Does every (complex) $ntimes n$ matrix have n linearly independent eigenvectors?












1












$begingroup$


Every $ntimes n$ matrix will produce an $n$-th order characteristic equation. This polynomial will have $n$ distinct roots or less than $n$ distinct roots (with some repeated).



In the former case, each eigenvalue corresponds to a distinct eigenvector, so you can form an eigenbasis of $n$ linearly independent eigenvectors.



In the latter case, I can see two possibilities:




  1. There are still $n$ linearly independent eigenvectors, but some share the same eigenvalue, so a $k$-fold root of the characteristic equation (an eigenvalue repeated $k$ times) will correspond to $k$ linearly independent eigenvectors.


  2. There are less than $n$ linearly independent eigenvectors, so a $k$-fold root may correspond to less than $k$ linearly independent eigenvectors.



Is one of these two possibilities always true, for all complex $ntimes n$ matrices, or could either possibility be true depending on the matrix in question?



Is there a proof of this, or perhaps a counter example?






linear-algebra matrices eigenvalues-eigenvectors







share|cite|improve this question











$endgroup$












  • $begingroup$
    $begin{bmatrix}1&1\0&1end{bmatrix}$
    $endgroup$
    – David C. Ullrich
    Dec 24 '18 at 17:23


















1












$begingroup$


Every $ntimes n$ matrix will produce an $n$-th order characteristic equation. This polynomial will have $n$ distinct roots or less than $n$ distinct roots (with some repeated).



In the former case, each eigenvalue corresponds to a distinct eigenvector, so you can form an eigenbasis of $n$ linearly independent eigenvectors.



In the latter case, I can see two possibilities:




  1. There are still $n$ linearly independent eigenvectors, but some share the same eigenvalue, so a $k$-fold root of the characteristic equation (an eigenvalue repeated $k$ times) will correspond to $k$ linearly independent eigenvectors.


  2. There are less than $n$ linearly independent eigenvectors, so a $k$-fold root may correspond to less than $k$ linearly independent eigenvectors.



Is one of these two possibilities always true, for all complex $ntimes n$ matrices, or could either possibility be true depending on the matrix in question?



Is there a proof of this, or perhaps a counter example?






linear-algebra matrices eigenvalues-eigenvectors







share|cite|improve this question











$endgroup$












  • $begingroup$
    $begin{bmatrix}1&1\0&1end{bmatrix}$
    $endgroup$
    – David C. Ullrich
    Dec 24 '18 at 17:23
















1












1








1





$begingroup$


Every $ntimes n$ matrix will produce an $n$-th order characteristic equation. This polynomial will have $n$ distinct roots or less than $n$ distinct roots (with some repeated).



In the former case, each eigenvalue corresponds to a distinct eigenvector, so you can form an eigenbasis of $n$ linearly independent eigenvectors.



In the latter case, I can see two possibilities:




  1. There are still $n$ linearly independent eigenvectors, but some share the same eigenvalue, so a $k$-fold root of the characteristic equation (an eigenvalue repeated $k$ times) will correspond to $k$ linearly independent eigenvectors.


  2. There are less than $n$ linearly independent eigenvectors, so a $k$-fold root may correspond to less than $k$ linearly independent eigenvectors.



Is one of these two possibilities always true, for all complex $ntimes n$ matrices, or could either possibility be true depending on the matrix in question?



Is there a proof of this, or perhaps a counter example?






linear-algebra matrices eigenvalues-eigenvectors







share|cite|improve this question











$endgroup$




Every $ntimes n$ matrix will produce an $n$-th order characteristic equation. This polynomial will have $n$ distinct roots or less than $n$ distinct roots (with some repeated).



In the former case, each eigenvalue corresponds to a distinct eigenvector, so you can form an eigenbasis of $n$ linearly independent eigenvectors.



In the latter case, I can see two possibilities:




  1. There are still $n$ linearly independent eigenvectors, but some share the same eigenvalue, so a $k$-fold root of the characteristic equation (an eigenvalue repeated $k$ times) will correspond to $k$ linearly independent eigenvectors.


  2. There are less than $n$ linearly independent eigenvectors, so a $k$-fold root may correspond to less than $k$ linearly independent eigenvectors.



Is one of these two possibilities always true, for all complex $ntimes n$ matrices, or could either possibility be true depending on the matrix in question?



