number of possible $ntimes n$ symmetric matrices with each entry either $o$ or unity












-1












$begingroup$


question:



number of different $ntimes n $ symmetric matrices with each element either 0 or 1 is



$(a)2^n$



$(b)2^{n^{2}}$



$(c)2^{frac{n(n+1)}{2}}$



$(d)2^{frac{n(n-1)}{2}}$



my attempt:



first every digonal matrix is symmetric so, answer should not be $2^n$ it should be greater than that . .....i also know if i choose elements of upper triangular matrix then ,since matrix is symmetric ,elements of lower triangular matrix is automatically fixed ...but i don't know how to do it using permuatations ....answer should be between $ b$ and $c $ . please help






combinatorics matrices permutations







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$endgroup$








  • 2




    $begingroup$
    Hint: a symmetric matrix is determined by $(n+1)n/2$ entries.
    $endgroup$
    – Jacky Chong
    Dec 16 '18 at 6:14










  • $begingroup$
    so, answer is $c $ but how symmetric matrix is determined by $dfrac{n(n+1)}{2}$ entries ....i want to understand it using permutational arguments ....help me with that please ........
    $endgroup$
    – deleteprofile
    Dec 16 '18 at 6:18








  • 1




    $begingroup$
    If symmetric is not there then there is $n^2$ places to put 0 or 1 .because of symmetry places below diagonal are already determined by elements above diagonal.
    $endgroup$
    – Cloud JR
    Dec 16 '18 at 6:21
















-1












$begingroup$


question:



number of different $ntimes n $ symmetric matrices with each element either 0 or 1 is



$(a)2^n$



$(b)2^{n^{2}}$



$(c)2^{frac{n(n+1)}{2}}$



$(d)2^{frac{n(n-1)}{2}}$



my attempt:



first every digonal matrix is symmetric so, answer should not be $2^n$ it should be greater than that . .....i also know if i choose elements of upper triangular matrix then ,since matrix is symmetric ,elements of lower triangular matrix is automatically fixed ...but i don't know how to do it using permuatations ....answer should be between $ b$ and $c $ . please help






combinatorics matrices permutations







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    Hint: a symmetric matrix is determined by $(n+1)n/2$ entries.
    $endgroup$
    – Jacky Chong
    Dec 16 '18 at 6:14










  • $begingroup$
    so, answer is $c $ but how symmetric matrix is determined by $dfrac{n(n+1)}{2}$ entries ....i want to understand it using permutational arguments ....help me with that please ........
    $endgroup$
    – deleteprofile
    Dec 16 '18 at 6:18








  • 1




    $begingroup$
    If symmetric is not there then there is $n^2$ places to put 0 or 1 .because of symmetry places below diagonal are already determined by elements above diagonal.
    $endgroup$
    – Cloud JR
    Dec 16 '18 at 6:21














-1












-1








-1





$begingroup$


question:



number of different $ntimes n $ symmetric matrices with each element either 0 or 1 is



$(a)2^n$



$(b)2^{n^{2}}$



$(c)2^{frac{n(n+1)}{2}}$



$(d)2^{frac{n(n-1)}{2}}$



my attempt:



first every digonal matrix is symmetric so, answer should not be $2^n$ it should be greater than that . .....i also know if i choose elements of upper triangular matrix then ,since matrix is symmetric ,elements of lower triangular matrix is automatically fixed ...but i don't know how to do it using permuatations ....answer should be between $ b$ and $c $ . please help






combinatorics matrices permutations







share|cite|improve this question









$endgroup$




question:



number of different $ntimes n $ symmetric matrices with each element either 0 or 1 is



$(a)2^n$



$(b)2^{n^{2}}$



$(c)2^{frac{n(n+1)}{2}}$



$(d)2^{frac{n(n-1)}{2}}$



my attempt:



first every digonal matrix is symmetric so, answer should not be $2^n$ it should be greater than that . .....i also know if i choose elements of upper triangular matrix then ,since matrix is symmetric ,elements of lower triangular matrix is automatically fixed ...but i don't know how to do it using permuatations ....answer should be between $ b$ and $c $ . please help






combinatorics matrices permutations




combinatorics matrices permutations






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 16 '18 at 6:13









deleteprofiledeleteprofile

1,155316




1,155316








  • 2




    $begingroup$
    Hint: a symmetric matrix is determined by $(n+1)n/2$ entries.
    $endgroup$
    – Jacky Chong
    Dec 16 '18 at 6:14










