    $endgroup$
    – Widawensen
    Dec 13 '18 at 16:31




    1




    1




    $begingroup$
    Nice observations, thanks!
    $endgroup$
    – user9077
    Dec 13 '18 at 16:48




    $begingroup$
    Nice observations, thanks!
    $endgroup$
    – user9077
    Dec 13 '18 at 16:48











    1












    $begingroup$

    With the hint of @user9077, and without need of commutativity of $A,B,$ assume $(lambda,v);$ is an eigenpair of $A.$
    Then $$Bv=(B-I)Av=(B-I)lambda v=lambda B v- lambda v$$ hence $$(1-lambda)Bv=-lambda v.$$ This means that $(frac{lambda}{lambda-1},v)$ is an eigenpair of $B.$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      With the hint of @user9077, and without need of commutativity of $A,B,$ assume $(lambda,v);$ is an eigenpair of $A.$
      Then $$Bv=(B-I)Av=(B-I)lambda v=lambda B v- lambda v$$ hence $$(1-lambda)Bv=-lambda v.$$ This means that $(frac{lambda}{lambda-1},v)$ is an eigenpair of $B.$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        With the hint of @user9077, and without need of commutativity of $A,B,$ assume $(lambda,v);$ is an eigenpair of $A.$
        Then $$Bv=(B-I)Av=(B-I)lambda v=lambda B v- lambda v$$ hence $$(1-lambda)Bv=-lambda v.$$ This means that $(frac{lambda}{lambda-1},v)$ is an eigenpair of $B.$






        share|cite|improve this answer









        $endgroup$



        With the hint of @user9077, and without need of commutativity of $A,B,$ assume $(lambda,v);$ is an eigenpair of $A.$
        Then $$Bv=(B-I)Av=(B-I)lambda v=lambda B v- lambda v$$ hence $$(1-lambda)Bv=-lambda v.$$ This means that $(frac{lambda}{lambda-1},v)$ is an eigenpair of $B.$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 13 '18 at 14:41









        user376343user376343

        3,7383827




        3,7383827






























            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.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3037923%2fquestion-about-matrices-and-eigenvectors%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