Author 
Topic: matrix (Read 1916 times) 

temporary
Full Member
Posts: 255

Construct a matrix where all numbers in it add up to 0, but with no negatives, no imaginary part greater than 0, and no 0s. This is not impossible, nor does it use infinity or something like that. Hint:there are infinite solutions

« Last Edit: Apr 16^{th}, 2008, 8:23pm by temporary » 
IP Logged 
My goal is to find what my goal is, once I find what my goal is, my goal will be complete.



Obob
Senior Riddler
Gender:
Posts: 489


Re: matrix
« Reply #1 on: Apr 9^{th}, 2008, 9:50am » 
Quote Modify

Yeah, no... You are asking for a series of positive numbers summing to zero. Even in nonstandard interpretations of "number," this is not possible, because those interpretations do not also have an ordering.


IP Logged 



Michael Dagg
Senior Riddler
Gender:
Posts: 500


Re: matrix
« Reply #2 on: Apr 9^{th}, 2008, 12:55pm » 
Quote Modify

[ 2 1 ] [ 1 2 ] mod 3 works.


IP Logged 
Regards, Michael Dagg



Michael Dagg
Senior Riddler
Gender:
Posts: 500


Re: matrix
« Reply #3 on: Apr 9^{th}, 2008, 1:19pm » 
Quote Modify

Since you might have also meant the sums along the diagnoals then [3 1 2] [1 2 3] [2 3 1] mod 6 works. Anyone got a general form?


IP Logged 
Regards, Michael Dagg



Obob
Senior Riddler
Gender:
Posts: 489


Re: matrix
« Reply #4 on: Apr 9^{th}, 2008, 2:47pm » 
Quote Modify

But is 2 > 0 mod 3?


IP Logged 



temporary
Full Member
Posts: 255


Re: matrix
« Reply #5 on: Apr 11^{th}, 2008, 11:18pm » 
Quote Modify

I have no idea what all this "mod" stuff is, but it is not part of the answer.


IP Logged 
My goal is to find what my goal is, once I find what my goal is, my goal will be complete.



Obob
Senior Riddler
Gender:
Posts: 489


Re: matrix
« Reply #6 on: Apr 11^{th}, 2008, 11:25pm » 
Quote Modify

Is it not part of the answer because there is no mod stuff in the answer, or is it not part of the answer because there is no answer? I suspect both are true...


IP Logged 



TJMann
Newbie
Posts: 18


Re: matrix
« Reply #7 on: Apr 12^{th}, 2008, 7:52am » 
Quote Modify

I get zero with both matrices. What makes this possible is the integers mod 3 and mod 6 wrap around. In mod 3, 2+1 = 3 = 0, so 2+1 + 1+2 = 6 = 0. In mod 6, 3+1+2 = 6, so (3+1+2) + (1+2+3) + (2+3+1) = 18 = 0.

« Last Edit: Apr 12^{th}, 2008, 7:57am by TJMann » 
IP Logged 



pex
Uberpuzzler
Gender:
Posts: 880


Re: matrix
« Reply #8 on: Apr 12^{th}, 2008, 8:14am » 
Quote Modify

It's a bit of a stretch, but the following 2x1 matrix seems to fit the criteria: [ 1i ] [ i1 ]  All numbers add up to zero: obvious.  No negatives: nonreal complex numbers are neither "positive" nor "negative".  No imaginary numbers: the numbers in this matrix are complex, but not imaginary, because the real parts are nonzero.  No zeros: obvious.  No "infinity or something like that": obvious.


IP Logged 



Michael Dagg
Senior Riddler
Gender:
Posts: 500


Re: matrix
« Reply #9 on: Apr 15^{th}, 2008, 10:42pm » 
Quote Modify

> But is 2 > 0 mod 3? If you are asking if there is order in the ring Z_3 then the answer is no. This in no way invalidates addition modulo 3, including multiplication; in fact, it is the other way around.

« Last Edit: Apr 15^{th}, 2008, 10:44pm by Michael Dagg » 
IP Logged 
Regards, Michael Dagg



Obob
Senior Riddler
Gender:
Posts: 489


Re: matrix
« Reply #10 on: Apr 16^{th}, 2008, 8:10am » 
Quote Modify

Yes, I slightly misread the original post. By reading "no negatives, no imaginary numbers, and no 0s" I automatically concluded that meant "positive numbers." You could either interpret imaginary numbers as "purely imaginary" numbers, as pex did, or do some modular arithmetic example and just say that the elements are not negative or imaginary, since those terms don't even make sense. At any rate, though, this riddle is mostly a wad of nonsense. At least the intended solution is.


IP Logged 



temporary
Full Member
Posts: 255


Re: matrix
« Reply #11 on: Apr 16^{th}, 2008, 8:17pm » 
Quote Modify

on Apr 12^{th}, 2008, 8:14am, pex wrote:It's a bit of a stretch, but the following 2x1 matrix seems to fit the criteria: [ 1i ] [ i1 ]  All numbers add up to zero: obvious.  No negatives: nonreal complex numbers are neither "positive" nor "negative".  No imaginary numbers: the numbers in this matrix are complex, but not imaginary, because the real parts are nonzero.  No zeros: obvious.  No "infinity or something like that": obvious. 
 I might have misphrased it, but you did kind of use negative numbers anyway. I'll fix the misphrase to avoid confusing anyone.

« Last Edit: Apr 16^{th}, 2008, 8:29pm by temporary » 
IP Logged 
My goal is to find what my goal is, once I find what my goal is, my goal will be complete.



pex
Uberpuzzler
Gender:
Posts: 880


Re: matrix
« Reply #12 on: Apr 17^{th}, 2008, 1:27am » 
Quote Modify

Okay... so we either need a nonstandard definition of addition, or a different set of numbers than real or complex numbers. For example, in the 10adic numbers, ...9999999 + 1 = 0. (The representation ...9999999 is infinite, but the number it represents isn't  does that count?)