  • $begingroup$
    so, answer is $c $ but how symmetric matrix is determined by $dfrac{n(n+1)}{2}$ entries ....i want to understand it using permutational arguments ....help me with that please ........
    $endgroup$
    – deleteprofile
    Dec 16 '18 at 6:18








  • 1




    $begingroup$
    If symmetric is not there then there is $n^2$ places to put 0 or 1 .because of symmetry places below diagonal are already determined by elements above diagonal.
    $endgroup$
    – Cloud JR
    Dec 16 '18 at 6:21














  • 2




    $begingroup$
    Hint: a symmetric matrix is determined by $(n+1)n/2$ entries.
    $endgroup$
    – Jacky Chong
    Dec 16 '18 at 6:14










  • $begingroup$
    so, answer is $c $ but how symmetric matrix is determined by $dfrac{n(n+1)}{2}$ entries ....i want to understand it using permutational arguments ....help me with that please ........
    $endgroup$
    – deleteprofile
    Dec 16 '18 at 6:18








  • 1




    $begingroup$
    If symmetric is not there then there is $n^2$ places to put 0 or 1 .because of symmetry places below diagonal are already determined by elements above diagonal.
    $endgroup$
    – Cloud JR
    Dec 16 '18 at 6:21








2




2




$begingroup$
Hint: a symmetric matrix is determined by $(n+1)n/2$ entries.
$endgroup$
– Jacky Chong
Dec 16 '18 at 6:14




$begingroup$
Hint: a symmetric matrix is determined by $(n+1)n/2$ entries.
$endgroup$
– Jacky Chong
Dec 16 '18 at 6:14












$begingroup$
so, answer is $c $ but how symmetric matrix is determined by $dfrac{n(n+1)}{2}$ entries ....i want to understand it using permutational arguments ....help me with that please ........
$endgroup$
– deleteprofile
Dec 16 '18 at 6:18






$begingroup$
so, answer is $c $ but how symmetric matrix is determined by $dfrac{n(n+1)}{2}$ entries ....i want to understand it using permutational arguments ....help me with that please ........
$endgroup$
– deleteprofile
Dec 16 '18 at 6:18






1




1




$begingroup$
If symmetric is not there then there is $n^2$ places to put 0 or 1 .because of symmetry places below diagonal are already determined by elements above diagonal.
$endgroup$
– Cloud JR
Dec 16 '18 at 6:21




$begingroup$
If symmetric is not there then there is $n^2$ places to put 0 or 1 .because of symmetry places below diagonal are already determined by elements above diagonal.
$endgroup$
– Cloud JR
Dec 16 '18 at 6:21










3 Answers
3






active

oldest

votes


















1












$begingroup$

As you say, there are $n$ elements along the diagonal. Each of them can be anything. That leaves $n^2-n$ elements, which occur in pairs symmetric across the diagonal. The entries of each pair must agree, but different pairs can have different values. There are $frac 12(n^2-n)$ pairs. That gives $n+frac 12(n^2-n)=frac 12n(n+1)$ free elements in an $n times n$ symmetric matrix.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    in case if in question matrix was skew symmetric then answer would be $2^{frac{n(n-1)}{2}}$
    $endgroup$
    – deleteprofile
    Dec 16 '18 at 6:28












  • $begingroup$
    You are right..
    $endgroup$
    – Cloud JR
    Dec 16 '18 at 6:29










  • $begingroup$
    @deleteprofile if the entries are only $0$ and $1$ it cannot be skew symmetric except if it's all $0$.
    $endgroup$
    – quid♦
    Dec 16 '18 at 23:01



















1












$begingroup$

Hint. What is the total number of entries on and above the principal diagonal of an $ntimes n$ matrix? You have $n$ entries on the diagonal, $n-1$ immediately above the diagonal, $n-2$ above that and so on till you only have $1$ entry in the upper-right corner of the matrix. What is the sum $1+2+...+(n-1)+n$?






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    For $ntimes n$ matrix there is $n^2$ places to put 0 or 1 .Because of symmetry places below diagonal are already determined by elements above diagonal.



