A pattern in determinants of Fibonacci numbers?
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
edited Dec 3 '18 at 9:02
Martin Sleziak
44.7k7115270
asked Dec 3 '18 at 1:55
YiFan
2,5801421
add a comment |
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
edited Dec 3 '18 at 9:02
Martin Sleziak
44.7k7115270
asked Dec 3 '18 at 1:55
YiFan
2,5801421
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
add a comment |
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
edited Dec 3 '18 at 9:02
Martin Sleziak
44.7k7115270
asked Dec 3 '18 at 1:55
YiFan
2,5801421
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
edited Dec 3 '18 at 9:02
Martin Sleziak
44.7k7115270
asked Dec 3 '18 at 1:55
YiFan
2,5801421
edited Dec 3 '18 at 9:02
Martin Sleziak
44.7k7115270
asked Dec 3 '18 at 1:55
YiFan
2,5801421
edited Dec 3 '18 at 9:02
Martin Sleziak
44.7k7115270
edited Dec 3 '18 at 9:02
Martin Sleziak
44.7k7115270
edited Dec 3 '18 at 9:02
Martin Sleziak
44.7k7115270
44.7k7115270
asked Dec 3 '18 at 1:55
YiFan
2,5801421
asked Dec 3 '18 at 1:55
YiFan
2,5801421
asked Dec 3 '18 at 1:55
YiFan
2,5801421
2,5801421
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
add a comment |
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
add a comment |
2 Answers
2
active
oldest
votes
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?
answered Dec 3 '18 at 1:58
obscurans
808211
add a comment |
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 $m$.
answered Dec 3 '18 at 2:06
YiFan
2,5801421
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
add a comment |
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2 Answers
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2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
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?
answered Dec 3 '18 at 1:58
obscurans
808211
add a comment |
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?
answered Dec 3 '18 at 1:58
obscurans
808211
add a comment |
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?
answered Dec 3 '18 at 1:58
obscurans
808211
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?
answered Dec 3 '18 at 1:58
obscurans
808211
answered Dec 3 '18 at 1:58
obscurans
808211
answered Dec 3 '18 at 1:58
obscurans
808211
answered Dec 3 '18 at 1:58
obscurans
808211
808211
add a comment |
add a comment |
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 $m$.
answered Dec 3 '18 at 2:06
YiFan
2,5801421
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
add a comment |
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 $m$.
answered Dec 3 '18 at 2:06
YiFan
2,5801421
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
add a comment |
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 $m$.
answered Dec 3 '18 at 2:06
YiFan
2,5801421
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 $m$.
answered Dec 3 '18 at 2:06
YiFan
2,5801421
edited Dec 3 '18 at 2:15
answered Dec 3 '18 at 2:06
YiFan
2,5801421
answered Dec 3 '18 at 2:06
YiFan
2,5801421
answered Dec 3 '18 at 2:06
YiFan
2,5801421
2,5801421
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
add a comment |
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
add a comment |
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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