Is there a proof of this, or perhaps a counter example?






linear-algebra matrices eigenvalues-eigenvectors




linear-algebra matrices eigenvalues-eigenvectors






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 24 '18 at 17:26









Namaste

1




1










asked Dec 24 '18 at 17:19









Pancake_SenpaiPancake_Senpai

25116




25116












  • $begingroup$
    $begin{bmatrix}1&1\0&1end{bmatrix}$
    $endgroup$
    – David C. Ullrich
    Dec 24 '18 at 17:23




















  • $begingroup$
    $begin{bmatrix}1&1\0&1end{bmatrix}$
    $endgroup$
    – David C. Ullrich
    Dec 24 '18 at 17:23


















$begingroup$
$begin{bmatrix}1&1\0&1end{bmatrix}$
$endgroup$
– David C. Ullrich
Dec 24 '18 at 17:23






$begingroup$
$begin{bmatrix}1&1\0&1end{bmatrix}$
$endgroup$
– David C. Ullrich
Dec 24 '18 at 17:23












2 Answers
2






active

oldest

votes


















1












$begingroup$

Both possibilities apply. Case (1) means the matrix is diagonalisable over $mathbb{C}$, of which there clearly are examples. However matrices like $$begin{bmatrix}1 & 1 \ 0 & 1end{bmatrix}$$ are ones that are not diagonalisable over $mathbb{C}$; this one would fit your case (2).



The reason why we would consider eigenvalues over $mathbb{C}$ rather than over $mathbb{R}$ is precisely because the characteristic polynomial will split as a products of its roots, which does not always happen over $mathbb{R}$. Hence we always find (counting multiplicities) $n$ eigenvalues, but it turns out this is not sufficient for diagonalisability.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks. I've also actually just managed to find $begin{bmatrix} 0 & 1 & 1 \ 1 & 0 & 1 \ 1 & 1 & 0 \ end{bmatrix}$ as an example of a matrix with one distinct eigenvalue and one degenerate eigenvalue, but three linearly independent eigenvectors.
    $endgroup$
    – Pancake_Senpai
    Dec 24 '18 at 17:38








  • 1




    $begingroup$
    @Pancake_Senpai If you want to look into this stuff in more detail, I suggest learning something about the Jordan canonical form of a matrix. The Jordan canonical form looks to the eye to almost be diagonalisation of it. Every matrix can be brought into a Jordan canonical form, even though not all matrices are diagonalisable. Moreover, you can read off from the Jordan canonical form whether a matrix is diagonalisable or not.
    $endgroup$
    – SvanN
    Dec 24 '18 at 18:49










  • $begingroup$
    Thanks for the info!
    $endgroup$
    – Pancake_Senpai
    Dec 25 '18 at 0:30



















0












$begingroup$

Standard example: $$ pmatrix{0 & 1cr 0 & 0cr}$$
with only one linearly independent eigenvector.






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%2f3051466%2fdoes-every-complex-n-times-n-matrix-have-n-linearly-independent-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









    1












    $begingroup$

    Both possibilities apply. Case (1) means the matrix is diagonalisable over $mathbb{C}$, of which there clearly are examples. However matrices like $$begin{bmatrix}1 & 1 \ 0 & 1end{bmatrix}$$ are ones that are not diagonalisable over $mathbb{C}$; this one would fit your case (2).



    The reason why we would consider eigenvalues over $mathbb{C}$ rather than over $mathbb{R}$ is precisely because the characteristic polynomial will split as a products of its roots, which does not always happen over $mathbb{R}$. Hence we always find (counting multiplicities) $n$ eigenvalues, but it turns out this is not sufficient for diagonalisability.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thanks. I've also actually just managed to find $begin{bmatrix} 0 & 1 & 1 \ 1 & 0 & 1 \ 1 & 1 & 0 \ end{bmatrix}$ as an example of a matrix with one distinct eigenvalue and one degenerate eigenvalue, but three linearly independent eigenvectors.
      $endgroup$
      – Pancake_Senpai
      Dec 24 '18 at 17:38








    • 1




      $begingroup$
      @Pancake_Senpai If you want to look into this stuff in more detail, I suggest learning something about the Jordan canonical form of a matrix. The Jordan canonical form looks to the eye to almost be diagonalisation of it. Every matrix can be brought into a Jordan canonical form, even though not all matrices are diagonalisable. Moreover, you can read off from the Jordan canonical form whether a matrix is diagonalisable or not.
      $endgroup$
      – SvanN
      Dec 24 '18 at 18:49










    • $begingroup$
      Thanks for the info!
      $endgroup$
      – Pancake_Senpai
      Dec 25 '18 at 0:30
















    1












    $begingroup$

    Both possibilities apply. Case (1) means the matrix is diagonalisable over $mathbb{C}$, of which there clearly are examples. However matrices like $$begin{bmatrix}1 & 1 \ 0 & 1end{bmatrix}$$ are ones that are not diagonalisable over $mathbb{C}$; this one would fit your case (2).



