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?










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  • 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 '18 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 '18 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?










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 '18 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 '18 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?










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






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edited Dec 3 '18 at 9:02









Martin Sleziak

44.7k7115270




44.7k7115270










asked Dec 3 '18 at 1:55









YiFan

2,5291421




2,5291421








  • 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 at 2:52











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    2 Answers
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    2 Answers
    2






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    active

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    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 '18 at 1:58









        obscurans

        808211




        808211























            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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 at 2:15

























            answered Dec 3 '18 at 2:06









            YiFan

            2,5291421




            2,5291421








            • 2




              One note: the matrix will have rank $leq$ the order of the linear recurrence, which is not necessarily 2.
              – obscurans
              Dec 3 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 at 2:52


















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