Mdtryhht

What does the writing on Poe's helmet say?

Why is it faster to reheat something than it is to cook it?

The Nth Gryphon Number

What does Turing mean by this statement?

Is multiple magic items in one inherently imbalanced?

What are the main differences between the original Stargate SG-1 and the Final Cut edition?

Does the Black Tentacles spell do damage twice at the start of turn to an already restrained creature?

White walkers, cemeteries and wights

How does light 'choose' between wave and particle behaviour?

Moving a wrapfig vertically to encroach partially on a subsection title

Where is the Next Backup Size entry on iOS 12?

In musical terms, what properties are varied by the human voice to produce different words / syllables?

Tannaka duality for semisimple groups

Did Mueller's report provide an evidentiary basis for the claim of Russian govt election interference via social media?

Relating to the President and obstruction, were Mueller's conclusions preordained?

Resize vertical bars (absolute-value symbols)

Why do early math courses focus on the cross sections of a cone and not on other 3D objects?

What is the "studentd" process?

Can two people see the same photon?

Was Kant an Intuitionist about mathematical objects?

Sally's older brother

Is it dangerous to install hacking tools on my private linux machine?

What order were files/directories output in dir?

One-one communication

Why is the change of basis formula counter-intuitive? [See details]

2

$begingroup$

The formula of change of basis $[T]_B' = P_B'leftarrow B[T]_BP_Bleftarrow B'$.

I don't understand why you need $P_Bleftarrow B'$? It seems to me that if you have the transformation expressed in $B$ already with $[T]_B$ you just need to translate to $B'$ by using $P_B'leftarrow B$ to get $[T]_B'$ rendering $P_Bleftarrow B'$ as useless. Can someone explain what I am missing here?

linear-algebra

share|cite|improve this question

edited 1 hour ago

Carmeister

2,8792924

asked 6 hours ago

Dr.StoneDr.Stone

676

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @littleO this is actually what I was looking for. Can you write it as real answer instead of a comment it might help others understand as well so I can approve it.
  $endgroup$
  – Dr.Stone
  5 hours ago
  
  

  
  
  
  

add a comment | 

2

$begingroup$

The formula of change of basis $[T]_B' = P_B'leftarrow B[T]_BP_Bleftarrow B'$.

I don't understand why you need $P_Bleftarrow B'$? It seems to me that if you have the transformation expressed in $B$ already with $[T]_B$ you just need to translate to $B'$ by using $P_B'leftarrow B$ to get $[T]_B'$ rendering $P_Bleftarrow B'$ as useless. Can someone explain what I am missing here?

linear-algebra

share|cite|improve this question

edited 1 hour ago

Carmeister

2,8792924

asked 6 hours ago

Dr.StoneDr.Stone

676

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @littleO this is actually what I was looking for. Can you write it as real answer instead of a comment it might help others understand as well so I can approve it.
  $endgroup$
  – Dr.Stone
  5 hours ago
  
  

  
  
  
  

add a comment | 

$begingroup$

The formula of change of basis $[T]_B' = P_B'leftarrow B[T]_BP_Bleftarrow B'$.

I don't understand why you need $P_Bleftarrow B'$? It seems to me that if you have the transformation expressed in $B$ already with $[T]_B$ you just need to translate to $B'$ by using $P_B'leftarrow B$ to get $[T]_B'$ rendering $P_Bleftarrow B'$ as useless. Can someone explain what I am missing here?

linear-algebra

share|cite|improve this question

edited 1 hour ago

Carmeister

2,8792924

asked 6 hours ago

Dr.StoneDr.Stone

676

$endgroup$

share|cite|improve this question

edited 1 hour ago

Carmeister

2,8792924

asked 6 hours ago

Dr.StoneDr.Stone

676

share|cite|improve this question

edited 1 hour ago

Carmeister

2,8792924

asked 6 hours ago

Dr.StoneDr.Stone

676

edited 1 hour ago

Carmeister

2,8792924

edited 1 hour ago

Carmeister

2,8792924

asked 6 hours ago

Dr.StoneDr.Stone

676

asked 6 hours ago

Dr.StoneDr.Stone

676

2 Answers
2

active

oldest

votes

3

$begingroup$

Imagine what you must do to a vector expressed in $B'$ coordinates in order to apply $T$ to it. First you switch from $B'$ coordinates to $B$ coordinates, then you multiply by the matrix of $T$ (with respect to $B$), then finally you switch back to $B'$ coordinates.

share|cite|improve this answer

answered 5 hours ago

littleOlittleO

30.6k649111

$endgroup$

add a comment | 

2

$begingroup$

Write $B=e_1,...,e_n, B' =e_1',...,e_n'$

If you have the first member of $B'$, $e_1'$, and you want to compute the effect of $T$ on it, then applying $[T]_B$ to $(1,0,...0)$ will be the effect of $T$ on the first member of the basis $B$, so $e_1$, written in the basis $B$ so it has nothing to do with the image of $e_1'$.