    Totally there are $n^2 $ entries,below principal diagonal entries are already determined by above. Entries above diagonal is $ cfrac{n^2-n}{2}$ .now add diagonal places which gives you $cfrac{n(n+1)}{2}$






    share|cite|improve this answer











    $endgroup$













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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      As you say, there are $n$ elements along the diagonal. Each of them can be anything. That leaves $n^2-n$ elements, which occur in pairs symmetric across the diagonal. The entries of each pair must agree, but different pairs can have different values. There are $frac 12(n^2-n)$ pairs. That gives $n+frac 12(n^2-n)=frac 12n(n+1)$ free elements in an $n times n$ symmetric matrix.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        in case if in question matrix was skew symmetric then answer would be $2^{frac{n(n-1)}{2}}$
        $endgroup$
        – deleteprofile
        Dec 16 '18 at 6:28












      • $begingroup$
        You are right..
        $endgroup$
        – Cloud JR
        Dec 16 '18 at 6:29










      • $begingroup$
        @deleteprofile if the entries are only $0$ and $1$ it cannot be skew symmetric except if it's all $0$.
        $endgroup$
        – quid♦
        Dec 16 '18 at 23:01
















      1












      $begingroup$

      As you say, there are $n$ elements along the diagonal. Each of them can be anything. That leaves $n^2-n$ elements, which occur in pairs symmetric across the diagonal. The entries of each pair must agree, but different pairs can have different values. There are $frac 12(n^2-n)$ pairs. That gives $n+frac 12(n^2-n)=frac 12n(n+1)$ free elements in an $n times n$ symmetric matrix.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        in case if in question matrix was skew symmetric then answer would be $2^{frac{n(n-1)}{2}}$
        $endgroup$
        – deleteprofile
        Dec 16 '18 at 6:28












      • $begingroup$
        You are right..
        $endgroup$
        – Cloud JR
        Dec 16 '18 at 6:29










      • $begingroup$
        @deleteprofile if the entries are only $0$ and $1$ it cannot be skew symmetric except if it's all $0$.
        $endgroup$
        – quid♦
        Dec 16 '18 at 23:01














      1












      1








      1





      $begingroup$

      As you say, there are $n$ elements along the diagonal. Each of them can be anything. That leaves $n^2-n$ elements, which occur in pairs symmetric across the diagonal. The entries of each pair must agree, but different pairs can have different values. There are $frac 12(n^2-n)$ pairs. That gives $n+frac 12(n^2-n)=frac 12n(n+1)$ free elements in an $n times n$ symmetric matrix.






      share|cite|improve this answer









      $endgroup$



      As you say, there are $n$ elements along the diagonal. Each of them can be anything. That leaves $n^2-n$ elements, which occur in pairs symmetric across the diagonal. The entries of each pair must agree, but different pairs can have different values. There are $frac 12(n^2-n)$ pairs. That gives $n+frac 12(n^2-n)=frac 12n(n+1)$ free elements in an $n times n$ symmetric matrix.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Dec 16 '18 at 6:24









      Ross MillikanRoss Millikan

      296k23198371




      296k23198371












      • $begingroup$
        in case if in question matrix was skew symmetric then answer would be $2^{frac{n(n-1)}{2}}$
        $endgroup$
        – deleteprofile
        Dec 16 '18 at 6:28












      • $begingroup$
        You are right..
        $endgroup$
        – Cloud JR
        Dec 16 '18 at 6:29










      • $begingroup$
        @deleteprofile if the entries are only $0$ and $1$ it cannot be skew symmetric except if it's all $0$.
        $endgroup$
        – quid♦
        Dec 16 '18 at 23:01


















      • $begingroup$
        in case if in question matrix was skew symmetric then answer would be $2^{frac{n(n-1)}{2}}$
        $endgroup$
        – deleteprofile
        Dec 16 '18 at 6:28












      • $begingroup$
        You are right..
        $endgroup$
        – Cloud JR
        Dec 16 '18 at 6:29










      • $begingroup$
        @deleteprofile if the entries are only $0$ and $1$ it cannot be skew symmetric except if it's all $0$.
        $endgroup$
        – quid♦
        Dec 16 '18 at 23:01
















