Exercise on number of walks in a graph












0












$begingroup$


The following is an exercise (Exercise #2 (a), Chapter 3, page 28) from Richard Stanley's Algebraic Combinatorics.




Let $G$ be a finite graph (allowing loops and multiple edges). Suppose that
there is some integer $l > 0$ such that the number of walks of length $l$ from
any fixed vertex u to any fixed vertex v is independent of u and v. Show that
G has the same number k of edges between any two vertices (including k
loops at each vertex).




My approach:



Let $A$ be the adjacency matrix of the graph $G$. The given condition implies that $exists l > 0$ such that $A^l = cJ$, where $c$ is a non-negative integer and $J$ is the all one matrix. In order to prove this exercise, it is equivalent to show that $A = d J$, for some non-negative integer $d$. The only way I am able to think of in order to prove this is via induction but that too I am not able to implement. Removing a vertex from a graph will still give some condition like $A^l = cJ$ is not clear to me.



Thanks in advance!!






matrices graph-theory random-walk algebraic-graph-theory algebraic-combinatorics







share|cite|improve this question











$endgroup$












  • $begingroup$
    I don't think induction will help. One approach is to show that $A$ must have rank equal to one, and then show that if any matrix $A$ is symmetric and rank-one, then $A^k=c^kA$ for any $k$.
    $endgroup$
    – Chris Godsil
    Mar 29 '17 at 22:18
















0












$begingroup$


The following is an exercise (Exercise #2 (a), Chapter 3, page 28) from Richard Stanley's Algebraic Combinatorics.




Let $G$ be a finite graph (allowing loops and multiple edges). Suppose that
there is some integer $l > 0$ such that the number of walks of length $l$ from
any fixed vertex u to any fixed vertex v is independent of u and v. Show that
G has the same number k of edges between any two vertices (including k
loops at each vertex).




My approach:



Let $A$ be the adjacency matrix of the graph $G$. The given condition implies that $exists l > 0$ such that $A^l = cJ$, where $c$ is a non-negative integer and $J$ is the all one matrix. In order to prove this exercise, it is equivalent to show that $A = d J$, for some non-negative integer $d$. The only way I am able to think of in order to prove this is via induction but that too I am not able to implement. Removing a vertex from a graph will still give some condition like $A^l = cJ$ is not clear to me.



Thanks in advance!!






matrices graph-theory random-walk algebraic-graph-theory algebraic-combinatorics







share|cite|improve this question











$endgroup$












  • $begingroup$
    I don't think induction will help. One approach is to show that $A$ must have rank equal to one, and then show that if any matrix $A$ is symmetric and rank-one, then $A^k=c^kA$ for any $k$.
    $endgroup$
    – Chris Godsil
    Mar 29 '17 at 22:18














0












0








0





$begingroup$


The following is an exercise (Exercise #2 (a), Chapter 3, page 28) from Richard Stanley's Algebraic Combinatorics.




Let $G$ be a finite graph (allowing loops and multiple edges). Suppose that
there is some integer $l > 0$ such that the number of walks of length $l$ from
any fixed vertex u to any fixed vertex v is independent of u and v. Show that
G has the same number k of edges between any two vertices (including k
loops at each vertex).




My approach:



Let $A$ be the adjacency matrix of the graph $G$. The given condition implies that $exists l > 0$ such that $A^l = cJ$, where $c$ is a non-negative integer and $J$ is the all one matrix. In order to prove this exercise, it is equivalent to show that $A = d J$, for some non-negative integer $d$. The only way I am able to think of in order to prove this is via induction but that too I am not able to implement. Removing a vertex from a graph will still give some condition like $A^l = cJ$ is not clear to me.



Thanks in advance!!






matrices graph-theory random-walk algebraic-graph-theory algebraic-combinatorics







share|cite|improve this question











$endgroup$




The following is an exercise (Exercise #2 (a), Chapter 3, page 28) from Richard Stanley's Algebraic Combinatorics.




Let $G$ be a finite graph (allowing loops and multiple edges). Suppose that
there is some integer $l > 0$ such that the number of walks of length $l$ from
any fixed vertex u to any fixed vertex v is independent of u and v. Show that
G has the same number k of edges between any two vertices (including k
loops at each vertex).




