Why is the Jacobi Matrix not a matrix?
In a textbook im reading about manifolds they write:
Let $W subset mathbb{ R } ^n $ be an open set and $ F : W rightarrow mathbb{ R } ^ m $ a $C ^ k $ function such that for every point $ p in M := F ^ { -1 }( 0 $ the derivative $DF( p ): mathbb{ R } ^ n rightarrow mathbb{ R } ^ m$ has rang $m $. Then $ M $ is a $C ^ k $-manifold with dimension $d = n - m$.
Shouldnt the derivate be a function from the type $ DF(p) : mathbb{R}^n rightarrow mathbb{R } ^ { m times n }$ ?
The first time I saw this I gussed it was a mistake in the textbook but it seems to appear multiple times afterwards. Do I get something wrong there?
calculus derivatives manifolds
asked Nov 28 '18 at 12:18
Arjihad
383111
|
show 2 more comments
In a textbook im reading about manifolds they write:
Let $W subset mathbb{ R } ^n $ be an open set and $ F : W rightarrow mathbb{ R } ^ m $ a $C ^ k $ function such that for every point $ p in M := F ^ { -1 }( 0 $ the derivative $DF( p ): mathbb{ R } ^ n rightarrow mathbb{ R } ^ m$ has rang $m $. Then $ M $ is a $C ^ k $-manifold with dimension $d = n - m$.
Shouldnt the derivate be a function from the type $ DF(p) : mathbb{R}^n rightarrow mathbb{R } ^ { m times n }$ ?
The first time I saw this I gussed it was a mistake in the textbook but it seems to appear multiple times afterwards. Do I get something wrong there?
calculus derivatives manifolds
asked Nov 28 '18 at 12:18
Arjihad
383111
1
No, a point $p$ has already been plugged in.
– Randall
Nov 28 '18 at 12:22
1
You are wrong. The derivative is a linear map $mathbb{R}^n to mathbb{R}^m$ and its matrix (for the usual bases) is the jacobian matrix.
– Math_QED
Nov 28 '18 at 12:25
1
The matrix is $mtimes n$, yes. $A$ say. We then take a vector $xin mathbb{R}^n$, and what we get is $Axin mathbb{R}^m$. So the derivative is a linear transformation from $mathbb{R}^n$ to $mathbb{R}^m$.
– Jakobian
Nov 28 '18 at 14:25
1
To repeat, yes, the derivative is a matrix--you're right--but only AFTER a point has been plugged in. Compare $DF(p)$ vs $DF$.
– Randall
Nov 28 '18 at 18:03
1
Oh okay. The difference between $DF(p)$ and $DF$ clearly makes a difference. Thanks for your help.
– Arjihad
Nov 28 '18 at 18:08
|
show 2 more comments
In a textbook im reading about manifolds they write:
Let $W subset mathbb{ R } ^n $ be an open set and $ F : W rightarrow mathbb{ R } ^ m $ a $C ^ k $ function such that for every point $ p in M := F ^ { -1 }( 0 $ the derivative $DF( p ): mathbb{ R } ^ n rightarrow mathbb{ R } ^ m$ has rang $m $. Then $ M $ is a $C ^ k $-manifold with dimension $d = n - m$.
Shouldnt the derivate be a function from the type $ DF(p) : mathbb{R}^n rightarrow mathbb{R } ^ { m times n }$ ?
The first time I saw this I gussed it was a mistake in the textbook but it seems to appear multiple times afterwards. Do I get something wrong there?
calculus derivatives manifolds
asked Nov 28 '18 at 12:18
Arjihad
383111
In a textbook im reading about manifolds they write:
Let $W subset mathbb{ R } ^n $ be an open set and $ F : W rightarrow mathbb{ R } ^ m $ a $C ^ k $ function such that for every point $ p in M := F ^ { -1 }( 0 $ the derivative $DF( p ): mathbb{ R } ^ n rightarrow mathbb{ R } ^ m$ has rang $m $. Then $ M $ is a $C ^ k $-manifold with dimension $d = n - m$.
Shouldnt the derivate be a function from the type $ DF(p) : mathbb{R}^n rightarrow mathbb{R } ^ { m times n }$ ?
The first time I saw this I gussed it was a mistake in the textbook but it seems to appear multiple times afterwards. Do I get something wrong there?
calculus derivatives manifolds
calculus derivatives manifolds
asked Nov 28 '18 at 12:18
Arjihad
383111
asked Nov 28 '18 at 12:18
Arjihad
383111
asked Nov 28 '18 at 12:18
Arjihad
383111
asked Nov 28 '18 at 12:18
Arjihad
383111
asked Nov 28 '18 at 12:18
Arjihad
383111
383111
1
No, a point $p$ has already been plugged in.
– Randall
Nov 28 '18 at 12:22
1
You are wrong. The derivative is a linear map $mathbb{R}^n to mathbb{R}^m$ and its matrix (for the usual bases) is the jacobian matrix.
