A pattern in determinants of Fibonacci numbers?












6














Let $F_n$ denote the $n$th Fibonacci number, adopting the convention $F_1=1$, $F_2=1$ and so on. Consider the $ntimes n$ matrix defined by



$$mathbf M_n:=begin{bmatrix}F_1&F_2&dots&F_n\F_{n+1}&F_{n+2}&dots&F_{2n}\vdots&vdots&ddots&vdots\F_{n^2-n+1}&F_{n^2-n+2}&dots&F_{n^2}end{bmatrix}.$$



I have the following conjecture:




Conjecture. For all integers $ngeq3$, $detmathbf M_n=0$.




I have used some Python code to test this conjecture for $n$ up to $9$, but I cannot go further. Note that $detmathbf M_1=detmathbf M_2=1$. Due to the elementary nature of this problem I have to assume that it has been discussed before, perhaps even on this site. But I couldn't find any reference on it, by Googling or searching here. Can someone shed light onto whether the conjecture is true, and a proof of it if so?






linear-algebra determinant fibonacci-numbers







share|cite|improve this question




















  • 1




    Fun question. Next time I teach matrix theory I might use this as an extra credit problem on a homework set involving determinants. As you know now, the solution isn't hard, but it would take a pretty clever student to spot it on their own. Thanks.
    – John Coleman
    Dec 3 at 12:19






  • 1




    Problem A3 from the 2009 Putnam competition has a very similar statement and a very similar solution, only replacing $F_k$ by $cos k$ (in radians).
    – Misha Lavrov
    Dec 3 at 15:46
















6














Let $F_n$ denote the $n$th Fibonacci number, adopting the convention $F_1=1$, $F_2=1$ and so on. Consider the $ntimes n$ matrix defined by



$$mathbf M_n:=begin{bmatrix}F_1&F_2&dots&F_n\F_{n+1}&F_{n+2}&dots&F_{2n}\vdots&vdots&ddots&vdots\F_{n^2-n+1}&F_{n^2-n+2}&dots&F_{n^2}end{bmatrix}.$$



I have the following conjecture:




Conjecture. For all integers $ngeq3$, $detmathbf M_n=0$.




I have used some Python code to test this conjecture for $n$ up to $9$, but I cannot go further. Note that $detmathbf M_1=detmathbf M_2=1$. Due to the elementary nature of this problem I have to assume that it has been discussed before, perhaps even on this site. But I couldn't find any reference on it, by Googling or searching here. Can someone shed light onto whether the conjecture is true, and a proof of it if so?






linear-algebra determinant fibonacci-numbers







share|cite|improve this question




















  • 1




    Fun question. Next time I teach matrix theory I might use this as an extra credit problem on a homework set involving determinants. As you know now, the solution isn't hard, but it would take a pretty clever student to spot it on their own. Thanks.
    – John Coleman
    Dec 3 at 12:19






  • 1




    Problem A3 from the 2009 Putnam competition has a very similar statement and a very similar solution, only replacing $F_k$ by $cos k$ (in radians).
    – Misha Lavrov
    Dec 3 at 15:46














6












6








6


2





Let $F_n$ denote the $n$th Fibonacci number, adopting the convention $F_1=1$, $F_2=1$ and so on. Consider the $ntimes n$ matrix defined by



$$mathbf M_n:=begin{bmatrix}F_1&F_2&dots&F_n\F_{n+1}&F_{n+2}&dots&F_{2n}\vdots&vdots&ddots&vdots\F_{n^2-n+1}&F_{n^2-n+2}&dots&F_{n^2}end{bmatrix}.$$



I have the following conjecture:




Conjecture. For all integers $ngeq3$, $detmathbf M_n=0$.




