Eigenvalues of the Laplacian of the directed De Bruijn graph Planned maintenance scheduled...



Eigenvalues of the Laplacian of the directed De Bruijn graph



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4












$begingroup$


We will denote by $DB(n,k)$ the directed De Bruijn graph, which is a digraph whose vertices are elements of ${0,1,dots,k-1}^n$, and $sigma_1cdots sigma_n$ is connected to $tau_1cdots tau_{n}$ if and only if $sigma_i=tau_{i+1}$ for every $i=1,dots,n-1$.



We would like to calculate the spectrum of the Laplacian matrix of $DB(n,k)$. Note that since $DB(n,k)$ is $k$-regular, then $lambda in Spec(L_{DB(n,k)})$ if and only if $k-lambda in Spec(A_{DB(n,k)})$, so we can calculate the eigenvalues of the Laplacian matrix from the eigenvalues of the Adjacency matrix.



We found the following article on the subject, but it only calculates the spectrum of the underlining undirected graph of the De Bruijn graph:



https://core.ac.uk/download/pdf/82810454.pdf



Thanks!






co.combinatorics graph-theory spectral-graph-theory enumerative-combinatorics







share|cite|improve this question











$endgroup$

















    4












    $begingroup$


    We will denote by $DB(n,k)$ the directed De Bruijn graph, which is a digraph whose vertices are elements of ${0,1,dots,k-1}^n$, and $sigma_1cdots sigma_n$ is connected to $tau_1cdots tau_{n}$ if and only if $sigma_i=tau_{i+1}$ for every $i=1,dots,n-1$.



    We would like to calculate the spectrum of the Laplacian matrix of $DB(n,k)$. Note that since $DB(n,k)$ is $k$-regular, then $lambda in Spec(L_{DB(n,k)})$ if and only if $k-lambda in Spec(A_{DB(n,k)})$, so we can calculate the eigenvalues of the Laplacian matrix from the eigenvalues of the Adjacency matrix.



    We found the following article on the subject, but it only calculates the spectrum of the underlining undirected graph of the De Bruijn graph:



    https://core.ac.uk/download/pdf/82810454.pdf



    Thanks!






    co.combinatorics graph-theory spectral-graph-theory enumerative-combinatorics







    share|cite|improve this question











    $endgroup$















      4












      4








      4





      $begingroup$


      We will denote by $DB(n,k)$ the directed De Bruijn graph, which is a digraph whose vertices are elements of ${0,1,dots,k-1}^n$, and $sigma_1cdots sigma_n$ is connected to $tau_1cdots tau_{n}$ if and only if $sigma_i=tau_{i+1}$ for every $i=1,dots,n-1$.



      We would like to calculate the spectrum of the Laplacian matrix of $DB(n,k)$. Note that since $DB(n,k)$ is $k$-regular, then $lambda in Spec(L_{DB(n,k)})$ if and only if $k-lambda in Spec(A_{DB(n,k)})$, so we can calculate the eigenvalues of the Laplacian matrix from the eigenvalues of the Adjacency matrix.



      We found the following article on the subject, but it only calculates the spectrum of the underlining undirected graph of the De Bruijn graph:



      https://core.ac.uk/download/pdf/82810454.pdf



      Thanks!






      co.combinatorics graph-theory spectral-graph-theory enumerative-combinatorics







      share|cite|improve this question











      $endgroup$




      We will denote by $DB(n,k)$ the directed De Bruijn graph, which is a digraph whose vertices are elements of ${0,1,dots,k-1}^n$, and $sigma_1cdots sigma_n$ is connected to $tau_1cdots tau_{n}$ if and only if $sigma_i=tau_{i+1}$ for every $i=1,dots,n-1$.



      We would like to calculate the spectrum of the Laplacian matrix of $DB(n,k)$. Note that since $DB(n,k)$ is $k$-regular, then $lambda in Spec(L_{DB(n,k)})$ if and only if $k-lambda in Spec(A_{DB(n,k)})$, so we can calculate the eigenvalues of the Laplacian matrix from the eigenvalues of the Adjacency matrix.