So if you only know $[T]_B$ and want to compute $Te_1'$, then you first have to write $e_1'$ in the basis $B$, so you compute $P_B'to B(1,0,...0)$, then compute $[T]_B$ times that, which yields $Te_1'$ but written in the basis $B$, so now you have to write it in the basis $B'$ to get the correct result, that's where $P_Bto B'$ comes from on the left. This gives the formula

share|cite|improve this answer

answered 6 hours ago

MaxMax

16.6k11144

$endgroup$

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
);



);

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%2f3195162%2fwhy-is-the-change-of-basis-formula-counter-intuitive-see-details%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

3

$begingroup$

Imagine what you must do to a vector expressed in $B'$ coordinates in order to apply $T$ to it. First you switch from $B'$ coordinates to $B$ coordinates, then you multiply by the matrix of $T$ (with respect to $B$), then finally you switch back to $B'$ coordinates.

share|cite|improve this answer

answered 5 hours ago

littleOlittleO

30.6k649111

$endgroup$

add a comment | 

3

$begingroup$

Imagine what you must do to a vector expressed in $B'$ coordinates in order to apply $T$ to it. First you switch from $B'$ coordinates to $B$ coordinates, then you multiply by the matrix of $T$ (with respect to $B$), then finally you switch back to $B'$ coordinates.

share|cite|improve this answer

answered 5 hours ago

littleOlittleO

30.6k649111

$endgroup$

add a comment | 

$begingroup$

Imagine what you must do to a vector expressed in $B'$ coordinates in order to apply $T$ to it. First you switch from $B'$ coordinates to $B$ coordinates, then you multiply by the matrix of $T$ (with respect to $B$), then finally you switch back to $B'$ coordinates.

share|cite|improve this answer

answered 5 hours ago

littleOlittleO

30.6k649111

$endgroup$

share|cite|improve this answer

answered 5 hours ago

littleOlittleO

30.6k649111

answered 5 hours ago

littleOlittleO

30.6k649111

answered 5 hours ago

littleOlittleO

30.6k649111

2

$begingroup$

Write $B=e_1,...,e_n, B' =e_1',...,e_n'$

If you have the first member of $B'$, $e_1'$, and you want to compute the effect of $T$ on it, then applying $[T]_B$ to $(1,0,...0)$ will be the effect of $T$ on the first member of the basis $B$, so $e_1$, written in the basis $B$ so it has nothing to do with the image of $e_1'$.

So if you only know $[T]_B$ and want to compute $Te_1'$, then you first have to write $e_1'$ in the basis $B$, so you compute $P_B'to B(1,0,...0)$, then compute $[T]_B$ times that, which yields $Te_1'$ but written in the basis $B$, so now you have to write it in the basis $B'$ to get the correct result, that's where $P_Bto B'$ comes from on the left. This gives the formula

share|cite|improve this answer

answered 6 hours ago

MaxMax

16.6k11144

$endgroup$

add a comment | 

2

$begingroup$

Write $B=e_1,...,e_n, B' =e_1',...,e_n'$

If you have the first member of $B'$, $e_1'$, and you want to compute the effect of $T$ on it, then applying $[T]_B$ to $(1,0,...0)$ will be the effect of $T$ on the first member of the basis $B$, so $e_1$, written in the basis $B$ so it has nothing to do with the image of $e_1'$.

So if you only know $[T]_B$ and want to compute $Te_1'$, then you first have to write $e_1'$ in the basis $B$, so you compute $P_B'to B(1,0,...0)$, then compute $[T]_B$ times that, which yields $Te_1'$ but written in the basis $B$, so now you have to write it in the basis $B'$ to get the correct result, that's where $P_Bto B'$ comes from on the left. This gives the formula

share|cite|improve this answer

answered 6 hours ago

MaxMax

16.6k11144

$endgroup$

add a comment | 

$begingroup$

Write $B=e_1,...,e_n, B' =e_1',...,e_n'$

If you have the first member of $B'$, $e_1'$, and you want to compute the effect of $T$ on it, then applying $[T]_B$ to $(1,0,...0)$ will be the effect of $T$ on the first member of the basis $B$, so $e_1$, written in the basis $B$ so it has nothing to do with the image of $e_1'$.

So if you only know $[T]_B$ and want to compute $Te_1'$, then you first have to write $e_1'$ in the basis $B$, so you compute $P_B'to B(1,0,...0)$, then compute $[T]_B$ times that, which yields $Te_1'$ but written in the basis $B$, so now you have to write it in the basis $B'$ to get the correct result, that's where $P_Bto B'$ comes from on the left. This gives the formula

share|cite|improve this answer

answered 6 hours ago

MaxMax

16.6k11144

$endgroup$

share|cite|improve this answer

answered 6 hours ago

MaxMax

16.6k11144

answered 6 hours ago

MaxMax

16.6k11144

answered 6 hours ago

MaxMax

16.6k11144