My approach:



Let $A$ be the adjacency matrix of the graph $G$. The given condition implies that $exists l > 0$ such that $A^l = cJ$, where $c$ is a non-negative integer and $J$ is the all one matrix. In order to prove this exercise, it is equivalent to show that $A = d J$, for some non-negative integer $d$. The only way I am able to think of in order to prove this is via induction but that too I am not able to implement. Removing a vertex from a graph will still give some condition like $A^l = cJ$ is not clear to me.



Thanks in advance!!






matrices graph-theory random-walk algebraic-graph-theory algebraic-combinatorics




matrices graph-theory random-walk algebraic-graph-theory algebraic-combinatorics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 29 '17 at 7:20







Shivani Goel

















asked Mar 29 '17 at 7:12









Shivani GoelShivani Goel

227115




227115












  • $begingroup$
    I don't think induction will help. One approach is to show that $A$ must have rank equal to one, and then show that if any matrix $A$ is symmetric and rank-one, then $A^k=c^kA$ for any $k$.
    $endgroup$
    – Chris Godsil
    Mar 29 '17 at 22:18


















  • $begingroup$
    I don't think induction will help. One approach is to show that $A$ must have rank equal to one, and then show that if any matrix $A$ is symmetric and rank-one, then $A^k=c^kA$ for any $k$.
    $endgroup$
    – Chris Godsil
    Mar 29 '17 at 22:18
















$begingroup$
I don't think induction will help. One approach is to show that $A$ must have rank equal to one, and then show that if any matrix $A$ is symmetric and rank-one, then $A^k=c^kA$ for any $k$.
$endgroup$
– Chris Godsil
Mar 29 '17 at 22:18




$begingroup$
I don't think induction will help. One approach is to show that $A$ must have rank equal to one, and then show that if any matrix $A$ is symmetric and rank-one, then $A^k=c^kA$ for any $k$.
$endgroup$
– Chris Godsil
Mar 29 '17 at 22:18










2 Answers
2






active

oldest

votes


















1












$begingroup$

Hint: Consider the formula for the number of triangles in a graph, $tr(A^3)/6$. The reason this holds is that, more generally, the entry $i,j$ in $A^k$ is constant (dependent only on $k$) multiple of the number of paths from $i$ to $j$.



When you think about the problem in these terms, you realize it's asking you to assume that every entry of $A^ell$ is the same, for some $ell$.



Can you take it from here?






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks for the suggestion!! But I am not familiar with much probability theory. If possible, I want to prove it in non probabilistic way.
    $endgroup$
    – Shivani Goel
    Mar 29 '17 at 7:24










  • $begingroup$
    @ShivaniGoel There's a way to do that same proof without probability theory. Are you aware of a formula for the number of triangles in a graph based on the adjacency matrix?
    $endgroup$
    – Stella Biderman
    Mar 29 '17 at 7:34












  • $begingroup$
    It is $Tr(A^3) / 6$?
    $endgroup$
    – Shivani Goel
    Mar 29 '17 at 7:44










  • $begingroup$
    @ShivaniGoel Correct. I've amended my answer to be based off of that fact.
    $endgroup$
    – Stella Biderman
    Mar 29 '17 at 7:51










  • $begingroup$
    I am not able to take it from here.
    $endgroup$
    – Shivani Goel
    Mar 29 '17 at 14:34





















0












$begingroup$

Since this exercise is in the chapter on random walks, I suspect the expected solution should also be in terms of those. It also follows some discussion on eigenvalues.



So I shall provide a solution that uses some elementary facts about the eigenvalues of adjacency matrices.



If $lambda_1, ldots, lambda_n$ are the $n$ eigenvalues of $A$ (of order $n$, the order of the graph), then the eigenvalues of $A^l$ are exactly $lambda_1^l, ldots, lambda_n^l$. Since $A^l = cJ$, whose eigenvalues are $cn, 0, ldots, 0$, we get
$lambda_1^l = cn$, $lambda_2^l = cdots = lambda_n^l = 0$.



This shows that $A$ is a matrix of rank $1$, and being symmetric, it can be written as $aa^T$, where $a$ is a (column) vector of length $n$.



Then $A^l = a(a^Ta)^{l-1}a = (a^Ta)^{l-1}A$ (since $a^Ta$ is scalar).
Thus, $A^l = cJ implies (a^Ta)^{l-1}A = cJ$, which shows that $A$ is a scalar multiple of $J$, as required.



