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Eigenvalues of a complete connected network
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The question is how to find a closed formula for the characteristic polynomial and the eigenvalues of the $N$*$N$ adjacency matrix $A$ of a fully connected network of size $N$:
$$ A =beginpmatrix
0 & 1 & 1 & cdots & 1
\ 1 & 0 & 1 & cdots & 1
\ vdots & vdots & vdots & ddots & vdots
\ 1 & 1 & 1 & cdots & 0
endpmatrix.$$
My attempt:
I tried to find a recursive formula for the characteristic polynomial $P(lambda)$. It seems that:
$$P(lambda)_N+1 = (-lambda)P(lambda)_N - NQ_N(lambda)$$ where $Q_N(lambda)$ is the determinant of the following matrix:
$$ beginpmatrix
1 & 1 & 1 & cdots & 1
\ 1 & -lambda & 1 & cdots & 1
\ vdots & vdots & vdots & ddots & vdots
\ 1 & 1 & 1 & cdots & -lambda
endpmatrix.$$
Again trying to find a recursion formula for $Q_N$, using gaussians elimination I found that:
$$Q_N(lambda) = (-1)^N-1(lambda+1)^N-1$$ and so:
$$P(lambda)_N+1 = (-lambda)P(lambda)_N +(-1)^NN(lambda+1)^N-1$$
but at this point I couldn't go on.
Looking for a pattern, starting from small N, it seems that the relation I'm looking for is:
$$P_N(lambda) = (lambda -N + 1)(lambda+1)^N-1$$
but I'm not sure and it seems in discord with the previous recurrence relation.
matrices eigenvalues-eigenvectors determinant network
asked Jul 31 at 14:34
Alex
1729
 |Â
show 1 more comment
up vote
0
down vote
favorite
The question is how to find a closed formula for the characteristic polynomial and the eigenvalues of the $N$*$N$ adjacency matrix $A$ of a fully connected network of size $N$:
$$ A =beginpmatrix
0 & 1 & 1 & cdots & 1
\ 1 & 0 & 1 & cdots & 1
\ vdots & vdots & vdots & ddots & vdots
\ 1 & 1 & 1 & cdots & 0
endpmatrix.$$
My attempt:
I tried to find a recursive formula for the characteristic polynomial $P(lambda)$. It seems that:
$$P(lambda)_N+1 = (-lambda)P(lambda)_N - NQ_N(lambda)$$ where $Q_N(lambda)$ is the determinant of the following matrix:
$$ beginpmatrix
1 & 1 & 1 & cdots & 1
\ 1 & -lambda & 1 & cdots & 1
\ vdots & vdots & vdots & ddots & vdots
\ 1 & 1 & 1 & cdots & -lambda
endpmatrix.$$
Again trying to find a recursion formula for $Q_N$, using gaussians elimination I found that:
$$Q_N(lambda) = (-1)^N-1(lambda+1)^N-1$$ and so:
$$P(lambda)_N+1 = (-lambda)P(lambda)_N +(-1)^NN(lambda+1)^N-1$$
but at this point I couldn't go on.
Looking for a pattern, starting from small N, it seems that the relation I'm looking for is:
$$P_N(lambda) = (lambda -N + 1)(lambda+1)^N-1$$
but I'm not sure and it seems in discord with the previous recurrence relation.
matrices eigenvalues-eigenvectors determinant network
asked Jul 31 at 14:34
Alex
1729
Possible duplicate: math.stackexchange.com/questions/1423024/…
– Ethan Bolker
Jul 31 at 14:42
Thanks! But I would like to have a close formula for the characteristic polynomial.
– Alex
Jul 31 at 14:53
Isn't it just $(x-n+1)(x+1)^n-1$ since you know the roots (eigenvalues)?
– Ethan Bolker
Jul 31 at 15:56
Yes! Thanks! but unfortunately I don’t know the “complete graphs theory†cited in the answer, so I still want an explanation using linear algebra if it is possible.
– Alex
Jul 31 at 17:40
1
That is a linear algebra explanation. It works by explicitly finding the eigenvectors. One of them is the vector of all $1$'s.
– Ethan Bolker
Jul 31 at 19:41
 |Â
show 1 more comment
up vote
0
down vote
favorite
up vote
0
down vote
favorite
The question is how to find a closed formula for the characteristic polynomial and the eigenvalues of the $N$*$N$ adjacency matrix $A$ of a fully connected network of size $N$:
$$ A =beginpmatrix
0 & 1 & 1 & cdots & 1
\ 1 & 0 & 1 & cdots & 1
\ vdots & vdots & vdots & ddots & vdots
\ 1 & 1 & 1 & cdots & 0
endpmatrix.$$
My attempt:
I tried to find a recursive formula for the characteristic polynomial $P(lambda)$. It seems that:
$$P(lambda)_N+1 = (-lambda)P(lambda)_N - NQ_N(lambda)$$ where $Q_N(lambda)$ is the determinant of the following matrix:
$$ beginpmatrix
1 & 1 & 1 & cdots & 1
\ 1 & -lambda & 1 & cdots & 1
\ vdots & vdots & vdots & ddots & vdots
\ 1 & 1 & 1 & cdots & -lambda
endpmatrix.$$
Again trying to find a recursion formula for $Q_N$, using gaussians elimination I found that:
$$Q_N(lambda) = (-1)^N-1(lambda+1)^N-1$$ and so:
$$P(lambda)_N+1 = (-lambda)P(lambda)_N +(-1)^NN(lambda+1)^N-1$$
but at this point I couldn't go on.