    The reason why we would consider eigenvalues over $mathbb{C}$ rather than over $mathbb{R}$ is precisely because the characteristic polynomial will split as a products of its roots, which does not always happen over $mathbb{R}$. Hence we always find (counting multiplicities) $n$ eigenvalues, but it turns out this is not sufficient for diagonalisability.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thanks. I've also actually just managed to find $begin{bmatrix} 0 & 1 & 1 \ 1 & 0 & 1 \ 1 & 1 & 0 \ end{bmatrix}$ as an example of a matrix with one distinct eigenvalue and one degenerate eigenvalue, but three linearly independent eigenvectors.
      $endgroup$
      – Pancake_Senpai
      Dec 24 '18 at 17:38








    • 1




      $begingroup$
      @Pancake_Senpai If you want to look into this stuff in more detail, I suggest learning something about the Jordan canonical form of a matrix. The Jordan canonical form looks to the eye to almost be diagonalisation of it. Every matrix can be brought into a Jordan canonical form, even though not all matrices are diagonalisable. Moreover, you can read off from the Jordan canonical form whether a matrix is diagonalisable or not.
      $endgroup$
      – SvanN
      Dec 24 '18 at 18:49










    • $begingroup$
      Thanks for the info!
      $endgroup$
      – Pancake_Senpai
      Dec 25 '18 at 0:30














    1












    1








    1





    $begingroup$

    Both possibilities apply. Case (1) means the matrix is diagonalisable over $mathbb{C}$, of which there clearly are examples. However matrices like $$begin{bmatrix}1 & 1 \ 0 & 1end{bmatrix}$$ are ones that are not diagonalisable over $mathbb{C}$; this one would fit your case (2).



    The reason why we would consider eigenvalues over $mathbb{C}$ rather than over $mathbb{R}$ is precisely because the characteristic polynomial will split as a products of its roots, which does not always happen over $mathbb{R}$. Hence we always find (counting multiplicities) $n$ eigenvalues, but it turns out this is not sufficient for diagonalisability.






    share|cite|improve this answer









    $endgroup$



    Both possibilities apply. Case (1) means the matrix is diagonalisable over $mathbb{C}$, of which there clearly are examples. However matrices like $$begin{bmatrix}1 & 1 \ 0 & 1end{bmatrix}$$ are ones that are not diagonalisable over $mathbb{C}$; this one would fit your case (2).



    The reason why we would consider eigenvalues over $mathbb{C}$ rather than over $mathbb{R}$ is precisely because the characteristic polynomial will split as a products of its roots, which does not always happen over $mathbb{R}$. Hence we always find (counting multiplicities) $n$ eigenvalues, but it turns out this is not sufficient for diagonalisability.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 24 '18 at 17:24









    SvanNSvanN

    2,0661422




    2,0661422












    • $begingroup$
      Thanks. I've also actually just managed to find $begin{bmatrix} 0 & 1 & 1 \ 1 & 0 & 1 \ 1 & 1 & 0 \ end{bmatrix}$ as an example of a matrix with one distinct eigenvalue and one degenerate eigenvalue, but three linearly independent eigenvectors.
      $endgroup$
      – Pancake_Senpai
      Dec 24 '18 at 17:38








    • 1




      $begingroup$
      @Pancake_Senpai If you want to look into this stuff in more detail, I suggest learning something about the Jordan canonical form of a matrix. The Jordan canonical form looks to the eye to almost be diagonalisation of it. Every matrix can be brought into a Jordan canonical form, even though not all matrices are diagonalisable. Moreover, you can read off from the Jordan canonical form whether a matrix is diagonalisable or not.
      $endgroup$
      – SvanN
      Dec 24 '18 at 18:49










    • $begingroup$
      Thanks for the info!
      $endgroup$
      – Pancake_Senpai
      Dec 25 '18 at 0:30


















    • $begingroup$
      Thanks. I've also actually just managed to find $begin{bmatrix} 0 & 1 & 1 \ 1 & 0 & 1 \ 1 & 1 & 0 \ end{bmatrix}$ as an example of a matrix with one distinct eigenvalue and one degenerate eigenvalue, but three linearly independent eigenvectors.
      $endgroup$
      – Pancake_Senpai
      Dec 24 '18 at 17:38








    • 1




      $begingroup$
      @Pancake_Senpai If you want to look into this stuff in more detail, I suggest learning something about the Jordan canonical form of a matrix. The Jordan canonical form looks to the eye to almost be diagonalisation of it. Every matrix can be brought into a Jordan canonical form, even though not all matrices are diagonalisable. Moreover, you can read off from the Jordan canonical form whether a matrix is diagonalisable or not.
      $endgroup$
      – SvanN
      Dec 24 '18 at 18:49










    • $begingroup$
      Thanks for the info!
      $endgroup$
      – Pancake_Senpai
      Dec 25 '18 at 0:30
















    $begingroup$
    Thanks. I've also actually just managed to find $begin{bmatrix} 0 & 1 & 1 \ 1 & 0 & 1 \ 1 & 1 & 0 \ end{bmatrix}$ as an example of a matrix with one distinct eigenvalue and one degenerate eigenvalue, but three linearly independent eigenvectors.
    $endgroup$
    – Pancake_Senpai
    Dec 24 '18 at 17:38