IP Logged 



pex
Uberpuzzler
Gender:
Posts: 880


Re: matrix
« Reply #13 on: Apr 17^{th}, 2008, 4:34am » 
Quote Modify

Just a thought that crossed my mind: is there a sensible way to define matrices of dimensions m x n, with m or n equal to zero? If yes, one might argue that the sum of its elements (an empty set) equals zero, and matrices of dimensions 0x0, 0x1, 1x0, 0x2, 2x0, 0x3, 3x0, ... would work.


IP Logged 



Michael Dagg
Senior Riddler
Gender:
Posts: 500


Re: matrix
« Reply #14 on: Apr 17^{th}, 2008, 3:24pm » 
Quote Modify

Of course I would have hardly guessed that those kind of matrices would be involved in solution a here, but some matrices like these can pop up when finite groups are wrote using Smith Normal Form, in particular, subgroups of G = (Z/p^2 Z) x (Z/pZ) for example.


IP Logged 
Regards, Michael Dagg



TJMann
Newbie
Posts: 18


Re: matrix
« Reply #15 on: Apr 17^{th}, 2008, 5:37pm » 
Quote Modify

Being curious but knowing a just little about groups, how do you make the product (Z/p^2Z) X (Z/pZ)? I looked my algebra book about it but I did not see an explicit way to actually do it or should I think that it is just a "product" through the use of notation?


IP Logged 



Obob
Senior Riddler
Gender:
Posts: 489


Re: matrix
« Reply #16 on: Apr 17^{th}, 2008, 7:58pm » 
Quote Modify

Given two groups G and H, you form the product G X H as the set of all pairs (g,h), where g is in G and h is in H. You multiply pairs (g_{1},h_{1}) and (g_{2},h_{2}) by the rule (g_{1},h_{1}) * (g_{2},h_{2}) = (g_{1} * g_{2},h_{1} * h_{2}). So this is totally explicit, but it is also essentially just taking the product through the use of notation.


IP Logged 



TJMann
Newbie
Posts: 18


Re: matrix
« Reply #17 on: Apr 18^{th}, 2008, 5:05pm » 
Quote Modify

Thx. I did know that and it seems to fit the idea that I thinking because if we have p = 2 then we have the product (Z/4Z) X (Z/2Z) which is more specific that the general form wrote but aren't both Z/4Z and Z/2Z finite fields? One is of length 4 and the other is of length 2 and now that means the product would effectively be a 4 x 2  tuple? Groups that I know about don't go that far as to require what I suppose are called lists.


IP Logged 



Obob
Senior Riddler
Gender:
Posts: 489


Re: matrix
« Reply #18 on: Apr 19^{th}, 2008, 7:46pm » 
Quote Modify

Z/4Z is not a field. 2 has no multiplicative inverse in Z/4Z. But you can view Z/4Z x Z/2Z as a twotuple of numbers, one coming from the set {0,1,2,3} and the other coming from {0,1}. The group Z/4Z x Z/2Z does in fact have 8 elements.


IP Logged 



temporary
Full Member
Posts: 255


Re: matrix
« Reply #19 on: Apr 22^{nd}, 2008, 5:11pm » 
Quote Modify

on Apr 17^{th}, 2008, 4:34am, pex wrote:Just a thought that crossed my mind: is there a sensible way to define matrices of dimensions m x n, with m or n equal to zero? If yes, one might argue that the sum of its elements (an empty set) equals zero, and matrices of dimensions 0x0, 0x1, 1x0, 0x2, 2x0, 0x3, 3x0, ... would work. 
 Yep. That pretty much hits the nail on the head.


IP Logged 
My goal is to find what my goal is, once I find what my goal is, my goal will be complete.



Michael Dagg
Senior Riddler
Gender:
Posts: 500


Re: matrix
« Reply #20 on: Apr 22^{nd}, 2008, 7:02pm » 
Quote Modify

> Yep. That pretty much hits the nail on the head. That's stated lightly  got a proof?


IP Logged 
Regards, Michael Dagg



temporary
Full Member
Posts: 255


Re: matrix
« Reply #21 on: Apr 24^{th}, 2008, 6:53pm » 
Quote Modify

Isn't it obvious and selfproven. A matrix with no numbers has to add up to 0. And I am prepared to bet 1ud8f7g6g6v7v7v7 dollars that it does. And the only time I bet something I don't have is when it doesn't matter because I win.


IP Logged 
My goal is to find what my goal is, once I find what my goal is, my goal will be complete.



Michael Dagg
Senior Riddler
Gender:
Posts: 500


Re: matrix
« Reply #22 on: Apr 24^{th}, 2008, 7:11pm » 
Quote Modify

That is not a solution to the problem you posted. It is technically clear, do you know why?


IP Logged 
Regards, Michael Dagg



temporary
Full Member
Posts: 255


Re: matrix
« Reply #23 on: Apr 24^{th}, 2008, 7:26pm » 
Quote Modify

There is an answer. A matrix with no numbers(like 0x3, 1x0, 9x0, 0x(pi+e), etc) Where have you been for the last several posts.


IP Logged 
My goal is to find what my goal is, once I find what my goal is, my goal will be complete.



Michael Dagg
Senior Riddler
Gender:
Posts: 500


Re: matrix
« Reply #24 on: Apr 24^{th}, 2008, 7:42pm » 
Quote Modify

A matrix with no numbers is not a solution to the problem you posted. It is true that a sum on an empty set is zero but it does not satisfy the hypothesis of your problem.


IP Logged 
Regards, Michael Dagg