I have used some Python code to test this conjecture for $n$ up to $9$, but I cannot go further. Note that $detmathbf M_1=detmathbf M_2=1$. Due to the elementary nature of this problem I have to assume that it has been discussed before, perhaps even on this site. But I couldn't find any reference on it, by Googling or searching here. Can someone shed light onto whether the conjecture is true, and a proof of it if so?






linear-algebra determinant fibonacci-numbers







share|cite|improve this question















Let $F_n$ denote the $n$th Fibonacci number, adopting the convention $F_1=1$, $F_2=1$ and so on. Consider the $ntimes n$ matrix defined by



$$mathbf M_n:=begin{bmatrix}F_1&F_2&dots&F_n\F_{n+1}&F_{n+2}&dots&F_{2n}\vdots&vdots&ddots&vdots\F_{n^2-n+1}&F_{n^2-n+2}&dots&F_{n^2}end{bmatrix}.$$



I have the following conjecture:




Conjecture. For all integers $ngeq3$, $detmathbf M_n=0$.




I have used some Python code to test this conjecture for $n$ up to $9$, but I cannot go further. Note that $detmathbf M_1=detmathbf M_2=1$. Due to the elementary nature of this problem I have to assume that it has been discussed before, perhaps even on this site. But I couldn't find any reference on it, by Googling or searching here. Can someone shed light onto whether the conjecture is true, and a proof of it if so?






linear-algebra determinant fibonacci-numbers




linear-algebra determinant fibonacci-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 3 at 9:02









Martin Sleziak

44.6k7115270




44.6k7115270










asked Dec 3 at 1:55









YiFan

2,3391421




2,3391421








  • 1




    Fun question. Next time I teach matrix theory I might use this as an extra credit problem on a homework set involving determinants. As you know now, the solution isn't hard, but it would take a pretty clever student to spot it on their own. Thanks.
    – John Coleman
    Dec 3 at 12:19






  • 1




    Problem A3 from the 2009 Putnam competition has a very similar statement and a very similar solution, only replacing $F_k$ by $cos k$ (in radians).
    – Misha Lavrov
    Dec 3 at 15:46














  • 1




    Fun question. Next time I teach matrix theory I might use this as an extra credit problem on a homework set involving determinants. As you know now, the solution isn't hard, but it would take a pretty clever student to spot it on their own. Thanks.
    – John Coleman
    Dec 3 at 12:19






  • 1




    Problem A3 from the 2009 Putnam competition has a very similar statement and a very similar solution, only replacing $F_k$ by $cos k$ (in radians).
    – Misha Lavrov
    Dec 3 at 15:46








1




1




Fun question. Next time I teach matrix theory I might use this as an extra credit problem on a homework set involving determinants. As you know now, the solution isn't hard, but it would take a pretty clever student to spot it on their own. Thanks.
– John Coleman
Dec 3 at 12:19




Fun question. Next time I teach matrix theory I might use this as an extra credit problem on a homework set involving determinants. As you know now, the solution isn't hard, but it would take a pretty clever student to spot it on their own. Thanks.
– John Coleman
Dec 3 at 12:19




1




1




Problem A3 from the 2009 Putnam competition has a very similar statement and a very similar solution, only replacing $F_k$ by $cos k$ (in radians).
– Misha Lavrov
Dec 3 at 15:46




Problem A3 from the 2009 Putnam competition has a very similar statement and a very similar solution, only replacing $F_k$ by $cos k$ (in radians).
– Misha Lavrov
Dec 3 at 15:46










2 Answers
2






active

oldest

votes


















23














Here's a hint: what's the relationship between $F_{k+1}+F_{k+2}$ and $F_{k+3}$? What does that say about the 1st, 2nd, and 3rd columns of this matrix?






share|cite|improve this answer





























    6














    The resolution is remarkably simple (many thanks to obscurans' answer for the hint!) By the definition of the Fibonacci numbers, $F_k+F_{k+1}=F_{k+2}$ for all $k$. If $ngeq3$ then these numbers are going to be in the first three columns of every row. Hence the first three rows are linearly dependent, so the determinant is $0$. It follows from this that any such sequence following a linear recurrence (of the form $F_{n}=aF_{n-1}+bF_{n-2}$, $a,b$ are constant), with possibly different starting terms, also satisfies the stated conjecture. In fact, this shows that all such matrices have rank $2$, with the only two linearly independent columns being the first two. If the linear recurrence is of higher order, say $m$, then the determinant is $0$ when $n>m$, and the rank of the matrix will be $m$.