(In fact, $a^Ta$ is nothing but the number of walks of length $1$ from the first vertex to itself, which is the number of loops on the first vertex)






share|cite|improve this answer









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    2 Answers
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    2 Answers
    2






    active

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    active

    oldest

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    active

    oldest

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    1












    $begingroup$

    Hint: Consider the formula for the number of triangles in a graph, $tr(A^3)/6$. The reason this holds is that, more generally, the entry $i,j$ in $A^k$ is constant (dependent only on $k$) multiple of the number of paths from $i$ to $j$.



    When you think about the problem in these terms, you realize it's asking you to assume that every entry of $A^ell$ is the same, for some $ell$.



    Can you take it from here?






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Thanks for the suggestion!! But I am not familiar with much probability theory. If possible, I want to prove it in non probabilistic way.
      $endgroup$
      – Shivani Goel
      Mar 29 '17 at 7:24










    • $begingroup$
      @ShivaniGoel There's a way to do that same proof without probability theory. Are you aware of a formula for the number of triangles in a graph based on the adjacency matrix?
      $endgroup$
      – Stella Biderman
      Mar 29 '17 at 7:34












    • $begingroup$
      It is $Tr(A^3) / 6$?
      $endgroup$
      – Shivani Goel
      Mar 29 '17 at 7:44










    • $begingroup$
      @ShivaniGoel Correct. I've amended my answer to be based off of that fact.
      $endgroup$
      – Stella Biderman
      Mar 29 '17 at 7:51










    • $begingroup$
      I am not able to take it from here.
      $endgroup$
      – Shivani Goel
      Mar 29 '17 at 14:34


















    1












    $begingroup$

    Hint: Consider the formula for the number of triangles in a graph, $tr(A^3)/6$. The reason this holds is that, more generally, the entry $i,j$ in $A^k$ is constant (dependent only on $k$) multiple of the number of paths from $i$ to $j$.



    When you think about the problem in these terms, you realize it's asking you to assume that every entry of $A^ell$ is the same, for some $ell$.



    Can you take it from here?






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Thanks for the suggestion!! But I am not familiar with much probability theory. If possible, I want to prove it in non probabilistic way.
      $endgroup$
      – Shivani Goel
      Mar 29 '17 at 7:24










    • $begingroup$
      @ShivaniGoel There's a way to do that same proof without probability theory. Are you aware of a formula for the number of triangles in a graph based on the adjacency matrix?
      $endgroup$
      – Stella Biderman
      Mar 29 '17 at 7:34












    • $begingroup$
      It is $Tr(A^3) / 6$?
      $endgroup$
      – Shivani Goel
      Mar 29 '17 at 7:44










    • $begingroup$
      @ShivaniGoel Correct. I've amended my answer to be based off of that fact.
      $endgroup$
      – Stella Biderman
      Mar 29 '17 at 7:51










    • $begingroup$
      I am not able to take it from here.
      $endgroup$
      – Shivani Goel
      Mar 29 '17 at 14:34
















    1












    1








    1





    $begingroup$

    Hint: Consider the formula for the number of triangles in a graph, $tr(A^3)/6$. The reason this holds is that, more generally, the entry $i,j$ in $A^k$ is constant (dependent only on $k$) multiple of the number of paths from $i$ to $j$.



    When you think about the problem in these terms, you realize it's asking you to assume that every entry of $A^ell$ is the same, for some $ell$.



    Can you take it from here?






    share|cite|improve this answer











    $endgroup$



    Hint: Consider the formula for the number of triangles in a graph, $tr(A^3)/6$. The reason this holds is that, more generally, the entry $i,j$ in $A^k$ is constant (dependent only on $k$) multiple of the number of paths from $i$ to $j$.



    When you think about the problem in these terms, you realize it's asking you to assume that every entry of $A^ell$ is the same, for some $ell$.



    Can you take it from here?