– Math_QED
Nov 28 '18 at 12:25
1
The matrix is $mtimes n$, yes. $A$ say. We then take a vector $xin mathbb{R}^n$, and what we get is $Axin mathbb{R}^m$. So the derivative is a linear transformation from $mathbb{R}^n$ to $mathbb{R}^m$.
– Jakobian
Nov 28 '18 at 14:25
1
To repeat, yes, the derivative is a matrix--you're right--but only AFTER a point has been plugged in. Compare $DF(p)$ vs $DF$.
– Randall
Nov 28 '18 at 18:03
1
Oh okay. The difference between $DF(p)$ and $DF$ clearly makes a difference. Thanks for your help.
– Arjihad
Nov 28 '18 at 18:08
|
show 2 more comments
1
No, a point $p$ has already been plugged in.
– Randall
Nov 28 '18 at 12:22
1
You are wrong. The derivative is a linear map $mathbb{R}^n to mathbb{R}^m$ and its matrix (for the usual bases) is the jacobian matrix.
– Math_QED
Nov 28 '18 at 12:25
1
The matrix is $mtimes n$, yes. $A$ say. We then take a vector $xin mathbb{R}^n$, and what we get is $Axin mathbb{R}^m$. So the derivative is a linear transformation from $mathbb{R}^n$ to $mathbb{R}^m$.
– Jakobian
Nov 28 '18 at 14:25
1
To repeat, yes, the derivative is a matrix--you're right--but only AFTER a point has been plugged in. Compare $DF(p)$ vs $DF$.
– Randall
Nov 28 '18 at 18:03
1
Oh okay. The difference between $DF(p)$ and $DF$ clearly makes a difference. Thanks for your help.
– Arjihad
Nov 28 '18 at 18:08
1
1
No, a point $p$ has already been plugged in.
– Randall
Nov 28 '18 at 12:22
No, a point $p$ has already been plugged in.
– Randall
Nov 28 '18 at 12:22
1
1
You are wrong. The derivative is a linear map $mathbb{R}^n to mathbb{R}^m$ and its matrix (for the usual bases) is the jacobian matrix.
– Math_QED
Nov 28 '18 at 12:25
You are wrong. The derivative is a linear map $mathbb{R}^n to mathbb{R}^m$ and its matrix (for the usual bases) is the jacobian matrix.
– Math_QED
Nov 28 '18 at 12:25
1
1
The matrix is $mtimes n$, yes. $A$ say. We then take a vector $xin mathbb{R}^n$, and what we get is $Axin mathbb{R}^m$. So the derivative is a linear transformation from $mathbb{R}^n$ to $mathbb{R}^m$.
– Jakobian
Nov 28 '18 at 14:25
The matrix is $mtimes n$, yes. $A$ say. We then take a vector $xin mathbb{R}^n$, and what we get is $Axin mathbb{R}^m$. So the derivative is a linear transformation from $mathbb{R}^n$ to $mathbb{R}^m$.
– Jakobian
Nov 28 '18 at 14:25
1
1
To repeat, yes, the derivative is a matrix--you're right--but only AFTER a point has been plugged in. Compare $DF(p)$ vs $DF$.
– Randall
Nov 28 '18 at 18:03
To repeat, yes, the derivative is a matrix--you're right--but only AFTER a point has been plugged in. Compare $DF(p)$ vs $DF$.
– Randall
Nov 28 '18 at 18:03
1
1
Oh okay. The difference between $DF(p)$ and $DF$ clearly makes a difference. Thanks for your help.
– Arjihad
Nov 28 '18 at 18:08
Oh okay. The difference between $DF(p)$ and $DF$ clearly makes a difference. Thanks for your help.
– Arjihad
Nov 28 '18 at 18:08
|
show 2 more comments
0
active
oldest
votes
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%2f3017080%2fwhy-is-the-jacobi-matrix-not-a-matrix%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3017080%2fwhy-is-the-jacobi-matrix-not-a-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
1
No, a point $p$ has already been plugged in.
– Randall
Nov 28 '18 at 12:22
1
You are wrong. The derivative is a linear map $mathbb{R}^n to mathbb{R}^m$ and its matrix (for the usual bases) is the jacobian matrix.
– Math_QED
Nov 28 '18 at 12:25
1
The matrix is $mtimes n$, yes. $A$ say. We then take a vector $xin mathbb{R}^n$, and what we get is $Axin mathbb{R}^m$. So the derivative is a linear transformation from $mathbb{R}^n$ to $mathbb{R}^m$.
– Jakobian
Nov 28 '18 at 14:25
1
To repeat, yes, the derivative is a matrix--you're right--but only AFTER a point has been plugged in. Compare $DF(p)$ vs $DF$.
– Randall
Nov 28 '18 at 18:03
1
Oh okay. The difference between $DF(p)$ and $DF$ clearly makes a difference. Thanks for your help.
– Arjihad
Nov 28 '18 at 18:08