    share|cite|improve this answer



















    • 2




      One note: the matrix will have rank $leq$ the order of the linear recurrence, which is not necessarily 2.
      – obscurans
      Dec 3 at 2:11












    • @obscurans Suppose that the matrix has rank $k$. Does that not mean that the linear recurrence can be rewritten as a linear recurrence of order $k$? I was under the impression that the order of a linear recurrence was the order of its simplest form, though I realize now that that may not be the case.
      – Spitemaster
      Dec 3 at 15:31










    • There are two different things: a particular linear recurrence, which is an equation with $n$ degrees of freedom of solutions, vs a particular fixed sequence of numbers generated by some linear recurrence. The matrix having rank $k$ does mean a linear recurrence of order $k$ can generate this sequence of numbers.
      – obscurans
      Dec 4 at 2:52











    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%2f3023500%2fa-pattern-in-determinants-of-fibonacci-numbers%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









    23














    Here's a hint: what's the relationship between $F_{k+1}+F_{k+2}$ and $F_{k+3}$? What does that say about the 1st, 2nd, and 3rd columns of this matrix?






    share|cite|improve this answer


























      23














      Here's a hint: what's the relationship between $F_{k+1}+F_{k+2}$ and $F_{k+3}$? What does that say about the 1st, 2nd, and 3rd columns of this matrix?






      share|cite|improve this answer
























        23












        23








        23






        Here's a hint: what's the relationship between $F_{k+1}+F_{k+2}$ and $F_{k+3}$? What does that say about the 1st, 2nd, and 3rd columns of this matrix?






        share|cite|improve this answer












        Here's a hint: what's the relationship between $F_{k+1}+F_{k+2}$ and $F_{k+3}$? What does that say about the 1st, 2nd, and 3rd columns of this matrix?







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 3 at 1:58









        obscurans

        65919




        65919























            6














            The resolution is remarkably simple (many thanks to obscurans' answer for the hint!) By the definition of the Fibonacci numbers, $F_k+F_{k+1}=F_{k+2}$ for all $k$. If $ngeq3$ then these numbers are going to be in the first three columns of every row. Hence the first three rows are linearly dependent, so the determinant is $0$. It follows from this that any such sequence following a linear recurrence (of the form $F_{n}=aF_{n-1}+bF_{n-2}$, $a,b$ are constant), with possibly different starting terms, also satisfies the stated conjecture. In fact, this shows that all such matrices have rank $2$, with the only two linearly independent columns being the first two. If the linear recurrence is of higher order, say $m$, then the determinant is $0$ when $n>m$, and the rank of the matrix will be $m$.






            share|cite|improve this answer



















            • 2




              One note: the matrix will have rank $leq$ the order of the linear recurrence, which is not necessarily 2.
              – obscurans
              Dec 3 at 2:11












            • @obscurans Suppose that the matrix has rank $k$. Does that not mean that the linear recurrence can be rewritten as a linear recurrence of order $k$? I was under the impression that the order of a linear recurrence was the order of its simplest form, though I realize now that that may not be the case.
              – Spitemaster
              Dec 3 at 15:31










            • There are two different things: a particular linear recurrence, which is an equation with $n$ degrees of freedom of solutions, vs a particular fixed sequence of numbers generated by some linear recurrence. The matrix having rank $k$ does mean a linear recurrence of order $k$ can generate this sequence of numbers.
              – obscurans
              Dec 4 at 2:52
















