Diagonally Dominant Matrix Preserved after Gaussian Elimination (with a modification)
$begingroup$
Prove or disprove: If a matrix has the property $0 neq |a_{ii}| geq sum_{substack{j=1 \ j neq i}} |a_{ij}| $ then Gaussian Elimination (without pivoting) will preserve this property.
I assume this to be true since I have seen other theorems state that "Gaussian elimination without pivoting preserves the diagonal dominance of a matrix" without much other qualification. I am not sure how the inequality to 0 would change this. However, most of those used the strong inequality not the weak, so perhaps that would change things as well. Could someone help me with a proof to verify this? Else provide a counterexample?
I have been trying to come up with one to no avail. I would assume showing that the first row iteration of Gaussian preserves the condition would mostly complete the proof so I have been trying to come at it with that in mind.
linear-algebra matrices
edited Jan 26 '14 at 14:54
Git Gud
28.9k1050101
asked Jan 26 '14 at 14:51
James SnyderJames Snyder
1479
$endgroup$
add a comment |
$begingroup$
Prove or disprove: If a matrix has the property $0 neq |a_{ii}| geq sum_{substack{j=1 \ j neq i}} |a_{ij}| $ then Gaussian Elimination (without pivoting) will preserve this property.
I assume this to be true since I have seen other theorems state that "Gaussian elimination without pivoting preserves the diagonal dominance of a matrix" without much other qualification. I am not sure how the inequality to 0 would change this. However, most of those used the strong inequality not the weak, so perhaps that would change things as well. Could someone help me with a proof to verify this? Else provide a counterexample?
I have been trying to come up with one to no avail. I would assume showing that the first row iteration of Gaussian preserves the condition would mostly complete the proof so I have been trying to come at it with that in mind.
linear-algebra matrices
edited Jan 26 '14 at 14:54
Git Gud
28.9k1050101
asked Jan 26 '14 at 14:51
James SnyderJames Snyder
1479
$endgroup$
1
$begingroup$
What is pivoting exactly?
$endgroup$
– Git Gud
Jan 26 '14 at 14:54
1
$begingroup$
After each iteration, switching the rows or columns such that the largest element is on the diagonal for the next set of row elimination in Gaussian.
$endgroup$
– James Snyder
Jan 26 '14 at 15:10
$begingroup$
I am a little confused as to why the answer that was here (while somewhat incorrect) went away?
$endgroup$
– James Snyder
Jan 26 '14 at 15:40
$begingroup$
Answers owners have the option (to some extent) to delete their answers. I chose to delete mine.
$endgroup$
– Git Gud
Jan 26 '14 at 15:41
add a comment |
$begingroup$
Prove or disprove: If a matrix has the property $0 neq |a_{ii}| geq sum_{substack{j=1 \ j neq i}} |a_{ij}| $ then Gaussian Elimination (without pivoting) will preserve this property.
I assume this to be true since I have seen other theorems state that "Gaussian elimination without pivoting preserves the diagonal dominance of a matrix" without much other qualification. I am not sure how the inequality to 0 would change this. However, most of those used the strong inequality not the weak, so perhaps that would change things as well. Could someone help me with a proof to verify this? Else provide a counterexample?
I have been trying to come up with one to no avail. I would assume showing that the first row iteration of Gaussian preserves the condition would mostly complete the proof so I have been trying to come at it with that in mind.
linear-algebra matrices
edited Jan 26 '14 at 14:54
Git Gud
28.9k1050101
asked Jan 26 '14 at 14:51
James SnyderJames Snyder
1479
$endgroup$
Prove or disprove: If a matrix has the property $0 neq |a_{ii}| geq sum_{substack{j=1 \ j neq i}} |a_{ij}| $ then Gaussian Elimination (without pivoting) will preserve this property.
I assume this to be true since I have seen other theorems state that "Gaussian elimination without pivoting preserves the diagonal dominance of a matrix" without much other qualification. I am not sure how the inequality to 0 would change this. However, most of those used the strong inequality not the weak, so perhaps that would change things as well. Could someone help me with a proof to verify this? Else provide a counterexample?
