Fdtxjr

Is there any way for the UK Prime Minister to make a motion directly dependent on Government confidence?

Is there a kind of relay only consumes power when switching?

How come Sam didn't become Lord of Horn Hill?

Can an alien society believe that their star system is the universe?

Does classifying an integer as a discrete log require it be part of a multiplicative group?

Generate an RGB colour grid

When the Haste spell ends on a creature, do attackers have advantage against that creature?

What is the meaning of the simile “quick as silk”?

Delete nth line from bottom

Is it a good idea to use CNN to classify 1D signal?

How do I make this wiring inside cabinet safer? (Pic)

Is there such thing as an Availability Group failover trigger?

8 Prisoners wearing hats

Where are Serre’s lectures at Collège de France to be found?

Can a party unilaterally change candidates in preparation for a General election?

Do I really need recursive chmod to restrict access to a folder?

How do I find out the mythology and history of my Fortress?

What's the meaning of "fortified infraction restraint"?

Has negative voting ever been officially implemented in elections, or seriously proposed, or even studied?

How to tell that you are a giant?

Why are both D and D# fitting into my E minor key?

Do jazz musicians improvise on the parent scale in addition to the chord-scales?

If a VARCHAR(MAX) column is included in an index, is the entire value always stored in the index page(s)?

Compare a given version number in the form major.minor.build.patch and see if one is less than the other

Spivak Change of Variable

0

$begingroup$

I was a bit confused about the following statement. On page 69, Spivak says "in fact, if T is the linear transformation $Dg(a)$, then $(T^-1circ g)'(a)=I $. I may be applying the chain rule incorrectly, but I do not immediately see why the result should be the identity matrix.

multivariable-calculus linear-transformations frechet-derivative

share|cite|improve this question

edited Mar 27 at 10:26

José Carlos Santos

175k24134243

asked Mar 27 at 3:46

user1848065user1848065

31

$endgroup$

add a comment | 

0

$begingroup$

I was a bit confused about the following statement. On page 69, Spivak says "in fact, if T is the linear transformation $Dg(a)$, then $(T^-1circ g)'(a)=I $. I may be applying the chain rule incorrectly, but I do not immediately see why the result should be the identity matrix.

multivariable-calculus linear-transformations frechet-derivative

share|cite|improve this question

edited Mar 27 at 10:26

José Carlos Santos

175k24134243

asked Mar 27 at 3:46

user1848065user1848065

31

$endgroup$

add a comment | 

$begingroup$

I was a bit confused about the following statement. On page 69, Spivak says "in fact, if T is the linear transformation $Dg(a)$, then $(T^-1circ g)'(a)=I $. I may be applying the chain rule incorrectly, but I do not immediately see why the result should be the identity matrix.

multivariable-calculus linear-transformations frechet-derivative

share|cite|improve this question

edited Mar 27 at 10:26

José Carlos Santos

175k24134243

asked Mar 27 at 3:46

user1848065user1848065

31

$endgroup$

share|cite|improve this question

edited Mar 27 at 10:26

José Carlos Santos

175k24134243

asked Mar 27 at 3:46

user1848065user1848065

31

share|cite|improve this question

edited Mar 27 at 10:26

José Carlos Santos

175k24134243

asked Mar 27 at 3:46

user1848065user1848065

31

edited Mar 27 at 10:26

José Carlos Santos

175k24134243

edited Mar 27 at 10:26

José Carlos Santos

175k24134243

asked Mar 27 at 3:46

user1848065user1848065

31

asked Mar 27 at 3:46

user1848065user1848065

31

1 Answer
1

active

oldest

votes

1

$begingroup$

Because $(T^-1circ g)'(a)=(T^-1)'bigl(g(a)bigr)circ g'(a)=T^-1circ Dg(a)=operatornameId$. Note that the equality $(T^-1)'bigl(g(a)bigr)=T^-1$ comes from the fact that $T$ is linear.

share|cite|improve this answer

answered Mar 27 at 3:50

José Carlos SantosJosé Carlos Santos

175k24134243

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you so much for making that clear!
  $endgroup$
  – user1848065
  Mar 27 at 12:21
  

  
  
  
  

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
);



);

draft saved

draft discarded

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

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3164074%2fspivak-change-of-variable%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

1

$begingroup$

Because $(T^-1circ g)'(a)=(T^-1)'bigl(g(a)bigr)circ g'(a)=T^-1circ Dg(a)=operatornameId$. Note that the equality $(T^-1)'bigl(g(a)bigr)=T^-1$ comes from the fact that $T$ is linear.

share|cite|improve this answer

answered Mar 27 at 3:50

José Carlos SantosJosé Carlos Santos

175k24134243

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you so much for making that clear!
  $endgroup$
  – user1848065
  Mar 27 at 12:21
  

  
  
  
  

add a comment | 

1

$begingroup$

Because $(T^-1circ g)'(a)=(T^-1)'bigl(g(a)bigr)circ g'(a)=T^-1circ Dg(a)=operatornameId$. Note that the equality $(T^-1)'bigl(g(a)bigr)=T^-1$ comes from the fact that $T$ is linear.

share|cite|improve this answer

answered Mar 27 at 3:50

José Carlos SantosJosé Carlos Santos

175k24134243

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you so much for making that clear!
  $endgroup$
  – user1848065
  Mar 27 at 12:21
  

  
  
  
  

add a comment | 

$begingroup$

Because $(T^-1circ g)'(a)=(T^-1)'bigl(g(a)bigr)circ g'(a)=T^-1circ Dg(a)=operatornameId$. Note that the equality $(T^-1)'bigl(g(a)bigr)=T^-1$ comes from the fact that $T$ is linear.

share|cite|improve this answer

answered Mar 27 at 3:50

José Carlos SantosJosé Carlos Santos

175k24134243

$endgroup$

share|cite|improve this answer

answered Mar 27 at 3:50

José Carlos SantosJosé Carlos Santos

175k24134243

answered Mar 27 at 3:50

José Carlos SantosJosé Carlos Santos

175k24134243

answered Mar 27 at 3:50

José Carlos SantosJosé Carlos Santos

175k24134243