            6














            The resolution is remarkably simple (many thanks to obscurans' answer for the hint!) By the definition of the Fibonacci numbers, $F_k+F_{k+1}=F_{k+2}$ for all $k$. If $ngeq3$ then these numbers are going to be in the first three columns of every row. Hence the first three rows are linearly dependent, so the determinant is $0$. It follows from this that any such sequence following a linear recurrence (of the form $F_{n}=aF_{n-1}+bF_{n-2}$, $a,b$ are constant), with possibly different starting terms, also satisfies the stated conjecture. In fact, this shows that all such matrices have rank $2$, with the only two linearly independent columns being the first two. If the linear recurrence is of higher order, say $m$, then the determinant is $0$ when $n>m$, and the rank of the matrix will be $m$.






            share|cite|improve this answer



















            • 2




              One note: the matrix will have rank $leq$ the order of the linear recurrence, which is not necessarily 2.
              – obscurans
              Dec 3 at 2:11












            • @obscurans Suppose that the matrix has rank $k$. Does that not mean that the linear recurrence can be rewritten as a linear recurrence of order $k$? I was under the impression that the order of a linear recurrence was the order of its simplest form, though I realize now that that may not be the case.
              – Spitemaster
              Dec 3 at 15:31










            • There are two different things: a particular linear recurrence, which is an equation with $n$ degrees of freedom of solutions, vs a particular fixed sequence of numbers generated by some linear recurrence. The matrix having rank $k$ does mean a linear recurrence of order $k$ can generate this sequence of numbers.
              – obscurans
              Dec 4 at 2:52














            6












            6








            6






            The resolution is remarkably simple (many thanks to obscurans' answer for the hint!) By the definition of the Fibonacci numbers, $F_k+F_{k+1}=F_{k+2}$ for all $k$. If $ngeq3$ then these numbers are going to be in the first three columns of every row. Hence the first three rows are linearly dependent, so the determinant is $0$. It follows from this that any such sequence following a linear recurrence (of the form $F_{n}=aF_{n-1}+bF_{n-2}$, $a,b$ are constant), with possibly different starting terms, also satisfies the stated conjecture. In fact, this shows that all such matrices have rank $2$, with the only two linearly independent columns being the first two. If the linear recurrence is of higher order, say $m$, then the determinant is $0$ when $n>m$, and the rank of the matrix will be $m$.






            share|cite|improve this answer














            The resolution is remarkably simple (many thanks to obscurans' answer for the hint!) By the definition of the Fibonacci numbers, $F_k+F_{k+1}=F_{k+2}$ for all $k$. If $ngeq3$ then these numbers are going to be in the first three columns of every row. Hence the first three rows are linearly dependent, so the determinant is $0$. It follows from this that any such sequence following a linear recurrence (of the form $F_{n}=aF_{n-1}+bF_{n-2}$, $a,b$ are constant), with possibly different starting terms, also satisfies the stated conjecture. In fact, this shows that all such matrices have rank $2$, with the only two linearly independent columns being the first two. If the linear recurrence is of higher order, say $m$, then the determinant is $0$ when $n>m$, and the rank of the matrix will be $m$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 3 at 2:15

























            answered Dec 3 at 2:06









            YiFan

            2,3391421




            2,3391421








            • 2




              One note: the matrix will have rank $leq$ the order of the linear recurrence, which is not necessarily 2.
              – obscurans
              Dec 3 at 2:11












            • @obscurans Suppose that the matrix has rank $k$. Does that not mean that the linear recurrence can be rewritten as a linear recurrence of order $k$? I was under the impression that the order of a linear recurrence was the order of its simplest form, though I realize now that that may not be the case.
              – Spitemaster
              Dec 3 at 15:31










            • There are two different things: a particular linear recurrence, which is an equation with $n$ degrees of freedom of solutions, vs a particular fixed sequence of numbers generated by some linear recurrence. The matrix having rank $k$ does mean a linear recurrence of order $k$ can generate this sequence of numbers.
              – obscurans
              Dec 4 at 2:52