I have been trying to come up with one to no avail. I would assume showing that the first row iteration of Gaussian preserves the condition would mostly complete the proof so I have been trying to come at it with that in mind.
linear-algebra matrices
linear-algebra matrices
edited Jan 26 '14 at 14:54
Git Gud
28.9k1050101
asked Jan 26 '14 at 14:51
James SnyderJames Snyder
1479
edited Jan 26 '14 at 14:54
Git Gud
28.9k1050101
asked Jan 26 '14 at 14:51
James SnyderJames Snyder
1479
edited Jan 26 '14 at 14:54
Git Gud
28.9k1050101
edited Jan 26 '14 at 14:54
Git Gud
28.9k1050101
edited Jan 26 '14 at 14:54
Git Gud
28.9k1050101
28.9k1050101
asked Jan 26 '14 at 14:51
James SnyderJames Snyder
1479
asked Jan 26 '14 at 14:51
James SnyderJames Snyder
1479
asked Jan 26 '14 at 14:51
James SnyderJames Snyder
1479
1479
1
$begingroup$
What is pivoting exactly?
$endgroup$
– Git Gud
Jan 26 '14 at 14:54
1
$begingroup$
After each iteration, switching the rows or columns such that the largest element is on the diagonal for the next set of row elimination in Gaussian.
$endgroup$
– James Snyder
Jan 26 '14 at 15:10
$begingroup$
I am a little confused as to why the answer that was here (while somewhat incorrect) went away?
$endgroup$
– James Snyder
Jan 26 '14 at 15:40
$begingroup$
Answers owners have the option (to some extent) to delete their answers. I chose to delete mine.
$endgroup$
– Git Gud
Jan 26 '14 at 15:41
add a comment |
1
$begingroup$
What is pivoting exactly?
$endgroup$
– Git Gud
Jan 26 '14 at 14:54
1
$begingroup$
After each iteration, switching the rows or columns such that the largest element is on the diagonal for the next set of row elimination in Gaussian.
$endgroup$
– James Snyder
Jan 26 '14 at 15:10
$begingroup$
I am a little confused as to why the answer that was here (while somewhat incorrect) went away?
$endgroup$
– James Snyder
Jan 26 '14 at 15:40
$begingroup$
Answers owners have the option (to some extent) to delete their answers. I chose to delete mine.
$endgroup$
– Git Gud
Jan 26 '14 at 15:41
1
1
$begingroup$
What is pivoting exactly?
$endgroup$
– Git Gud
Jan 26 '14 at 14:54
$begingroup$
What is pivoting exactly?
$endgroup$
– Git Gud
Jan 26 '14 at 14:54
1
1
$begingroup$
After each iteration, switching the rows or columns such that the largest element is on the diagonal for the next set of row elimination in Gaussian.
$endgroup$
– James Snyder
Jan 26 '14 at 15:10
$begingroup$
After each iteration, switching the rows or columns such that the largest element is on the diagonal for the next set of row elimination in Gaussian.
$endgroup$
– James Snyder
Jan 26 '14 at 15:10
$begingroup$
I am a little confused as to why the answer that was here (while somewhat incorrect) went away?
$endgroup$
– James Snyder
Jan 26 '14 at 15:40
$begingroup$
I am a little confused as to why the answer that was here (while somewhat incorrect) went away?
$endgroup$
– James Snyder
Jan 26 '14 at 15:40
$begingroup$
Answers owners have the option (to some extent) to delete their answers. I chose to delete mine.
$endgroup$
– Git Gud
Jan 26 '14 at 15:41
$begingroup$
Answers owners have the option (to some extent) to delete their answers. I chose to delete mine.