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Mar 29 '17 at 7:50

























    answered Mar 29 '17 at 7:22









    Stella BidermanStella Biderman

    26.6k63275




    26.6k63275












    • $begingroup$
      Thanks for the suggestion!! But I am not familiar with much probability theory. If possible, I want to prove it in non probabilistic way.
      $endgroup$
      – Shivani Goel
      Mar 29 '17 at 7:24










    • $begingroup$
      @ShivaniGoel There's a way to do that same proof without probability theory. Are you aware of a formula for the number of triangles in a graph based on the adjacency matrix?
      $endgroup$
      – Stella Biderman
      Mar 29 '17 at 7:34












    • $begingroup$
      It is $Tr(A^3) / 6$?
      $endgroup$
      – Shivani Goel
      Mar 29 '17 at 7:44










    • $begingroup$
      @ShivaniGoel Correct. I've amended my answer to be based off of that fact.
      $endgroup$
      – Stella Biderman
      Mar 29 '17 at 7:51










    • $begingroup$
      I am not able to take it from here.
      $endgroup$
      – Shivani Goel
      Mar 29 '17 at 14:34




















    • $begingroup$
      Thanks for the suggestion!! But I am not familiar with much probability theory. If possible, I want to prove it in non probabilistic way.
      $endgroup$
      – Shivani Goel
      Mar 29 '17 at 7:24










    • $begingroup$
      @ShivaniGoel There's a way to do that same proof without probability theory. Are you aware of a formula for the number of triangles in a graph based on the adjacency matrix?
      $endgroup$
      – Stella Biderman
      Mar 29 '17 at 7:34












    • $begingroup$
      It is $Tr(A^3) / 6$?
      $endgroup$
      – Shivani Goel
      Mar 29 '17 at 7:44










    • $begingroup$
      @ShivaniGoel Correct. I've amended my answer to be based off of that fact.
      $endgroup$
      – Stella Biderman
      Mar 29 '17 at 7:51










    • $begingroup$
      I am not able to take it from here.
      $endgroup$
      – Shivani Goel
      Mar 29 '17 at 14:34


















    $begingroup$
    Thanks for the suggestion!! But I am not familiar with much probability theory. If possible, I want to prove it in non probabilistic way.
    $endgroup$
    – Shivani Goel
    Mar 29 '17 at 7:24




    $begingroup$
    Thanks for the suggestion!! But I am not familiar with much probability theory. If possible, I want to prove it in non probabilistic way.
    $endgroup$
    – Shivani Goel
    Mar 29 '17 at 7:24












    $begingroup$
    @ShivaniGoel There's a way to do that same proof without probability theory. Are you aware of a formula for the number of triangles in a graph based on the adjacency matrix?
    $endgroup$
    – Stella Biderman
    Mar 29 '17 at 7:34






    $begingroup$
    @ShivaniGoel There's a way to do that same proof without probability theory. Are you aware of a formula for the number of triangles in a graph based on the adjacency matrix?
    $endgroup$
    – Stella Biderman
    Mar 29 '17 at 7:34














    $begingroup$
    It is $Tr(A^3) / 6$?
    $endgroup$
    – Shivani Goel
    Mar 29 '17 at 7:44




    $begingroup$
    It is $Tr(A^3) / 6$?
    $endgroup$
    – Shivani Goel
    Mar 29 '17 at 7:44












    $begingroup$
    @ShivaniGoel Correct. I've amended my answer to be based off of that fact.
    $endgroup$
    – Stella Biderman
    Mar 29 '17 at 7:51




    $begingroup$
    @ShivaniGoel Correct. I've amended my answer to be based off of that fact.
    $endgroup$
    – Stella Biderman
    Mar 29 '17 at 7:51












    $begingroup$
    I am not able to take it from here.
    $endgroup$
    – Shivani Goel
    Mar 29 '17 at 14:34






    $begingroup$
    I am not able to take it from here.
    $endgroup$
    – Shivani Goel
    Mar 29 '17 at 14:34













    0












    $begingroup$

    Since this exercise is in the chapter on random walks, I suspect the expected solution should also be in terms of those. It also follows some discussion on eigenvalues.



    So I shall provide a solution that uses some elementary facts about the eigenvalues of adjacency matrices.



    If $lambda_1, ldots, lambda_n$ are the $n$ eigenvalues of $A$ (of order $n$, the order of the graph), then the eigenvalues of $A^l$ are exactly $lambda_1^l, ldots, lambda_n^l$. Since $A^l = cJ$, whose eigenvalues are $cn, 0, ldots, 0$, we get
    $lambda_1^l = cn$, $lambda_2^l = cdots = lambda_n^l = 0$.



    This shows that $A$ is a matrix of rank $1$, and being symmetric, it can be written as $aa^T$, where $a$ is a (column) vector of length $n$.



