Not able to solve a matrix problem












1












$begingroup$


This is my first question of SO so sorry if I made a mistake somewhere.



I have this matrices problem which I'm not able to solve. I asked many of my friends but they just don't get the question.
This is the question:



$text{Let n} geq 3. text{Let A} = ((a_{ij}))_{1 leq i, j leq n} text{be an n} times text{n matrix such that a}_{ij} in {1, -1} text{for all 0} leq i, j leq n.
text{Suppose a}_{k1} = text{1 for all 1} leq k leq text{n. } \ text{And} sum_{k = 1}^{n} a_{ki}cdot a_{kj}text{= 0. For all i} neq j.\ text{Show that n is a multiple of 4.}$



(sorry I can't add an image and I'm not so good at latex too)



I think I've understood the summation part: when k = 1, the argument of summation becomes $a_{1i} cdot a_{1j}$. Which is the sum of product of elements of 1st row taken 2 at a time. (that's what I interpreted $a_{1i} cdot a_{1j}$ as. Is this right?)
I'm not able to utilize the summation to get an equation in n.



Thanx in advance!






matrices summation







share|cite|improve this question









$endgroup$












  • $begingroup$
    You are looking at Hadamard matrices, so now you know what to look for.
    $endgroup$
    – Gerry Myerson
    Nov 30 '18 at 11:54










  • $begingroup$
    See, e.g., math.stackexchange.com/questions/753867/…
    $endgroup$
    – Gerry Myerson
    Nov 30 '18 at 11:59
















1












$begingroup$


This is my first question of SO so sorry if I made a mistake somewhere.



I have this matrices problem which I'm not able to solve. I asked many of my friends but they just don't get the question.
This is the question:



$text{Let n} geq 3. text{Let A} = ((a_{ij}))_{1 leq i, j leq n} text{be an n} times text{n matrix such that a}_{ij} in {1, -1} text{for all 0} leq i, j leq n.
text{Suppose a}_{k1} = text{1 for all 1} leq k leq text{n. } \ text{And} sum_{k = 1}^{n} a_{ki}cdot a_{kj}text{= 0. For all i} neq j.\ text{Show that n is a multiple of 4.}$



(sorry I can't add an image and I'm not so good at latex too)



I think I've understood the summation part: when k = 1, the argument of summation becomes $a_{1i} cdot a_{1j}$. Which is the sum of product of elements of 1st row taken 2 at a time. (that's what I interpreted $a_{1i} cdot a_{1j}$ as. Is this right?)
I'm not able to utilize the summation to get an equation in n.



Thanx in advance!






matrices summation







share|cite|improve this question









$endgroup$












  • $begingroup$
    You are looking at Hadamard matrices, so now you know what to look for.
    $endgroup$
    – Gerry Myerson
    Nov 30 '18 at 11:54










  • $begingroup$
    See, e.g., math.stackexchange.com/questions/753867/…
    $endgroup$
    – Gerry Myerson
    Nov 30 '18 at 11:59














1












1








1


1



$begingroup$


This is my first question of SO so sorry if I made a mistake somewhere.



I have this matrices problem which I'm not able to solve. I asked many of my friends but they just don't get the question.
This is the question:



$text{Let n} geq 3. text{Let A} = ((a_{ij}))_{1 leq i, j leq n} text{be an n} times text{n matrix such that a}_{ij} in {1, -1} text{for all 0} leq i, j leq n.
text{Suppose a}_{k1} = text{1 for all 1} leq k leq text{n. } \ text{And} sum_{k = 1}^{n} a_{ki}cdot a_{kj}text{= 0. For all i} neq j.\ text{Show that n is a multiple of 4.}$



(sorry I can't add an image and I'm not so good at latex too)



I think I've understood the summation part: when k = 1, the argument of summation becomes $a_{1i} cdot a_{1j}$. Which is the sum of product of elements of 1st row taken 2 at a time. (that's what I interpreted $a_{1i} cdot a_{1j}$ as. Is this right?)
I'm not able to utilize the summation to get an equation in n.



Thanx in advance!






matrices summation







share|cite|improve this question









$endgroup$




This is my first question of SO so sorry if I made a mistake somewhere.



I have this matrices problem which I'm not able to solve. I asked many of my friends but they just don't get the question.
This is the question:



$text{Let n} geq 3. text{Let A} = ((a_{ij}))_{1 leq i, j leq n} text{be an n} times text{n matrix such that a}_{ij} in {1, -1} text{for all 0} leq i, j leq n.
text{Suppose a}_{k1} = text{1 for all 1} leq k leq text{n. } \ text{And} sum_{k = 1}^{n} a_{ki}cdot a_{kj}text{= 0. For all i} neq j.\ text{Show that n is a multiple of 4.}$



(sorry I can't add an image and I'm not so good at latex too)



I think I've understood the summation part: when k = 1, the argument of summation becomes $a_{1i} cdot a_{1j}$. Which is the sum of product of elements of 1st row taken 2 at a time. (that's what I interpreted $a_{1i} cdot a_{1j}$ as. Is this right?)
I'm not able to utilize the summation to get an equation in n.