$endgroup$
– Git Gud
Jan 26 '14 at 15:41
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I answered this question in another post. Here it is:
We only need to show that after eliminating $a_{2,1}$, diagonal dominance is preserved, i.e.,
$$
left|a_{2,2}-a_{1,2}{a_{2,1}over a_{1,1}}right|gesum_{i=3}^nleft|a_{2,i}-a_{1,i}{a_{2,1}over a_{1,1}}right|,
$$
which is equivalent to
$$
|a_{2,2}a_{1,1}-a_{1,2}a_{2,1}|gesum_{i=3}^n|a_{2,i}a_{1,1}-a_{1,i}a_{2,1}|.
$$
But this is true:
begin{eqnarray*}
sum_{i=3}^n|a_{2,i}a_{1,1}-a_{1,i}a_{2,1}|&le&
|a_{1,1}|sum_{i=3}^n|a_{2,i}|+|a_{2,1}|sum_{i=3}^n|a_{1,i}| \
&le& |a_{1,1}|(|a_{2,2}|-|a_{2,1}|)+|a_{2,1}|(|a_{1,1}|-|a_{1,2}|) \
&=&|a_{1,1}||a_{2,2}|-|a_{2,1}||a_{1,2}|\
&le& |a_{1,1}a_{2,2}-a_{2,1}a_{1,2}|
end{eqnarray*}
answered May 22 '14 at 20:39
user32828user32828
17210
$endgroup$
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%2f652011%2fdiagonally-dominant-matrix-preserved-after-gaussian-elimination-with-a-modifica%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
$begingroup$
I answered this question in another post. Here it is:
We only need to show that after eliminating $a_{2,1}$, diagonal dominance is preserved, i.e.,
$$
left|a_{2,2}-a_{1,2}{a_{2,1}over a_{1,1}}right|gesum_{i=3}^nleft|a_{2,i}-a_{1,i}{a_{2,1}over a_{1,1}}right|,
$$
which is equivalent to
$$
|a_{2,2}a_{1,1}-a_{1,2}a_{2,1}|gesum_{i=3}^n|a_{2,i}a_{1,1}-a_{1,i}a_{2,1}|.
$$
But this is true:
begin{eqnarray*}
sum_{i=3}^n|a_{2,i}a_{1,1}-a_{1,i}a_{2,1}|&le&
|a_{1,1}|sum_{i=3}^n|a_{2,i}|+|a_{2,1}|sum_{i=3}^n|a_{1,i}| \
&le& |a_{1,1}|(|a_{2,2}|-|a_{2,1}|)+|a_{2,1}|(|a_{1,1}|-|a_{1,2}|) \
&=&|a_{1,1}||a_{2,2}|-|a_{2,1}||a_{1,2}|\
&le& |a_{1,1}a_{2,2}-a_{2,1}a_{1,2}|
end{eqnarray*}
answered May 22 '14 at 20:39
user32828user32828
17210
$endgroup$
add a comment |
$begingroup$
I answered this question in another post. Here it is:
We only need to show that after eliminating $a_{2,1}$, diagonal dominance is preserved, i.e.,
$$
left|a_{2,2}-a_{1,2}{a_{2,1}over a_{1,1}}right|gesum_{i=3}^nleft|a_{2,i}-a_{1,i}{a_{2,1}over a_{1,1}}right|,
$$
which is equivalent to
$$
|a_{2,2}a_{1,1}-a_{1,2}a_{2,1}|gesum_{i=3}^n|a_{2,i}a_{1,1}-a_{1,i}a_{2,1}|.
