Determinant of a matrix that contains the first $n^2$ primes.
22
$begingroup$
Let $n$ be an integer and $p_1,ldots,p_{n^2}$ be the first prime numbers. Writing them down in a matrix
$$
left(begin{matrix}
p_1 & p_2 & cdots & p_n \
p_{n+1} & p_{n+2} & cdots & p_{2n} \
vdots & vdots & ddots & vdots \
cdots & cdots & cdots & p_{n^2}
end{matrix}
right)
$$
we can take the determinant. How to prove that determinant is not zero for every $n$?
linear-algebra number-theory prime-numbers determinant
edited Jul 12 '17 at 20:22
Davide Giraudo
127k16153268
asked Jul 12 '17 at 6:55
Rofl UkulusRofl Ukulus
1646
$endgroup$
|
show 14 more comments
22
$begingroup$
Let $n$ be an integer and $p_1,ldots,p_{n^2}$ be the first prime numbers. Writing them down in a matrix
$$
left(begin{matrix}
p_1 & p_2 & cdots & p_n \
p_{n+1} & p_{n+2} & cdots & p_{2n} \
vdots & vdots & ddots & vdots \
cdots & cdots & cdots & p_{n^2}
end{matrix}
right)
$$
we can take the determinant. How to prove that determinant is not zero for every $n$?
linear-algebra number-theory prime-numbers determinant
edited Jul 12 '17 at 20:22
Davide Giraudo
127k16153268
asked Jul 12 '17 at 6:55
Rofl UkulusRofl Ukulus
1646
$endgroup$
4
$begingroup$
The sequence of determinants is OEIS sequence A067276. Not that this helps...
$endgroup$
– Robert Israel
Jul 12 '17 at 7:11
10
$begingroup$
I think this is going to be an intractable problem. Note that there are square matrices with determinant $0$ made up of distinct primes, e.g. $$pmatrix{2 & 3 & 5cr 7 & 11 & 13cr 19 & 23 & 97cr}$$ Thus you somehow have to depend on the fact that you're using the consecutive primes. And those just don't have enough regularity.
$endgroup$
– Robert Israel
Jul 12 '17 at 15:03
4
$begingroup$
Also with determinant $0$: $$ pmatrix{2 & 3 & 5 & 7cr 11 & 13 & 17 & 19cr 23 & 29 & 31 & 37cr 41 & 47 & 67 & 73cr }$$
$endgroup$
– Robert Israel
Jul 12 '17 at 15:12
1
$begingroup$
@Robert Israel thank you for these examples. Indeed, this makes it harder to prove and there might be some evil matrix around which determinant goes to zero :-) I looked up the prime factors of the determinants but did not find any pattern. There are large powers of 2 appearing in the factorization, but there not even increasing monotonely.
$endgroup$
– Rofl Ukulus
Jul 12 '17 at 17:01
1
$begingroup$
No answer: You might look at "Integral Matrices" by Morris Newman. In particular the "Smith Normal Form". It discusses divisibility and such. As I recall some of the results were surprising. The class is just a distance memory (but I have the book) so it might lead nowhere.
$endgroup$
– rrogers
Jul 18 '17 at 18:37
|
show 14 more comments
22
22
22
8
$begingroup$
Let $n$ be an integer and $p_1,ldots,p_{n^2}$ be the first prime numbers. Writing them down in a matrix
$$
left(begin{matrix}
p_1 & p_2 & cdots & p_n \
p_{n+1} & p_{n+2} & cdots & p_{2n} \
vdots & vdots & ddots & vdots \
cdots & cdots & cdots & p_{n^2}
end{matrix}
right)
$$
we can take the determinant. How to prove that determinant is not zero for every $n$?
linear-algebra number-theory prime-numbers determinant
edited Jul 12 '17 at 20:22
Davide Giraudo
127k16153268
asked Jul 12 '17 at 6:55
Rofl UkulusRofl Ukulus
1646
$endgroup$
Let $n$ be an integer and $p_1,ldots,p_{n^2}$ be the first prime numbers. Writing them down in a matrix
$$
left(begin{matrix}
p_1 & p_2 & cdots & p_n \
p_{n+1} & p_{n+2} & cdots & p_{2n} \
vdots & vdots & ddots & vdots \
cdots & cdots & cdots & p_{n^2}
end{matrix}
right)
$$
we can take the determinant. How to prove that determinant is not zero for every $n$?