    Then $A^l = a(a^Ta)^{l-1}a = (a^Ta)^{l-1}A$ (since $a^Ta$ is scalar).
    Thus, $A^l = cJ implies (a^Ta)^{l-1}A = cJ$, which shows that $A$ is a scalar multiple of $J$, as required.



    (In fact, $a^Ta$ is nothing but the number of walks of length $1$ from the first vertex to itself, which is the number of loops on the first vertex)






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Since this exercise is in the chapter on random walks, I suspect the expected solution should also be in terms of those. It also follows some discussion on eigenvalues.



      So I shall provide a solution that uses some elementary facts about the eigenvalues of adjacency matrices.



      If $lambda_1, ldots, lambda_n$ are the $n$ eigenvalues of $A$ (of order $n$, the order of the graph), then the eigenvalues of $A^l$ are exactly $lambda_1^l, ldots, lambda_n^l$. Since $A^l = cJ$, whose eigenvalues are $cn, 0, ldots, 0$, we get
      $lambda_1^l = cn$, $lambda_2^l = cdots = lambda_n^l = 0$.



      This shows that $A$ is a matrix of rank $1$, and being symmetric, it can be written as $aa^T$, where $a$ is a (column) vector of length $n$.



      Then $A^l = a(a^Ta)^{l-1}a = (a^Ta)^{l-1}A$ (since $a^Ta$ is scalar).
      Thus, $A^l = cJ implies (a^Ta)^{l-1}A = cJ$, which shows that $A$ is a scalar multiple of $J$, as required.



      (In fact, $a^Ta$ is nothing but the number of walks of length $1$ from the first vertex to itself, which is the number of loops on the first vertex)






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Since this exercise is in the chapter on random walks, I suspect the expected solution should also be in terms of those. It also follows some discussion on eigenvalues.



        So I shall provide a solution that uses some elementary facts about the eigenvalues of adjacency matrices.



        If $lambda_1, ldots, lambda_n$ are the $n$ eigenvalues of $A$ (of order $n$, the order of the graph), then the eigenvalues of $A^l$ are exactly $lambda_1^l, ldots, lambda_n^l$. Since $A^l = cJ$, whose eigenvalues are $cn, 0, ldots, 0$, we get
        $lambda_1^l = cn$, $lambda_2^l = cdots = lambda_n^l = 0$.



        This shows that $A$ is a matrix of rank $1$, and being symmetric, it can be written as $aa^T$, where $a$ is a (column) vector of length $n$.



        Then $A^l = a(a^Ta)^{l-1}a = (a^Ta)^{l-1}A$ (since $a^Ta$ is scalar).
        Thus, $A^l = cJ implies (a^Ta)^{l-1}A = cJ$, which shows that $A$ is a scalar multiple of $J$, as required.



        (In fact, $a^Ta$ is nothing but the number of walks of length $1$ from the first vertex to itself, which is the number of loops on the first vertex)






        share|cite|improve this answer









        $endgroup$



        Since this exercise is in the chapter on random walks, I suspect the expected solution should also be in terms of those. It also follows some discussion on eigenvalues.



        So I shall provide a solution that uses some elementary facts about the eigenvalues of adjacency matrices.



        If $lambda_1, ldots, lambda_n$ are the $n$ eigenvalues of $A$ (of order $n$, the order of the graph), then the eigenvalues of $A^l$ are exactly $lambda_1^l, ldots, lambda_n^l$. Since $A^l = cJ$, whose eigenvalues are $cn, 0, ldots, 0$, we get
        $lambda_1^l = cn$, $lambda_2^l = cdots = lambda_n^l = 0$.



        This shows that $A$ is a matrix of rank $1$, and being symmetric, it can be written as $aa^T$, where $a$ is a (column) vector of length $n$.



        Then $A^l = a(a^Ta)^{l-1}a = (a^Ta)^{l-1}A$ (since $a^Ta$ is scalar).
        Thus, $A^l = cJ implies (a^Ta)^{l-1}A = cJ$, which shows that $A$ is a scalar multiple of $J$, as required.



        (In fact, $a^Ta$ is nothing but the number of walks of length $1$ from the first vertex to itself, which is the number of loops on the first vertex)







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 17 '18 at 11:51









        M. VinayM. Vinay

        6,76822135




        6,76822135






























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            Cabo Verde

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              Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º)
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            Gyllenstierna

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