Looking for a pattern, starting from small N, it seems that the relation I'm looking for is:
$$P_N(lambda) = (lambda -N + 1)(lambda+1)^N-1$$
but I'm not sure and it seems in discord with the previous recurrence relation.
matrices eigenvalues-eigenvectors determinant network
asked Jul 31 at 14:34
Alex
1729
The question is how to find a closed formula for the characteristic polynomial and the eigenvalues of the $N$*$N$ adjacency matrix $A$ of a fully connected network of size $N$:
$$ A =beginpmatrix
0 & 1 & 1 & cdots & 1
\ 1 & 0 & 1 & cdots & 1
\ vdots & vdots & vdots & ddots & vdots
\ 1 & 1 & 1 & cdots & 0
endpmatrix.$$
My attempt:
I tried to find a recursive formula for the characteristic polynomial $P(lambda)$. It seems that:
$$P(lambda)_N+1 = (-lambda)P(lambda)_N - NQ_N(lambda)$$ where $Q_N(lambda)$ is the determinant of the following matrix:
$$ beginpmatrix
1 & 1 & 1 & cdots & 1
\ 1 & -lambda & 1 & cdots & 1
\ vdots & vdots & vdots & ddots & vdots
\ 1 & 1 & 1 & cdots & -lambda
endpmatrix.$$
Again trying to find a recursion formula for $Q_N$, using gaussians elimination I found that:
$$Q_N(lambda) = (-1)^N-1(lambda+1)^N-1$$ and so:
$$P(lambda)_N+1 = (-lambda)P(lambda)_N +(-1)^NN(lambda+1)^N-1$$
but at this point I couldn't go on.
Looking for a pattern, starting from small N, it seems that the relation I'm looking for is:
$$P_N(lambda) = (lambda -N + 1)(lambda+1)^N-1$$
but I'm not sure and it seems in discord with the previous recurrence relation.
matrices eigenvalues-eigenvectors determinant network
asked Jul 31 at 14:34
Alex
1729
edited Jul 31 at 17:38
asked Jul 31 at 14:34
Alex
1729
asked Jul 31 at 14:34
Alex
1729
asked Jul 31 at 14:34
Alex
1729
1729
Possible duplicate: math.stackexchange.com/questions/1423024/…
– Ethan Bolker
Jul 31 at 14:42
Thanks! But I would like to have a close formula for the characteristic polynomial.
– Alex
Jul 31 at 14:53
Isn't it just $(x-n+1)(x+1)^n-1$ since you know the roots (eigenvalues)?
– Ethan Bolker
Jul 31 at 15:56
Yes! Thanks! but unfortunately I don’t know the “complete graphs theory†cited in the answer, so I still want an explanation using linear algebra if it is possible.
– Alex
Jul 31 at 17:40
1
That is a linear algebra explanation. It works by explicitly finding the eigenvectors. One of them is the vector of all $1$'s.
– Ethan Bolker
Jul 31 at 19:41
 |Â
show 1 more comment
Possible duplicate: math.stackexchange.com/questions/1423024/…
– Ethan Bolker
Jul 31 at 14:42
Thanks! But I would like to have a close formula for the characteristic polynomial.
– Alex
Jul 31 at 14:53
Isn't it just $(x-n+1)(x+1)^n-1$ since you know the roots (eigenvalues)?
– Ethan Bolker
Jul 31 at 15:56
Yes! Thanks! but unfortunately I don’t know the “complete graphs theory†cited in the answer, so I still want an explanation using linear algebra if it is possible.
– Alex
Jul 31 at 17:40
1
That is a linear algebra explanation. It works by explicitly finding the eigenvectors. One of them is the vector of all $1$'s.
– Ethan Bolker
Jul 31 at 19:41
Possible duplicate: math.stackexchange.com/questions/1423024/…
– Ethan Bolker
Jul 31 at 14:42
Possible duplicate: math.stackexchange.com/questions/1423024/…
– Ethan Bolker
Jul 31 at 14:42
Thanks! But I would like to have a close formula for the characteristic polynomial.
– Alex
Jul 31 at 14:53
Thanks! But I would like to have a close formula for the characteristic polynomial.
– Alex
Jul 31 at 14:53
Isn't it just $(x-n+1)(x+1)^n-1$ since you know the roots (eigenvalues)?
– Ethan Bolker
Jul 31 at 15:56
Isn't it just $(x-n+1)(x+1)^n-1$ since you know the roots (eigenvalues)?
– Ethan Bolker
Jul 31 at 15:56
Yes! Thanks! but unfortunately I don’t know the “complete graphs theory†cited in the answer, so I still want an explanation using linear algebra if it is possible.
– Alex
Jul 31 at 17:40
Yes! Thanks! but unfortunately I don’t know the “complete graphs theory†cited in the answer, so I still want an explanation using linear algebra if it is possible.
– Alex
Jul 31 at 17:40
1
1
That is a linear algebra explanation. It works by explicitly finding the eigenvectors. One of them is the vector of all $1$'s.
– Ethan Bolker
Jul 31 at 19:41
That is a linear algebra explanation. It works by explicitly finding the eigenvectors. One of them is the vector of all $1$'s.
– Ethan Bolker
Jul 31 at 19:41
 |Â
show 1 more comment
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Possible duplicate: math.stackexchange.com/questions/1423024/…
– Ethan Bolker
Jul 31 at 14:42
Thanks! But I would like to have a close formula for the characteristic polynomial.
– Alex
Jul 31 at 14:53
Isn't it just $(x-n+1)(x+1)^n-1$ since you know the roots (eigenvalues)?
– Ethan Bolker
Jul 31 at 15:56
Yes! Thanks! but unfortunately I don’t know the “complete graphs theory†cited in the answer, so I still want an explanation using linear algebra if it is possible.
– Alex
Jul 31 at 17:40
1
That is a linear algebra explanation. It works by explicitly finding the eigenvectors. One of them is the vector of all $1$'s.
– Ethan Bolker
Jul 31 at 19:41