Thanx in advance!






matrices summation




matrices summation






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 30 '18 at 10:58









YashKandalkarYashKandalkar

82




82












  • $begingroup$
    You are looking at Hadamard matrices, so now you know what to look for.
    $endgroup$
    – Gerry Myerson
    Nov 30 '18 at 11:54










  • $begingroup$
    See, e.g., math.stackexchange.com/questions/753867/…
    $endgroup$
    – Gerry Myerson
    Nov 30 '18 at 11:59


















  • $begingroup$
    You are looking at Hadamard matrices, so now you know what to look for.
    $endgroup$
    – Gerry Myerson
    Nov 30 '18 at 11:54










  • $begingroup$
    See, e.g., math.stackexchange.com/questions/753867/…
    $endgroup$
    – Gerry Myerson
    Nov 30 '18 at 11:59
















$begingroup$
You are looking at Hadamard matrices, so now you know what to look for.
$endgroup$
– Gerry Myerson
Nov 30 '18 at 11:54




$begingroup$
You are looking at Hadamard matrices, so now you know what to look for.
$endgroup$
– Gerry Myerson
Nov 30 '18 at 11:54












$begingroup$
See, e.g., math.stackexchange.com/questions/753867/…
$endgroup$
– Gerry Myerson
Nov 30 '18 at 11:59




$begingroup$
See, e.g., math.stackexchange.com/questions/753867/…
$endgroup$
– Gerry Myerson
Nov 30 '18 at 11:59










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$begingroup$

Clearly $n$ can't be odd as the inner product of the first column, which solely consists of ones, must be zero. Hence in every other column we must have an equal number of ones and minus ones, let's call that number $m$. (That is $n=2m$.)



Now consider two column except the first one. Have a look at the rows in the first column with element one (there are $m$ such rows). Let the number of ones in the corresponding rows in the second column be $k$, clearly $0leq kleq m$. Their product equals obviously $k-(m-k)=2k-m$. Similarly, the sum of the product of the other rows is $2k-m$ as well.



So in total the sum of all products is $4k-2m=2(2k-m)$. As that sum is zero, $m$ must be even, hence $n$ is divisible by $4$.






share|cite|improve this answer









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    $begingroup$

    Clearly $n$ can't be odd as the inner product of the first column, which solely consists of ones, must be zero. Hence in every other column we must have an equal number of ones and minus ones, let's call that number $m$. (That is $n=2m$.)



    Now consider two column except the first one. Have a look at the rows in the first column with element one (there are $m$ such rows). Let the number of ones in the corresponding rows in the second column be $k$, clearly $0leq kleq m$. Their product equals obviously $k-(m-k)=2k-m$. Similarly, the sum of the product of the other rows is $2k-m$ as well.



    So in total the sum of all products is $4k-2m=2(2k-m)$. As that sum is zero, $m$ must be even, hence $n$ is divisible by $4$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Clearly $n$ can't be odd as the inner product of the first column, which solely consists of ones, must be zero. Hence in every other column we must have an equal number of ones and minus ones, let's call that number $m$. (That is $n=2m$.)



      Now consider two column except the first one. Have a look at the rows in the first column with element one (there are $m$ such rows). Let the number of ones in the corresponding rows in the second column be $k$, clearly $0leq kleq m$. Their product equals obviously $k-(m-k)=2k-m$. Similarly, the sum of the product of the other rows is $2k-m$ as well.



      So in total the sum of all products is $4k-2m=2(2k-m)$. As that sum is zero, $m$ must be even, hence $n$ is divisible by $4$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Clearly $n$ can't be odd as the inner product of the first column, which solely consists of ones, must be zero. Hence in every other column we must have an equal number of ones and minus ones, let's call that number $m$. (That is $n=2m$.)



        Now consider two column except the first one. Have a look at the rows in the first column with element one (there are $m$ such rows). Let the number of ones in the corresponding rows in the second column be $k$, clearly $0leq kleq m$. Their product equals obviously $k-(m-k)=2k-m$. Similarly, the sum of the product of the other rows is $2k-m$ as well.



        So in total the sum of all products is $4k-2m=2(2k-m)$. As that sum is zero, $m$ must be even, hence $n$ is divisible by $4$.






        share|cite|improve this answer









        $endgroup$



        Clearly $n$ can't be odd as the inner product of the first column, which solely consists of ones, must be zero. Hence in every other column we must have an equal number of ones and minus ones, let's call that number $m$. (That is $n=2m$.)



        Now consider two column except the first one. Have a look at the rows in the first column with element one (there are $m$ such rows). Let the number of ones in the corresponding rows in the second column be $k$, clearly $0leq kleq m$. Their product equals obviously $k-(m-k)=2k-m$. Similarly, the sum of the product of the other rows is $2k-m$ as well.



        So in total the sum of all products is $4k-2m=2(2k-m)$. As that sum is zero, $m$ must be even, hence $n$ is divisible by $4$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 30 '18 at 12:40









        Michael HoppeMichael Hoppe

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        10.8k31834






























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