Showing Eigenvalues are non negative in a graph problem
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
edited Jan 4 at 13:57
amWhy
192k28225439
asked Oct 29 '18 at 20:46
I suck at MathsI suck at Maths
179
add a comment |
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
edited Jan 4 at 13:57
amWhy
192k28225439
asked Oct 29 '18 at 20:46
I suck at MathsI suck at Maths
179
add a comment |
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
edited Jan 4 at 13:57
amWhy
192k28225439
asked Oct 29 '18 at 20:46
I suck at MathsI suck at Maths
179
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
edited Jan 4 at 13:57
amWhy
192k28225439
asked Oct 29 '18 at 20:46
I suck at MathsI suck at Maths
179
edited Jan 4 at 13:57
amWhy
192k28225439
asked Oct 29 '18 at 20:46
I suck at MathsI suck at Maths
179
edited Jan 4 at 13:57
amWhy
192k28225439
edited Jan 4 at 13:57
amWhy
192k28225439
edited Jan 4 at 13:57
amWhy
192k28225439
192k28225439
asked Oct 29 '18 at 20:46
I suck at MathsI suck at Maths
179
asked Oct 29 '18 at 20:46
I suck at MathsI suck at Maths
179
asked Oct 29 '18 at 20:46
I suck at MathsI suck at Maths
179
179
add a comment |
add a comment |
1 Answer
1
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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.
answered Oct 29 '18 at 21:07
Ramiro ScorolliRamiro Scorolli
655113
Thank you, I need to get better at recognising things like this:)
– I suck at Maths
Oct 30 '18 at 8:50
add a comment |
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1 Answer
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active
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votes
1 Answer
1
active
oldest
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active
oldest
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active
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votes
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.
answered Oct 29 '18 at 21:07
Ramiro ScorolliRamiro Scorolli
655113
Thank you, I need to get better at recognising things like this:)
– I suck at Maths
Oct 30 '18 at 8:50
add a comment |
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.
answered Oct 29 '18 at 21:07
Ramiro ScorolliRamiro Scorolli
655113
Thank you, I need to get better at recognising things like this:)
– I suck at Maths
Oct 30 '18 at 8:50
add a comment |
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.
answered Oct 29 '18 at 21:07
Ramiro ScorolliRamiro Scorolli
655113
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.
answered Oct 29 '18 at 21:07
Ramiro ScorolliRamiro Scorolli
655113
answered Oct 29 '18 at 21:07
Ramiro ScorolliRamiro Scorolli
655113
answered Oct 29 '18 at 21:07
Ramiro ScorolliRamiro Scorolli
655113
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
add a comment |
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
add a comment |
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