Infinite norm of a vector
$begingroup$
While reading the book Numerical Linear Algebra by Trefethen and Bau, I came across the following example.
The authors indicate that $|J|_{infty} = 2$,
however if I recall the definition of $|cdot|_{infty}$ correctly, $|[1, -1]|_{infty} = max(|1|, |{-1}|)$,
which is obviously $1$.
Is this just a typo (the argument still holds if $|J|_{infty} = 1$) or am I misinterpreting something?
linear-algebra norm jacobian condition-number
asked Jan 19 at 14:39
PeiffapPeiffap
487
$endgroup$
add a comment |
$begingroup$
While reading the book Numerical Linear Algebra by Trefethen and Bau, I came across the following example.
The authors indicate that $|J|_{infty} = 2$,
however if I recall the definition of $|cdot|_{infty}$ correctly, $|[1, -1]|_{infty} = max(|1|, |{-1}|)$,
which is obviously $1$.
Is this just a typo (the argument still holds if $|J|_{infty} = 1$) or am I misinterpreting something?
linear-algebra norm jacobian condition-number
asked Jan 19 at 14:39
PeiffapPeiffap
487
$endgroup$
add a comment |
$begingroup$
While reading the book Numerical Linear Algebra by Trefethen and Bau, I came across the following example.
The authors indicate that $|J|_{infty} = 2$,
however if I recall the definition of $|cdot|_{infty}$ correctly, $|[1, -1]|_{infty} = max(|1|, |{-1}|)$,
which is obviously $1$.
Is this just a typo (the argument still holds if $|J|_{infty} = 1$) or am I misinterpreting something?
linear-algebra norm jacobian condition-number
asked Jan 19 at 14:39
PeiffapPeiffap
487
$endgroup$
While reading the book Numerical Linear Algebra by Trefethen and Bau, I came across the following example.
The authors indicate that $|J|_{infty} = 2$,
however if I recall the definition of $|cdot|_{infty}$ correctly, $|[1, -1]|_{infty} = max(|1|, |{-1}|)$,
which is obviously $1$.
Is this just a typo (the argument still holds if $|J|_{infty} = 1$) or am I misinterpreting something?
linear-algebra norm jacobian condition-number
linear-algebra norm jacobian condition-number
asked Jan 19 at 14:39
PeiffapPeiffap
487
asked Jan 19 at 14:39
PeiffapPeiffap
487
asked Jan 19 at 14:39
PeiffapPeiffap
487
asked Jan 19 at 14:39
PeiffapPeiffap
487
asked Jan 19 at 14:39
PeiffapPeiffap
487
487
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It is the subordinate matrix infinity norm defined as:
$$|A|_{infty} =max_{1 leq i leq m}sum_{j=1}^{n}|a_{ij}|,$$
for the matrix $$A=left(
begin{array}{ccc}
a_{11}&cdots&a_{1n}\
vdots & ddots & vdots \
a_{m1}&cdots&a_{mn}
end{array}
right). $$
answered Jan 19 at 14:49
Thomas ShelbyThomas Shelby
3,3571524
$endgroup$
$begingroup$
Why are we using the subordinate matrix infinity norm, instead of the vector infinity norm?
$endgroup$
– Peiffap
Jan 19 at 14:51
1
$begingroup$
Is it because the Jacobian is a $1 times 2$ matrix, and not a vector?
$endgroup$
– Peiffap
Jan 19 at 14:52
$begingroup$
Refer matrix norms induced by vector norms here.
$endgroup$
– Thomas Shelby
Jan 19 at 14:55
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
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%2f3079412%2finfinite-norm-of-a-vector%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
$begingroup$
It is the subordinate matrix infinity norm defined as:
$$|A|_{infty} =max_{1 leq i leq m}sum_{j=1}^{n}|a_{ij}|,$$
for the matrix $$A=left(
begin{array}{ccc}
a_{11}&cdots&a_{1n}\
vdots & ddots & vdots \
a_{m1}&cdots&a_{mn}
end{array}
right). $$
answered Jan 19 at 14:49
Thomas ShelbyThomas Shelby
3,3571524
$endgroup$
$begingroup$
Why are we using the subordinate matrix infinity norm, instead of the vector infinity norm?
$endgroup$
– Peiffap
Jan 19 at 14:51
1
$begingroup$
Is it because the Jacobian is a $1 times 2$ matrix, and not a vector?
