Showing Eigenvalues are non negative in a graph problem












2














This is a graph problem, related to one I posted earlier as this is for a later part of the same question. I've tried a few things but haven't made much progress so far.



$mathcal{E}$ is a set of edges such that $(x,y)$ is an edge between nodes $x$ and $y$.



$w_{(x,y)}$ is the weight of edge $(x,y)$.



$d_y=sum_{x text{such that} (x,y) in mathcal{E}}w_{(x,y)}$



$D=text{diag}{d_y}$



Which means the diagonal matrix containing vector $mathbf{d}$ on its leading diagonal which has elements which are the sums of rows of $w(x,y)$. But I am not really sure.



$W_{x,y}=begin{cases}w_{(x,y)},& text{for } (x,y) in mathcal{E}\ 0, & text{otherwise}end{cases}$



$M=D-W$



Show that eigenvalues of $M$ are non-negative.



If anyone has any advice it would be much appreciated.






linear-algebra matrices graph-theory







share|cite|improve this question





























    2














    This is a graph problem, related to one I posted earlier as this is for a later part of the same question. I've tried a few things but haven't made much progress so far.



    $mathcal{E}$ is a set of edges such that $(x,y)$ is an edge between nodes $x$ and $y$.



    $w_{(x,y)}$ is the weight of edge $(x,y)$.



    $d_y=sum_{x text{such that} (x,y) in mathcal{E}}w_{(x,y)}$



    $D=text{diag}{d_y}$



    Which means the diagonal matrix containing vector $mathbf{d}$ on its leading diagonal which has elements which are the sums of rows of $w(x,y)$. But I am not really sure.



    $W_{x,y}=begin{cases}w_{(x,y)},& text{for } (x,y) in mathcal{E}\ 0, & text{otherwise}end{cases}$



    $M=D-W$



    Show that eigenvalues of $M$ are non-negative.



    If anyone has any advice it would be much appreciated.






    linear-algebra matrices graph-theory







    share|cite|improve this question



























      2












      2








      2







      This is a graph problem, related to one I posted earlier as this is for a later part of the same question. I've tried a few things but haven't made much progress so far.



      $mathcal{E}$ is a set of edges such that $(x,y)$ is an edge between nodes $x$ and $y$.



      $w_{(x,y)}$ is the weight of edge $(x,y)$.



      $d_y=sum_{x text{such that} (x,y) in mathcal{E}}w_{(x,y)}$



      $D=text{diag}{d_y}$



      Which means the diagonal matrix containing vector $mathbf{d}$ on its leading diagonal which has elements which are the sums of rows of $w(x,y)$. But I am not really sure.



      $W_{x,y}=begin{cases}w_{(x,y)},& text{for } (x,y) in mathcal{E}\ 0, & text{otherwise}end{cases}$



      $M=D-W$



      Show that eigenvalues of $M$ are non-negative.



      If anyone has any advice it would be much appreciated.






      linear-algebra matrices graph-theory







      share|cite|improve this question















      This is a graph problem, related to one I posted earlier as this is for a later part of the same question. I've tried a few things but haven't made much progress so far.



      $mathcal{E}$ is a set of edges such that $(x,y)$ is an edge between nodes $x$ and $y$.



      $w_{(x,y)}$ is the weight of edge $(x,y)$.



      $d_y=sum_{x text{such that} (x,y) in mathcal{E}}w_{(x,y)}$



      $D=text{diag}{d_y}$



      Which means the diagonal matrix containing vector $mathbf{d}$ on its leading diagonal which has elements which are the sums of rows of $w(x,y)$. But I am not really sure.



      $W_{x,y}=begin{cases}w_{(x,y)},& text{for } (x,y) in mathcal{E}\ 0, & text{otherwise}end{cases}$



      $M=D-W$



      Show that eigenvalues of $M$ are non-negative.