    $begingroup$
    Thanks. I've also actually just managed to find $begin{bmatrix} 0 & 1 & 1 \ 1 & 0 & 1 \ 1 & 1 & 0 \ end{bmatrix}$ as an example of a matrix with one distinct eigenvalue and one degenerate eigenvalue, but three linearly independent eigenvectors.
    $endgroup$
    – Pancake_Senpai
    Dec 24 '18 at 17:38






    1




    1




    $begingroup$
    @Pancake_Senpai If you want to look into this stuff in more detail, I suggest learning something about the Jordan canonical form of a matrix. The Jordan canonical form looks to the eye to almost be diagonalisation of it. Every matrix can be brought into a Jordan canonical form, even though not all matrices are diagonalisable. Moreover, you can read off from the Jordan canonical form whether a matrix is diagonalisable or not.
    $endgroup$
    – SvanN
    Dec 24 '18 at 18:49




    $begingroup$
    @Pancake_Senpai If you want to look into this stuff in more detail, I suggest learning something about the Jordan canonical form of a matrix. The Jordan canonical form looks to the eye to almost be diagonalisation of it. Every matrix can be brought into a Jordan canonical form, even though not all matrices are diagonalisable. Moreover, you can read off from the Jordan canonical form whether a matrix is diagonalisable or not.
    $endgroup$
    – SvanN
    Dec 24 '18 at 18:49












    $begingroup$
    Thanks for the info!
    $endgroup$
    – Pancake_Senpai
    Dec 25 '18 at 0:30




    $begingroup$
    Thanks for the info!
    $endgroup$
    – Pancake_Senpai
    Dec 25 '18 at 0:30











    0












    $begingroup$

    Standard example: $$ pmatrix{0 & 1cr 0 & 0cr}$$
    with only one linearly independent eigenvector.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Standard example: $$ pmatrix{0 & 1cr 0 & 0cr}$$
      with only one linearly independent eigenvector.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Standard example: $$ pmatrix{0 & 1cr 0 & 0cr}$$
        with only one linearly independent eigenvector.






        share|cite|improve this answer









        $endgroup$



        Standard example: $$ pmatrix{0 & 1cr 0 & 0cr}$$
        with only one linearly independent eigenvector.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 24 '18 at 17:23









        Robert IsraelRobert Israel

        327k23216470




        327k23216470






























            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%2f3051466%2fdoes-every-complex-n-times-n-matrix-have-n-linearly-independent-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

            How do I know what Microsoft account the skydrive app is syncing to?

            Image
            up vote 4 down vote favorite 1 Some time ago, I set up skydrive for my wife on her mac. But now she wants to access the files online, I cannot remember what Microsoft account I setup and linked her skydive app to and nor can I see anything in the skydrive app settings to tell me either? How can I see the name of the Microsoft account the skydive app is currently linked to? I would like to know this for skydrive app for Mac and PC onedrive share | improve this question asked Feb 25 '13 at 4:52 Gary Barrett 260 2 5 12
            Read more

            When does type information flow backwards in C++?

            Image
            up vote 62 down vote favorite 19 I just watched Stephan T. Lavavej talk at CppCon 2018 on "Class Template Argument Deduction", where at some point he incidentally says: In C++ type information almost never flows backwards ... I had to say "almost" because there's one or two cases, possibly more but very few . Despite trying to figure out which cases he might be referring to, I couldn't come up with anything. Hence the question: In which cases the C++17 standard mandates that type information propagate backwards? c++ language-lawyer c++17 share | improve this question edited 7 hours ago scohe0
            Read more

            Grease: Live!

            Image
            A tradução deste artigo está abaixo da qualidade média aceitável. É possível que tenha sido feita por um tradutor automático ou por alguém que não conhece bem o português ou a língua original do texto. Caso queira colaborar com a Wikipédia, tente encontrar a página original e melhore este verbete conforme o guia de tradução. Grease: Live! Grease: Live!  (PRT/BRA)  Estados Unidos 2016 •   cor •   130 min. min   Direção Thomas Kail Alex Rudzinski Produção Marc Platt Adam Siegel Roteiro Allan Carr Robert Cary Warren Casey Jim Jacobs Jonathan Tolins Bronte Woodard Baseado em Grease (musical) Grease (filme) Elenco Julianne Hough Aaron Tveit Vanessa Hudgens Carlos Pena Jr. Keke Palmer Carly Rae Jepsen Kether Donohue Jordan Fisher David Del Rio Andrew Call Gênero Musical Música Tom Kitt Figurino William Ivey Long Companhia(s) produtora(s) Warner Bros. Studios Distribuição Fox Lançamento 31 de Janei
            Read more