linear-algebra number-theory prime-numbers determinant
linear-algebra number-theory prime-numbers determinant
edited Jul 12 '17 at 20:22
Davide Giraudo
127k16153268
asked Jul 12 '17 at 6:55
Rofl UkulusRofl Ukulus
1646
edited Jul 12 '17 at 20:22
Davide Giraudo
127k16153268
asked Jul 12 '17 at 6:55
Rofl UkulusRofl Ukulus
1646
edited Jul 12 '17 at 20:22
Davide Giraudo
127k16153268
edited Jul 12 '17 at 20:22
Davide Giraudo
127k16153268
edited Jul 12 '17 at 20:22
Davide Giraudo
127k16153268
127k16153268
asked Jul 12 '17 at 6:55
Rofl UkulusRofl Ukulus
1646
asked Jul 12 '17 at 6:55
Rofl UkulusRofl Ukulus
1646
asked Jul 12 '17 at 6:55
Rofl UkulusRofl Ukulus
1646
1646
4
$begingroup$
The sequence of determinants is OEIS sequence A067276. Not that this helps...
$endgroup$
– Robert Israel
Jul 12 '17 at 7:11
10
$begingroup$
I think this is going to be an intractable problem. Note that there are square matrices with determinant $0$ made up of distinct primes, e.g. $$pmatrix{2 & 3 & 5cr 7 & 11 & 13cr 19 & 23 & 97cr}$$ Thus you somehow have to depend on the fact that you're using the consecutive primes. And those just don't have enough regularity.
$endgroup$
– Robert Israel
Jul 12 '17 at 15:03
4
$begingroup$
Also with determinant $0$: $$ pmatrix{2 & 3 & 5 & 7cr 11 & 13 & 17 & 19cr 23 & 29 & 31 & 37cr 41 & 47 & 67 & 73cr }$$
$endgroup$
– Robert Israel
Jul 12 '17 at 15:12
1
$begingroup$
@Robert Israel thank you for these examples. Indeed, this makes it harder to prove and there might be some evil matrix around which determinant goes to zero :-) I looked up the prime factors of the determinants but did not find any pattern. There are large powers of 2 appearing in the factorization, but there not even increasing monotonely.
$endgroup$
– Rofl Ukulus
Jul 12 '17 at 17:01
1
$begingroup$
No answer: You might look at "Integral Matrices" by Morris Newman. In particular the "Smith Normal Form". It discusses divisibility and such. As I recall some of the results were surprising. The class is just a distance memory (but I have the book) so it might lead nowhere.
$endgroup$
– rrogers
Jul 18 '17 at 18:37
|
show 14 more comments
4
$begingroup$
The sequence of determinants is OEIS sequence A067276. Not that this helps...
$endgroup$
– Robert Israel
Jul 12 '17 at 7:11
10
$begingroup$
I think this is going to be an intractable problem. Note that there are square matrices with determinant $0$ made up of distinct primes, e.g. $$pmatrix{2 & 3 & 5cr 7 & 11 & 13cr 19 & 23 & 97cr}$$ Thus you somehow have to depend on the fact that you're using the consecutive primes. And those just don't have enough regularity.
$endgroup$
– Robert Israel
Jul 12 '17 at 15:03
4
$begingroup$
Also with determinant $0$: $$ pmatrix{2 & 3 & 5 & 7cr 11 & 13 & 17 & 19cr 23 & 29 & 31 & 37cr 41 & 47 & 67 & 73cr }$$
$endgroup$
– Robert Israel
Jul 12 '17 at 15:12
1
$begingroup$
@Robert Israel thank you for these examples. Indeed, this makes it harder to prove and there might be some evil matrix around which determinant goes to zero :-) I looked up the prime factors of the determinants but did not find any pattern. There are large powers of 2 appearing in the factorization, but there not even increasing monotonely.
$endgroup$
– Rofl Ukulus
Jul 12 '17 at 17:01
1
$begingroup$
No answer: You might look at "Integral Matrices" by Morris Newman. In particular the "Smith Normal Form". It discusses divisibility and such. As I recall some of the results were surprising. The class is just a distance memory (but I have the book) so it might lead nowhere.
$endgroup$
– rrogers
Jul 18 '17 at 18:37
4
4
$begingroup$
The sequence of determinants is OEIS sequence A067276. Not that this helps...
$endgroup$
– Robert Israel
Jul 12 '17 at 7:11
$begingroup$
The sequence of determinants is OEIS sequence A067276. Not that this helps...
