Monotone eigenvector











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Definition: We say that a vector $xinmathbb{R}^n$ is monotone if, $x_1geq x_2 geq cdots geq x_n$ or $x_1 leq x_2 leq cdots leq x_n$.



Let $M=(m_{ij})_{1leq i,j leq n}$ be a matrix defined by $m_{ij}=1$ if $|i-j|leq frac{n}{2}$ and $m_{ij}=0$ otherwise. Show that the eigenvector corresponding to the second largest eigenvalue is monotone. (i.e., if $lambda_1geqlambda_2cdotsgeqlambda_n$ are the eigenvalues in non-increasing order and $v_1,v_2,ldots,v_n$ are the corresponding eigenvectors then $v_2$ is monotone).



We can prove this by finding a closed-form expression of the eigenvector, but I am trying to figure out if there is an other way, in fact one can show the same for the matrix defined by $m_{ij}=n-|i-j|$. Note that both are Toeplitz matrices, and Robinsonian matrices!



In this article the authors show that : if $M$ is a Robinsonian matrix and $L_M$ is the corresponding Laplacian, then the eigenvector corresponding to the second smallest eigenvalue (called the Fiedler vector) is monotone. I am wondering if there is a link?










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    up vote
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    down vote

    favorite












    Definition: We say that a vector $xinmathbb{R}^n$ is monotone if, $x_1geq x_2 geq cdots geq x_n$ or $x_1 leq x_2 leq cdots leq x_n$.



    Let $M=(m_{ij})_{1leq i,j leq n}$ be a matrix defined by $m_{ij}=1$ if $|i-j|leq frac{n}{2}$ and $m_{ij}=0$ otherwise. Show that the eigenvector corresponding to the second largest eigenvalue is monotone. (i.e., if $lambda_1geqlambda_2cdotsgeqlambda_n$ are the eigenvalues in non-increasing order and $v_1,v_2,ldots,v_n$ are the corresponding eigenvectors then $v_2$ is monotone).



    We can prove this by finding a closed-form expression of the eigenvector, but I am trying to figure out if there is an other way, in fact one can show the same for the matrix defined by $m_{ij}=n-|i-j|$. Note that both are Toeplitz matrices, and Robinsonian matrices!



    In this article the authors show that : if $M$ is a Robinsonian matrix and $L_M$ is the corresponding Laplacian, then the eigenvector corresponding to the second smallest eigenvalue (called the Fiedler vector) is monotone. I am wondering if there is a link?










    share|cite|improve this question
























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      Definition: We say that a vector $xinmathbb{R}^n$ is monotone if, $x_1geq x_2 geq cdots geq x_n$ or $x_1 leq x_2 leq cdots leq x_n$.



      Let $M=(m_{ij})_{1leq i,j leq n}$ be a matrix defined by $m_{ij}=1$ if $|i-j|leq frac{n}{2}$ and $m_{ij}=0$ otherwise. Show that the eigenvector corresponding to the second largest eigenvalue is monotone. (i.e., if $lambda_1geqlambda_2cdotsgeqlambda_n$ are the eigenvalues in non-increasing order and $v_1,v_2,ldots,v_n$ are the corresponding eigenvectors then $v_2$ is monotone).



      We can prove this by finding a closed-form expression of the eigenvector, but I am trying to figure out if there is an other way, in fact one can show the same for the matrix defined by $m_{ij}=n-|i-j|$. Note that both are Toeplitz matrices, and Robinsonian matrices!



      In this article the authors show that : if $M$ is a Robinsonian matrix and $L_M$ is the corresponding Laplacian, then the eigenvector corresponding to the second smallest eigenvalue (called the Fiedler vector) is monotone. I am wondering if there is a link?










      share|cite|improve this question













      Definition: We say that a vector $xinmathbb{R}^n$ is monotone if, $x_1geq x_2 geq cdots geq x_n$ or $x_1 leq x_2 leq cdots leq x_n$.



      Let $M=(m_{ij})_{1leq i,j leq n}$ be a matrix defined by $m_{ij}=1$ if $|i-j|leq frac{n}{2}$ and $m_{ij}=0$ otherwise. Show that the eigenvector corresponding to the second largest eigenvalue is monotone. (i.e., if $lambda_1geqlambda_2cdotsgeqlambda_n$ are the eigenvalues in non-increasing order and $v_1,v_2,ldots,v_n$ are the corresponding eigenvectors then $v_2$ is monotone).



      We can prove this by finding a closed-form expression of the eigenvector, but I am trying to figure out if there is an other way, in fact one can show the same for the matrix defined by $m_{ij}=n-|i-j|$. Note that both are Toeplitz matrices, and Robinsonian matrices!



      In this article the authors show that : if $M$ is a Robinsonian matrix and $L_M$ is the corresponding Laplacian, then the eigenvector corresponding to the second smallest eigenvalue (called the Fiedler vector) is monotone. I am wondering if there is a link?







      linear-algebra graph-theory eigenvalues-eigenvectors laplacian graph-laplacian






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      asked Nov 19 at 13:52









      Samovem

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