            • 2




              One note: the matrix will have rank $leq$ the order of the linear recurrence, which is not necessarily 2.
              – obscurans
              Dec 3 at 2:11












            • @obscurans Suppose that the matrix has rank $k$. Does that not mean that the linear recurrence can be rewritten as a linear recurrence of order $k$? I was under the impression that the order of a linear recurrence was the order of its simplest form, though I realize now that that may not be the case.
              – Spitemaster
              Dec 3 at 15:31










            • There are two different things: a particular linear recurrence, which is an equation with $n$ degrees of freedom of solutions, vs a particular fixed sequence of numbers generated by some linear recurrence. The matrix having rank $k$ does mean a linear recurrence of order $k$ can generate this sequence of numbers.
              – obscurans
              Dec 4 at 2:52








            2




            2




            One note: the matrix will have rank $leq$ the order of the linear recurrence, which is not necessarily 2.
            – obscurans
            Dec 3 at 2:11






            One note: the matrix will have rank $leq$ the order of the linear recurrence, which is not necessarily 2.
            – obscurans
            Dec 3 at 2:11














            @obscurans Suppose that the matrix has rank $k$. Does that not mean that the linear recurrence can be rewritten as a linear recurrence of order $k$? I was under the impression that the order of a linear recurrence was the order of its simplest form, though I realize now that that may not be the case.
            – Spitemaster
            Dec 3 at 15:31




            @obscurans Suppose that the matrix has rank $k$. Does that not mean that the linear recurrence can be rewritten as a linear recurrence of order $k$? I was under the impression that the order of a linear recurrence was the order of its simplest form, though I realize now that that may not be the case.
            – Spitemaster
            Dec 3 at 15:31












            There are two different things: a particular linear recurrence, which is an equation with $n$ degrees of freedom of solutions, vs a particular fixed sequence of numbers generated by some linear recurrence. The matrix having rank $k$ does mean a linear recurrence of order $k$ can generate this sequence of numbers.
            – obscurans
            Dec 4 at 2:52




            There are two different things: a particular linear recurrence, which is an equation with $n$ degrees of freedom of solutions, vs a particular fixed sequence of numbers generated by some linear recurrence. The matrix having rank $k$ does mean a linear recurrence of order $k$ can generate this sequence of numbers.
            – obscurans
            Dec 4 at 2:52


















            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%2f3023500%2fa-pattern-in-determinants-of-fibonacci-numbers%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

            Plaza Victoria

            Image
            Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l
            Read more

            In PowerPoint, is there a keyboard shortcut for bulleted / numbered list?

            Image
            .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 2 OneNote has two useful keyboard shortcuts for bulleted and numbered lists, namely Ctrl + . and Ctrl + - . Is there something similar for PowerPoint where the work with lists is also very common? Haven't found anything here: Use keyboard shortcuts in PowerPoint. keyboard-shortcuts microsoft-powerpoint share | improve this question edited Aug 14 '14 at 14:49 Linger 2,826 10 28 40
            Read more

            How to put 3 figures in Latex with 2 figures side by side and 1 below these side by side images but in...

            Image
            1 0 I want to put 3 images in LateX such as 2 figures are side by side horizotally and 3rd figure below these 2 side by side figures but in middle. I have following code: begin{figure}[H] centering begin{minipage}{.5textwidth} centering includegraphics[width=.50linewidth]{Text/Images/Genelec_8010_AP.jpg} captionof{figure}{Genelec 8010 AP} label{fig:Genelec 8010} end{minipage}% begin{minipage}{.5textwidth} centering includegraphics[width=.69linewidth]{Text/Images/Genelec_8020_CPM.jpg} captionof{figure}{Genelec 8020 CPM} label{fig:Genelec 8020} end{minipage}% end{figure} %%Here is my 3rd picture code begin{figure} centering includegraphics[width=5cm]{Text/Images/Genelec_8030_BPM.jpg} caption{Genelec 8030 BPM} label{fig:Genelec 8030} end{figure} I want to ask ho
            Read more