$endgroup$
– Robert Israel
Jul 12 '17 at 7:11
10
10
$begingroup$
I think this is going to be an intractable problem. Note that there are square matrices with determinant $0$ made up of distinct primes, e.g. $$pmatrix{2 & 3 & 5cr 7 & 11 & 13cr 19 & 23 & 97cr}$$ Thus you somehow have to depend on the fact that you're using the consecutive primes. And those just don't have enough regularity.
$endgroup$
– Robert Israel
Jul 12 '17 at 15:03
$begingroup$
I think this is going to be an intractable problem. Note that there are square matrices with determinant $0$ made up of distinct primes, e.g. $$pmatrix{2 & 3 & 5cr 7 & 11 & 13cr 19 & 23 & 97cr}$$ Thus you somehow have to depend on the fact that you're using the consecutive primes. And those just don't have enough regularity.
$endgroup$
– Robert Israel
Jul 12 '17 at 15:03
4
4
$begingroup$
Also with determinant $0$: $$ pmatrix{2 & 3 & 5 & 7cr 11 & 13 & 17 & 19cr 23 & 29 & 31 & 37cr 41 & 47 & 67 & 73cr }$$
$endgroup$
– Robert Israel
Jul 12 '17 at 15:12
$begingroup$
Also with determinant $0$: $$ pmatrix{2 & 3 & 5 & 7cr 11 & 13 & 17 & 19cr 23 & 29 & 31 & 37cr 41 & 47 & 67 & 73cr }$$
$endgroup$
– Robert Israel
Jul 12 '17 at 15:12
1
1
$begingroup$
@Robert Israel thank you for these examples. Indeed, this makes it harder to prove and there might be some evil matrix around which determinant goes to zero :-) I looked up the prime factors of the determinants but did not find any pattern. There are large powers of 2 appearing in the factorization, but there not even increasing monotonely.
$endgroup$
– Rofl Ukulus
Jul 12 '17 at 17:01
$begingroup$
@Robert Israel thank you for these examples. Indeed, this makes it harder to prove and there might be some evil matrix around which determinant goes to zero :-) I looked up the prime factors of the determinants but did not find any pattern. There are large powers of 2 appearing in the factorization, but there not even increasing monotonely.
$endgroup$
– Rofl Ukulus
Jul 12 '17 at 17:01
1
1
$begingroup$
No answer: You might look at "Integral Matrices" by Morris Newman. In particular the "Smith Normal Form". It discusses divisibility and such. As I recall some of the results were surprising. The class is just a distance memory (but I have the book) so it might lead nowhere.
$endgroup$
– rrogers
Jul 18 '17 at 18:37
$begingroup$
No answer: You might look at "Integral Matrices" by Morris Newman. In particular the "Smith Normal Form". It discusses divisibility and such. As I recall some of the results were surprising. The class is just a distance memory (but I have the book) so it might lead nowhere.
$endgroup$
– rrogers
Jul 18 '17 at 18:37
|
show 14 more comments
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4
$begingroup$
The sequence of determinants is OEIS sequence A067276. Not that this helps...
$endgroup$
– Robert Israel
Jul 12 '17 at 7:11
10
$begingroup$
I think this is going to be an intractable problem. Note that there are square matrices with determinant $0$ made up of distinct primes, e.g. $$pmatrix{2 & 3 & 5cr 7 & 11 & 13cr 19 & 23 & 97cr}$$ Thus you somehow have to depend on the fact that you're using the consecutive primes. And those just don't have enough regularity.
$endgroup$
– Robert Israel
Jul 12 '17 at 15:03
4
$begingroup$
Also with determinant $0$: $$ pmatrix{2 & 3 & 5 & 7cr 11 & 13 & 17 & 19cr 23 & 29 & 31 & 37cr 41 & 47 & 67 & 73cr }$$
$endgroup$
– Robert Israel
Jul 12 '17 at 15:12
1
$begingroup$
@Robert Israel thank you for these examples. Indeed, this makes it harder to prove and there might be some evil matrix around which determinant goes to zero :-) I looked up the prime factors of the determinants but did not find any pattern. There are large powers of 2 appearing in the factorization, but there not even increasing monotonely.
$endgroup$
– Rofl Ukulus
Jul 12 '17 at 17:01
1
$begingroup$
No answer: You might look at "Integral Matrices" by Morris Newman. In particular the "Smith Normal Form". It discusses divisibility and such. As I recall some of the results were surprising. The class is just a distance memory (but I have the book) so it might lead nowhere.
$endgroup$
– rrogers
Jul 18 '17 at 18:37