      We found the following article on the subject, but it only calculates the spectrum of the underlining undirected graph of the De Bruijn graph:



      https://core.ac.uk/download/pdf/82810454.pdf



      Thanks!






      co.combinatorics graph-theory spectral-graph-theory enumerative-combinatorics




      co.combinatorics graph-theory spectral-graph-theory enumerative-combinatorics






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




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      edited 3 hours ago









      Fedor Petrov

      52.5k6122241




      52.5k6122241










      asked 4 hours ago









      Serge the ToasterSerge the Toaster

      542




      542






















          2 Answers
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          2












          $begingroup$

          The computation for $k=2$ appears (among other places) on pp. 158--159 of my book Algebraic Combinatorics, second edition. The computation for arbitrary $k$ is completely analogous. Dima Pasechnik is correct that there is one eigenvalue equal to 0, and all the others are equal to $k$.






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            I believe that for $p$ prime $DB(m,p)$ has one 0 eigenvalue (this is easy to check), and the other eigenvalues are all equal to $p$. This is something that we observed while working on https://arxiv.org/abs/1405.0113, but didn't try to prove (should not be too hard I guess, using a decription of $DB(m,p)$ as the line graph of $DB(m-1,p)$, and then induction).






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              2 Answers
              2






              active

              oldest

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              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              2












              $begingroup$

              The computation for $k=2$ appears (among other places) on pp. 158--159 of my book Algebraic Combinatorics, second edition. The computation for arbitrary $k$ is completely analogous. Dima Pasechnik is correct that there is one eigenvalue equal to 0, and all the others are equal to $k$.






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                The computation for $k=2$ appears (among other places) on pp. 158--159 of my book Algebraic Combinatorics, second edition. The computation for arbitrary $k$ is completely analogous. Dima Pasechnik is correct that there is one eigenvalue equal to 0, and all the others are equal to $k$.






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  The computation for $k=2$ appears (among other places) on pp. 158--159 of my book Algebraic Combinatorics, second edition. The computation for arbitrary $k$ is completely analogous. Dima Pasechnik is correct that there is one eigenvalue equal to 0, and all the others are equal to $k$.






                  share|cite|improve this answer









                  $endgroup$



                  The computation for $k=2$ appears (among other places) on pp. 158--159 of my book Algebraic Combinatorics, second edition. The computation for arbitrary $k$ is completely analogous. Dima Pasechnik is correct that there is one eigenvalue equal to 0, and all the others are equal to $k$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 1 hour ago









                  Richard StanleyRichard Stanley

                  29.4k9118191




                  29.4k9118191























                      1












                      $begingroup$

                      I believe that for $p$ prime $DB(m,p)$ has one 0 eigenvalue (this is easy to check), and the other eigenvalues are all equal to $p$. This is something that we observed while working on https://arxiv.org/abs/1405.0113, but didn't try to prove (should not be too hard I guess, using a decription of $DB(m,p)$ as the line graph of $DB(m-1,p)$, and then induction).






                      share|cite|improve this answer











                      $endgroup$


















                        1












                        $begingroup$

                        I believe that for $p$ prime $DB(m,p)$ has one 0 eigenvalue (this is easy to check), and the other eigenvalues are all equal to $p$. This is something that we observed while working on https://arxiv.org/abs/1405.0113, but didn't try to prove (should not be too hard I guess, using a decription of $DB(m,p)$ as the line graph of $DB(m-1,p)$, and then induction).






                        share|cite|improve this answer











                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          I believe that for $p$ prime $DB(m,p)$ has one 0 eigenvalue (this is easy to check), and the other eigenvalues are all equal to $p$. This is something that we observed while working on https://arxiv.org/abs/1405.0113, but didn't try to prove (should not be too hard I guess, using a decription of $DB(m,p)$ as the line graph of $DB(m-1,p)$, and then induction).






                          share|cite|improve this answer











                          $endgroup$



                          I believe that for $p$ prime $DB(m,p)$ has one 0 eigenvalue (this is easy to check), and the other eigenvalues are all equal to $p$. This is something that we observed while working on https://arxiv.org/abs/1405.0113, but didn't try to prove (should not be too hard I guess, using a decription of $DB(m,p)$ as the line graph of $DB(m-1,p)$, and then induction).







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited 3 hours ago

























                          answered 3 hours ago









                          Dima PasechnikDima Pasechnik

                          9,43311952




                          9,43311952






























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