$$
But this is true:
begin{eqnarray*}
sum_{i=3}^n|a_{2,i}a_{1,1}-a_{1,i}a_{2,1}|&le&
|a_{1,1}|sum_{i=3}^n|a_{2,i}|+|a_{2,1}|sum_{i=3}^n|a_{1,i}| \
&le& |a_{1,1}|(|a_{2,2}|-|a_{2,1}|)+|a_{2,1}|(|a_{1,1}|-|a_{1,2}|) \
&=&|a_{1,1}||a_{2,2}|-|a_{2,1}||a_{1,2}|\
&le& |a_{1,1}a_{2,2}-a_{2,1}a_{1,2}|
end{eqnarray*}
answered May 22 '14 at 20:39
user32828user32828
17210
$endgroup$
add a comment |
$begingroup$
I answered this question in another post. Here it is:
We only need to show that after eliminating $a_{2,1}$, diagonal dominance is preserved, i.e.,
$$
left|a_{2,2}-a_{1,2}{a_{2,1}over a_{1,1}}right|gesum_{i=3}^nleft|a_{2,i}-a_{1,i}{a_{2,1}over a_{1,1}}right|,
$$
which is equivalent to
$$
|a_{2,2}a_{1,1}-a_{1,2}a_{2,1}|gesum_{i=3}^n|a_{2,i}a_{1,1}-a_{1,i}a_{2,1}|.
$$
But this is true:
begin{eqnarray*}
sum_{i=3}^n|a_{2,i}a_{1,1}-a_{1,i}a_{2,1}|&le&
|a_{1,1}|sum_{i=3}^n|a_{2,i}|+|a_{2,1}|sum_{i=3}^n|a_{1,i}| \
&le& |a_{1,1}|(|a_{2,2}|-|a_{2,1}|)+|a_{2,1}|(|a_{1,1}|-|a_{1,2}|) \
&=&|a_{1,1}||a_{2,2}|-|a_{2,1}||a_{1,2}|\
&le& |a_{1,1}a_{2,2}-a_{2,1}a_{1,2}|
end{eqnarray*}
answered May 22 '14 at 20:39
user32828user32828
17210
$endgroup$
I answered this question in another post. Here it is:
We only need to show that after eliminating $a_{2,1}$, diagonal dominance is preserved, i.e.,
$$
left|a_{2,2}-a_{1,2}{a_{2,1}over a_{1,1}}right|gesum_{i=3}^nleft|a_{2,i}-a_{1,i}{a_{2,1}over a_{1,1}}right|,
$$
which is equivalent to
$$
|a_{2,2}a_{1,1}-a_{1,2}a_{2,1}|gesum_{i=3}^n|a_{2,i}a_{1,1}-a_{1,i}a_{2,1}|.
$$
But this is true:
begin{eqnarray*}
sum_{i=3}^n|a_{2,i}a_{1,1}-a_{1,i}a_{2,1}|&le&
|a_{1,1}|sum_{i=3}^n|a_{2,i}|+|a_{2,1}|sum_{i=3}^n|a_{1,i}| \
&le& |a_{1,1}|(|a_{2,2}|-|a_{2,1}|)+|a_{2,1}|(|a_{1,1}|-|a_{1,2}|) \
&=&|a_{1,1}||a_{2,2}|-|a_{2,1}||a_{1,2}|\
&le& |a_{1,1}a_{2,2}-a_{2,1}a_{1,2}|
end{eqnarray*}
answered May 22 '14 at 20:39
user32828user32828
17210
answered May 22 '14 at 20:39
user32828user32828
17210
answered May 22 '14 at 20:39
user32828user32828
17210
answered May 22 '14 at 20:39
user32828user32828
17210
17210
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%2f652011%2fdiagonally-dominant-matrix-preserved-after-gaussian-elimination-with-a-modifica%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
1
$begingroup$
What is pivoting exactly?
$endgroup$
– Git Gud
Jan 26 '14 at 14:54
1
$begingroup$
After each iteration, switching the rows or columns such that the largest element is on the diagonal for the next set of row elimination in Gaussian.
$endgroup$
– James Snyder
Jan 26 '14 at 15:10
$begingroup$
I am a little confused as to why the answer that was here (while somewhat incorrect) went away?
$endgroup$
– James Snyder
Jan 26 '14 at 15:40
$begingroup$
Answers owners have the option (to some extent) to delete their answers. I chose to delete mine.
$endgroup$
– Git Gud
Jan 26 '14 at 15:41