      $begingroup$
      in case if in question matrix was skew symmetric then answer would be $2^{frac{n(n-1)}{2}}$
      $endgroup$
      – deleteprofile
      Dec 16 '18 at 6:28






      $begingroup$
      in case if in question matrix was skew symmetric then answer would be $2^{frac{n(n-1)}{2}}$
      $endgroup$
      – deleteprofile
      Dec 16 '18 at 6:28














      $begingroup$
      You are right..
      $endgroup$
      – Cloud JR
      Dec 16 '18 at 6:29




      $begingroup$
      You are right..
      $endgroup$
      – Cloud JR
      Dec 16 '18 at 6:29












      $begingroup$
      @deleteprofile if the entries are only $0$ and $1$ it cannot be skew symmetric except if it's all $0$.
      $endgroup$
      – quid♦
      Dec 16 '18 at 23:01




      $begingroup$
      @deleteprofile if the entries are only $0$ and $1$ it cannot be skew symmetric except if it's all $0$.
      $endgroup$
      – quid♦
      Dec 16 '18 at 23:01











      1












      $begingroup$

      Hint. What is the total number of entries on and above the principal diagonal of an $ntimes n$ matrix? You have $n$ entries on the diagonal, $n-1$ immediately above the diagonal, $n-2$ above that and so on till you only have $1$ entry in the upper-right corner of the matrix. What is the sum $1+2+...+(n-1)+n$?






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        Hint. What is the total number of entries on and above the principal diagonal of an $ntimes n$ matrix? You have $n$ entries on the diagonal, $n-1$ immediately above the diagonal, $n-2$ above that and so on till you only have $1$ entry in the upper-right corner of the matrix. What is the sum $1+2+...+(n-1)+n$?






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          Hint. What is the total number of entries on and above the principal diagonal of an $ntimes n$ matrix? You have $n$ entries on the diagonal, $n-1$ immediately above the diagonal, $n-2$ above that and so on till you only have $1$ entry in the upper-right corner of the matrix. What is the sum $1+2+...+(n-1)+n$?






          share|cite|improve this answer









          $endgroup$



          Hint. What is the total number of entries on and above the principal diagonal of an $ntimes n$ matrix? You have $n$ entries on the diagonal, $n-1$ immediately above the diagonal, $n-2$ above that and so on till you only have $1$ entry in the upper-right corner of the matrix. What is the sum $1+2+...+(n-1)+n$?







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 16 '18 at 6:23









          Shubham JohriShubham Johri

          5,172717




          5,172717























              1












              $begingroup$

              For $ntimes n$ matrix there is $n^2$ places to put 0 or 1 .Because of symmetry places below diagonal are already determined by elements above diagonal.



              Totally there are $n^2 $ entries,below principal diagonal entries are already determined by above. Entries above diagonal is $ cfrac{n^2-n}{2}$ .now add diagonal places which gives you $cfrac{n(n+1)}{2}$






              share|cite|improve this answer











              $endgroup$


















                1












                $begingroup$

                For $ntimes n$ matrix there is $n^2$ places to put 0 or 1 .Because of symmetry places below diagonal are already determined by elements above diagonal.



                Totally there are $n^2 $ entries,below principal diagonal entries are already determined by above. Entries above diagonal is $ cfrac{n^2-n}{2}$ .now add diagonal places which gives you $cfrac{n(n+1)}{2}$






                share|cite|improve this answer











                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  For $ntimes n$ matrix there is $n^2$ places to put 0 or 1 .Because of symmetry places below diagonal are already determined by elements above diagonal.



                  Totally there are $n^2 $ entries,below principal diagonal entries are already determined by above. Entries above diagonal is $ cfrac{n^2-n}{2}$ .now add diagonal places which gives you $cfrac{n(n+1)}{2}$






                  share|cite|improve this answer











                  $endgroup$



                  For $ntimes n$ matrix there is $n^2$ places to put 0 or 1 .Because of symmetry places below diagonal are already determined by elements above diagonal.



                  Totally there are $n^2 $ entries,below principal diagonal entries are already determined by above. Entries above diagonal is $ cfrac{n^2-n}{2}$ .now add diagonal places which gives you $cfrac{n(n+1)}{2}$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 16 '18 at 6:28

























                  answered Dec 16 '18 at 6:23









                  Cloud JRCloud JR

                  909518




                  909518






























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