$endgroup$
– Peiffap
Jan 19 at 14:52
$begingroup$
Refer matrix norms induced by vector norms here.
$endgroup$
– Thomas Shelby
Jan 19 at 14:55
add a comment |
$begingroup$
It is the subordinate matrix infinity norm defined as:
$$|A|_{infty} =max_{1 leq i leq m}sum_{j=1}^{n}|a_{ij}|,$$
for the matrix $$A=left(
begin{array}{ccc}
a_{11}&cdots&a_{1n}\
vdots & ddots & vdots \
a_{m1}&cdots&a_{mn}
end{array}
right). $$
answered Jan 19 at 14:49
Thomas ShelbyThomas Shelby
3,3571524
$endgroup$
$begingroup$
Why are we using the subordinate matrix infinity norm, instead of the vector infinity norm?
$endgroup$
– Peiffap
Jan 19 at 14:51
1
$begingroup$
Is it because the Jacobian is a $1 times 2$ matrix, and not a vector?
$endgroup$
– Peiffap
Jan 19 at 14:52
$begingroup$
Refer matrix norms induced by vector norms here.
$endgroup$
– Thomas Shelby
Jan 19 at 14:55
add a comment |
$begingroup$
It is the subordinate matrix infinity norm defined as:
$$|A|_{infty} =max_{1 leq i leq m}sum_{j=1}^{n}|a_{ij}|,$$
for the matrix $$A=left(
begin{array}{ccc}
a_{11}&cdots&a_{1n}\
vdots & ddots & vdots \
a_{m1}&cdots&a_{mn}
end{array}
right). $$
answered Jan 19 at 14:49
Thomas ShelbyThomas Shelby
3,3571524
$endgroup$
It is the subordinate matrix infinity norm defined as:
$$|A|_{infty} =max_{1 leq i leq m}sum_{j=1}^{n}|a_{ij}|,$$
for the matrix $$A=left(
begin{array}{ccc}
a_{11}&cdots&a_{1n}\
vdots & ddots & vdots \
a_{m1}&cdots&a_{mn}
end{array}
right). $$
answered Jan 19 at 14:49
Thomas ShelbyThomas Shelby
3,3571524
edited Jan 19 at 15:14
answered Jan 19 at 14:49
Thomas ShelbyThomas Shelby
3,3571524
answered Jan 19 at 14:49
Thomas ShelbyThomas Shelby
3,3571524
answered Jan 19 at 14:49
Thomas ShelbyThomas Shelby
3,3571524
3,3571524
$begingroup$
Why are we using the subordinate matrix infinity norm, instead of the vector infinity norm?
$endgroup$
– Peiffap
Jan 19 at 14:51
1
$begingroup$
Is it because the Jacobian is a $1 times 2$ matrix, and not a vector?
$endgroup$
– Peiffap
Jan 19 at 14:52
$begingroup$
Refer matrix norms induced by vector norms here.
$endgroup$
– Thomas Shelby
Jan 19 at 14:55
add a comment |
$begingroup$
Why are we using the subordinate matrix infinity norm, instead of the vector infinity norm?
$endgroup$
– Peiffap
Jan 19 at 14:51
1
$begingroup$
Is it because the Jacobian is a $1 times 2$ matrix, and not a vector?
$endgroup$
– Peiffap
Jan 19 at 14:52
$begingroup$
Refer matrix norms induced by vector norms here.
$endgroup$
– Thomas Shelby
Jan 19 at 14:55
$begingroup$
Why are we using the subordinate matrix infinity norm, instead of the vector infinity norm?
$endgroup$
– Peiffap
Jan 19 at 14:51
$begingroup$
Why are we using the subordinate matrix infinity norm, instead of the vector infinity norm?
$endgroup$
– Peiffap
Jan 19 at 14:51
1
1
$begingroup$
Is it because the Jacobian is a $1 times 2$ matrix, and not a vector?
$endgroup$
– Peiffap
Jan 19 at 14:52
$begingroup$
Is it because the Jacobian is a $1 times 2$ matrix, and not a vector?
$endgroup$
– Peiffap
Jan 19 at 14:52
$begingroup$
Refer matrix norms induced by vector norms here.
$endgroup$
– Thomas Shelby
Jan 19 at 14:55
$begingroup$
Refer matrix norms induced by vector norms here.
$endgroup$
– Thomas Shelby
Jan 19 at 14:55
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%2f3079412%2finfinite-norm-of-a-vector%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
