Why does the normalized graph Laplacian give negative square root at off-diagonal cells?
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The normalized graph Laplacian fits the relationship $L=D^{-1/2}(D-A)D^{-1/2}=I-D^{-1/2}AD^{-1/2}$, where $I$ is the identity matrix, $D^{-1/2}$ is the diagonal matrix with $D(i,i)=frac{-1}{sqrt{n_{i}}}$, and $A$ is the $n times n$ adjacency matrix. The Laplacian then takes the form:
$L(i,j)=1$ if $i=j$, and $frac{-1}{sqrt{n_{i}n_{j}}}$ if $ineq j$.
Why, for $i neq j$, is it not $frac{-1}{ n_{i}n_{j}}$?
Thanks for any help.
graph-laplacian
edited Dec 10 '18 at 18:28
Zach Langley
9731019
asked Oct 22 '18 at 19:14
user607205user607205
61
$endgroup$
add a comment |
$begingroup$
The normalized graph Laplacian fits the relationship $L=D^{-1/2}(D-A)D^{-1/2}=I-D^{-1/2}AD^{-1/2}$, where $I$ is the identity matrix, $D^{-1/2}$ is the diagonal matrix with $D(i,i)=frac{-1}{sqrt{n_{i}}}$, and $A$ is the $n times n$ adjacency matrix. The Laplacian then takes the form:
$L(i,j)=1$ if $i=j$, and $frac{-1}{sqrt{n_{i}n_{j}}}$ if $ineq j$.
Why, for $i neq j$, is it not $frac{-1}{ n_{i}n_{j}}$?
Thanks for any help.
graph-laplacian
edited Dec 10 '18 at 18:28
Zach Langley
9731019
asked Oct 22 '18 at 19:14
user607205user607205
61
$endgroup$
$begingroup$
I'm not sure that there is a good reason other than "it works nicely". Both are valid descriptions of some linear operator. Some notes that develop theory using this definition of the Laplacian are here.
$endgroup$
– Joppy
Oct 23 '18 at 2:08
$begingroup$
OK, thanks. Can you point towards somewhere that shows how we get this operator from the relationship above? (Chung's book glosses over this.)
$endgroup$
– user607205
Oct 23 '18 at 14:25
add a comment |
$begingroup$
The normalized graph Laplacian fits the relationship $L=D^{-1/2}(D-A)D^{-1/2}=I-D^{-1/2}AD^{-1/2}$, where $I$ is the identity matrix, $D^{-1/2}$ is the diagonal matrix with $D(i,i)=frac{-1}{sqrt{n_{i}}}$, and $A$ is the $n times n$ adjacency matrix. The Laplacian then takes the form:
$L(i,j)=1$ if $i=j$, and $frac{-1}{sqrt{n_{i}n_{j}}}$ if $ineq j$.
Why, for $i neq j$, is it not $frac{-1}{ n_{i}n_{j}}$?
Thanks for any help.
graph-laplacian
edited Dec 10 '18 at 18:28
Zach Langley
9731019
asked Oct 22 '18 at 19:14
user607205user607205
61
$endgroup$
The normalized graph Laplacian fits the relationship $L=D^{-1/2}(D-A)D^{-1/2}=I-D^{-1/2}AD^{-1/2}$, where $I$ is the identity matrix, $D^{-1/2}$ is the diagonal matrix with $D(i,i)=frac{-1}{sqrt{n_{i}}}$, and $A$ is the $n times n$ adjacency matrix. The Laplacian then takes the form:
$L(i,j)=1$ if $i=j$, and $frac{-1}{sqrt{n_{i}n_{j}}}$ if $ineq j$.
Why, for $i neq j$, is it not $frac{-1}{ n_{i}n_{j}}$?
Thanks for any help.
graph-laplacian
graph-laplacian
edited Dec 10 '18 at 18:28
Zach Langley
9731019
asked Oct 22 '18 at 19:14
user607205user607205
61
edited Dec 10 '18 at 18:28
Zach Langley
9731019
asked Oct 22 '18 at 19:14
user607205user607205
61
edited Dec 10 '18 at 18:28
Zach Langley
9731019
edited Dec 10 '18 at 18:28
Zach Langley
9731019
edited Dec 10 '18 at 18:28
Zach Langley
9731019
9731019
asked Oct 22 '18 at 19:14
user607205user607205
61
asked Oct 22 '18 at 19:14
user607205user607205
61
asked Oct 22 '18 at 19:14
user607205user607205
61
61
$begingroup$
I'm not sure that there is a good reason other than "it works nicely". Both are valid descriptions of some linear operator. Some notes that develop theory using this definition of the Laplacian are here.
$endgroup$
– Joppy
Oct 23 '18 at 2:08
$begingroup$
OK, thanks. Can you point towards somewhere that shows how we get this operator from the relationship above? (Chung's book glosses over this.)
$endgroup$
– user607205
Oct 23 '18 at 14:25
add a comment |
$begingroup$
I'm not sure that there is a good reason other than "it works nicely". Both are valid descriptions of some linear operator. Some notes that develop theory using this definition of the Laplacian are here.
$endgroup$
– Joppy
Oct 23 '18 at 2:08
$begingroup$
OK, thanks. Can you point towards somewhere that shows how we get this operator from the relationship above? (Chung's book glosses over this.)
$endgroup$
– user607205
Oct 23 '18 at 14:25
$begingroup$
I'm not sure that there is a good reason other than "it works nicely". Both are valid descriptions of some linear operator. Some notes that develop theory using this definition of the Laplacian are here.
$endgroup$
– Joppy
Oct 23 '18 at 2:08
$begingroup$
I'm not sure that there is a good reason other than "it works nicely". Both are valid descriptions of some linear operator. Some notes that develop theory using this definition of the Laplacian are here.
$endgroup$
– Joppy
Oct 23 '18 at 2:08
$begingroup$
OK, thanks. Can you point towards somewhere that shows how we get this operator from the relationship above? (Chung's book glosses over this.)
$endgroup$
– user607205
Oct 23 '18 at 14:25
$begingroup$
OK, thanks. Can you point towards somewhere that shows how we get this operator from the relationship above? (Chung's book glosses over this.)
$endgroup$
– user607205
Oct 23 '18 at 14:25
add a comment |
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$begingroup$
I'm not sure that there is a good reason other than "it works nicely". Both are valid descriptions of some linear operator. Some notes that develop theory using this definition of the Laplacian are here.
$endgroup$
– Joppy
Oct 23 '18 at 2:08
$begingroup$
OK, thanks. Can you point towards somewhere that shows how we get this operator from the relationship above? (Chung's book glosses over this.)
$endgroup$
– user607205
Oct 23 '18 at 14:25