Mixing
Rank of Laplacian matrix
$begingroup$
Let $ L $ be a Laplacian matrix of a balanced and strongly connected digraph having $n $ nodes.
$ L[r]$ is a submatrix of $L$ which is obtained by deleting $rth$ row and $rth$ column of Laplacian matrix $L$. It is observed for any $r$ the rank of $ L[r]$ is full$(n-1)$. How can I prove this fact?
Thanks in advance for your suggestions.
matrices graph-theory graphing-functions matrix-rank algebraic-graph-theory
asked May 18 '18 at 10:52
sinasina
62
$endgroup$
add a comment |
$begingroup$
Let $ L $ be a Laplacian matrix of a balanced and strongly connected digraph having $n $ nodes.
$ L[r]$ is a submatrix of $L$ which is obtained by deleting $rth$ row and $rth$ column of Laplacian matrix $L$. It is observed for any $r$ the rank of $ L[r]$ is full$(n-1)$. How can I prove this fact?
Thanks in advance for your suggestions.
matrices graph-theory graphing-functions matrix-rank algebraic-graph-theory
asked May 18 '18 at 10:52
sinasina
62
$endgroup$
$begingroup$
I am not sure that it is true. Take a strongly connected digraph which can be broken in to two strongly connected graphs by removing a single node $r$. The Laplacian of the new graph must have a zero of multiplicity $2$ because it is not connected.
$endgroup$
– Daniel Beale
May 18 '18 at 13:48
add a comment |
$begingroup$
Let $ L $ be a Laplacian matrix of a balanced and strongly connected digraph having $n $ nodes.
$ L[r]$ is a submatrix of $L$ which is obtained by deleting $rth$ row and $rth$ column of Laplacian matrix $L$. It is observed for any $r$ the rank of $ L[r]$ is full$(n-1)$. How can I prove this fact?
Thanks in advance for your suggestions.
matrices graph-theory graphing-functions matrix-rank algebraic-graph-theory
asked May 18 '18 at 10:52
sinasina
62
$endgroup$
Let $ L $ be a Laplacian matrix of a balanced and strongly connected digraph having $n $ nodes.
$ L[r]$ is a submatrix of $L$ which is obtained by deleting $rth$ row and $rth$ column of Laplacian matrix $L$. It is observed for any $r$ the rank of $ L[r]$ is full$(n-1)$. How can I prove this fact?
Thanks in advance for your suggestions.
matrices graph-theory graphing-functions matrix-rank algebraic-graph-theory
matrices graph-theory graphing-functions matrix-rank algebraic-graph-theory
asked May 18 '18 at 10:52
sinasina
62
asked May 18 '18 at 10:52
sinasina
62
asked May 18 '18 at 10:52
sinasina
62
asked May 18 '18 at 10:52
sinasina
62
asked May 18 '18 at 10:52
sinasina
62
62
$begingroup$
I am not sure that it is true. Take a strongly connected digraph which can be broken in to two strongly connected graphs by removing a single node $r$. The Laplacian of the new graph must have a zero of multiplicity $2$ because it is not connected.
$endgroup$
– Daniel Beale
May 18 '18 at 13:48
add a comment |
$begingroup$
I am not sure that it is true. Take a strongly connected digraph which can be broken in to two strongly connected graphs by removing a single node $r$. The Laplacian of the new graph must have a zero of multiplicity $2$ because it is not connected.
$endgroup$
– Daniel Beale
May 18 '18 at 13:48
$begingroup$
I am not sure that it is true. Take a strongly connected digraph which can be broken in to two strongly connected graphs by removing a single node $r$. The Laplacian of the new graph must have a zero of multiplicity $2$ because it is not connected.
$endgroup$
– Daniel Beale
May 18 '18 at 13:48
$begingroup$
I am not sure that it is true. Take a strongly connected digraph which can be broken in to two strongly connected graphs by removing a single node $r$. The Laplacian of the new graph must have a zero of multiplicity $2$ because it is not connected.
$endgroup$
– Daniel Beale
May 18 '18 at 13:48
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This is in fact true. What's more, the celebrated matrix-tree-theorem (or Kirchhoff theorem) states that the determinant of the reduced matrix $L[r]$ is precisely the number of spanning trees in the underlying graph (so surprisingly it's also the same number for any $1leq rleq n$). The number of spanning trees in a connected graph is a positive number, so $det(L[r])neq 0$. And hence, $L[r]$ is regular.
answered Feb 2 at 17:10
IPPBerlinIPPBerlin
311
$endgroup$
add a comment |
Your Answer
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2786151%2frank-of-laplacian-matrix%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This is in fact true. What's more, the celebrated matrix-tree-theorem (or Kirchhoff theorem) states that the determinant of the reduced matrix $L[r]$ is precisely the number of spanning trees in the underlying graph (so surprisingly it's also the same number for any $1leq rleq n$). The number of spanning trees in a connected graph is a positive number, so $det(L[r])neq 0$. And hence, $L[r]$ is regular.
answered Feb 2 at 17:10
IPPBerlinIPPBerlin
311
$endgroup$
add a comment |
$begingroup$
This is in fact true. What's more, the celebrated matrix-tree-theorem (or Kirchhoff theorem) states that the determinant of the reduced matrix $L[r]$ is precisely the number of spanning trees in the underlying graph (so surprisingly it's also the same number for any $1leq rleq n$). The number of spanning trees in a connected graph is a positive number, so $det(L[r])neq 0$. And hence, $L[r]$ is regular.
answered Feb 2 at 17:10
IPPBerlinIPPBerlin
311
$endgroup$
add a comment |
$begingroup$
This is in fact true. What's more, the celebrated matrix-tree-theorem (or Kirchhoff theorem) states that the determinant of the reduced matrix $L[r]$ is precisely the number of spanning trees in the underlying graph (so surprisingly it's also the same number for any $1leq rleq n$). The number of spanning trees in a connected graph is a positive number, so $det(L[r])neq 0$. And hence, $L[r]$ is regular.
answered Feb 2 at 17:10
IPPBerlinIPPBerlin
311
$endgroup$
This is in fact true. What's more, the celebrated matrix-tree-theorem (or Kirchhoff theorem) states that the determinant of the reduced matrix $L[r]$ is precisely the number of spanning trees in the underlying graph (so surprisingly it's also the same number for any $1leq rleq n$). The number of spanning trees in a connected graph is a positive number, so $det(L[r])neq 0$. And hence, $L[r]$ is regular.
answered Feb 2 at 17:10
IPPBerlinIPPBerlin
311
answered Feb 2 at 17:10
IPPBerlinIPPBerlin
311
answered Feb 2 at 17:10
IPPBerlinIPPBerlin
311
answered Feb 2 at 17:10
IPPBerlinIPPBerlin
311
311
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2786151%2frank-of-laplacian-matrix%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
I am not sure that it is true. Take a strongly connected digraph which can be broken in to two strongly connected graphs by removing a single node $r$. The Laplacian of the new graph must have a zero of multiplicity $2$ because it is not connected.
$endgroup$
– Daniel Beale
May 18 '18 at 13:48