      If anyone has any advice it would be much appreciated.






      linear-algebra matrices graph-theory




      linear-algebra matrices graph-theory






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 4 at 13:57









      amWhy

      192k28225439




      192k28225439










      asked Oct 29 '18 at 20:46









      I suck at MathsI suck at Maths

      179




      179






















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          For every vector $f$ in $R^n$ we have:



          $$f'Mf=f'Df-f'Wf=sum_{i=1}^nd_if_i^2-sum_{i,j=1}^nf_if_jw_{ij}$$
          $$frac{1}2(sum_{i=1}^n d_if_i^2+sum_{j=1}^nd_jf_j^2-2sum_{i,j=1}^nf_if_jw_{ij})=frac{1}2sum_{i,j=1}^nw_{ij}(f_i-f_j)^2geq0$$



          This shows you that M is a positive semi-definite matrix, and hence the eigenvalues are non-negative.






          share|cite|improve this answer





















          • Thank you, I need to get better at recognising things like this:)
            – I suck at Maths
            Oct 30 '18 at 8:50











          Your Answer





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          1 Answer
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          active

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          1 Answer
          1






          active

          oldest

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          active

          oldest

          votes






          active

          oldest

          votes









          4














          For every vector $f$ in $R^n$ we have:



          $$f'Mf=f'Df-f'Wf=sum_{i=1}^nd_if_i^2-sum_{i,j=1}^nf_if_jw_{ij}$$
          $$frac{1}2(sum_{i=1}^n d_if_i^2+sum_{j=1}^nd_jf_j^2-2sum_{i,j=1}^nf_if_jw_{ij})=frac{1}2sum_{i,j=1}^nw_{ij}(f_i-f_j)^2geq0$$



          This shows you that M is a positive semi-definite matrix, and hence the eigenvalues are non-negative.






          share|cite|improve this answer





















          • Thank you, I need to get better at recognising things like this:)
            – I suck at Maths
            Oct 30 '18 at 8:50
















          4














          For every vector $f$ in $R^n$ we have:



          $$f'Mf=f'Df-f'Wf=sum_{i=1}^nd_if_i^2-sum_{i,j=1}^nf_if_jw_{ij}$$
          $$frac{1}2(sum_{i=1}^n d_if_i^2+sum_{j=1}^nd_jf_j^2-2sum_{i,j=1}^nf_if_jw_{ij})=frac{1}2sum_{i,j=1}^nw_{ij}(f_i-f_j)^2geq0$$



          This shows you that M is a positive semi-definite matrix, and hence the eigenvalues are non-negative.






          share|cite|improve this answer





















          • Thank you, I need to get better at recognising things like this:)
            – I suck at Maths
            Oct 30 '18 at 8:50














          4












          4








          4






          For every vector $f$ in $R^n$ we have:



          $$f'Mf=f'Df-f'Wf=sum_{i=1}^nd_if_i^2-sum_{i,j=1}^nf_if_jw_{ij}$$
          $$frac{1}2(sum_{i=1}^n d_if_i^2+sum_{j=1}^nd_jf_j^2-2sum_{i,j=1}^nf_if_jw_{ij})=frac{1}2sum_{i,j=1}^nw_{ij}(f_i-f_j)^2geq0$$



          This shows you that M is a positive semi-definite matrix, and hence the eigenvalues are non-negative.






          share|cite|improve this answer












          For every vector $f$ in $R^n$ we have:



          $$f'Mf=f'Df-f'Wf=sum_{i=1}^nd_if_i^2-sum_{i,j=1}^nf_if_jw_{ij}$$
          $$frac{1}2(sum_{i=1}^n d_if_i^2+sum_{j=1}^nd_jf_j^2-2sum_{i,j=1}^nf_if_jw_{ij})=frac{1}2sum_{i,j=1}^nw_{ij}(f_i-f_j)^2geq0$$



          This shows you that M is a positive semi-definite matrix, and hence the eigenvalues are non-negative.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Oct 29 '18 at 21:07









          Ramiro ScorolliRamiro Scorolli

          655113




          655113












          • Thank you, I need to get better at recognising things like this:)
            – I suck at Maths
            Oct 30 '18 at 8:50


















          • Thank you, I need to get better at recognising things like this:)
            – I suck at Maths
            Oct 30 '18 at 8:50
















          Thank you, I need to get better at recognising things like this:)
          – I suck at Maths
          Oct 30 '18 at 8:50




          Thank you, I need to get better at recognising things like this:)
          – I suck at Maths
          Oct 30 '18 at 8:50


















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