Zrjydtj

What does it mean that physics no longer uses mechanical models to describe phenomena?

Delete free apps from library

Sally's older brother

Does silver oxide react with hydrogen sulfide?

Should a wizard buy fine inks every time he want to copy spells into his spellbook?

How can a team of shapeshifters communicate?

Special flights

How do living politicians protect their readily obtainable signatures from misuse?

Google .dev domain strangely redirects to https

A proverb that is used to imply that you have unexpectedly faced a big problem

How to ask rejected full-time candidates to apply to teach individual courses?

Is CEO the "profession" with the most psychopaths?

Moving a wrapfig vertically to encroach partially on a subsection title

A term for a woman complaining about things/begging in a cute/childish way

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

Trying to understand entropy as a novice in thermodynamics

Tannaka duality for semisimple groups

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

Why is std::move not [[nodiscard]] in C++20?

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

Universal covering space of the real projective line?

Was Kant an Intuitionist about mathematical objects?

Resize vertical bars (absolute-value symbols)

If Windows 7 doesn't support WSL, then what is "Subsystem for UNIX-based Applications"?

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

1

$begingroup$

The formula of change of basis $[T]_B' = P_B' <-B[T]_BP_B <- B'$.

I don't understand why you need $P_B <- B'$? 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 $P_B' <-B$ to get $[T]_B'$ rendering $P_B <- B'$ as useless. Can someone explain what I am missing here?

linear-algebra

share|cite|improve this question

asked 4 hours ago

Dr.StoneDr.Stone

626

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @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
  4 hours ago
  
  

  
  
  
  

add a comment | 

1

$begingroup$

The formula of change of basis $[T]_B' = P_B' <-B[T]_BP_B <- B'$.

I don't understand why you need $P_B <- B'$? 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 $P_B' <-B$ to get $[T]_B'$ rendering $P_B <- B'$ as useless. Can someone explain what I am missing here?

linear-algebra

share|cite|improve this question

asked 4 hours ago

Dr.StoneDr.Stone

626

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @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
  4 hours ago
  
  

  
  
  
  

add a comment | 

$begingroup$

The formula of change of basis $[T]_B' = P_B' <-B[T]_BP_B <- B'$.

I don't understand why you need $P_B <- B'$? 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 $P_B' <-B$ to get $[T]_B'$ rendering $P_B <- B'$ as useless. Can someone explain what I am missing here?

linear-algebra

share|cite|improve this question

asked 4 hours ago

Dr.StoneDr.Stone

626

$endgroup$

share|cite|improve this question

asked 4 hours ago

Dr.StoneDr.Stone

626

share|cite|improve this question

asked 4 hours ago

Dr.StoneDr.Stone

626

asked 4 hours ago

Dr.StoneDr.Stone

626

asked 4 hours ago

Dr.StoneDr.Stone

626

2 Answers
2

active

oldest

votes

2

$begingroup$

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 3 hours ago

littleOlittleO

30.6k649111

$endgroup$

add a comment | 

1

$begingroup$

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 4 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

2

$begingroup$

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 3 hours ago

littleOlittleO

30.6k649111

$endgroup$

add a comment | 

2

$begingroup$

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 3 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 3 hours ago

littleOlittleO

30.6k649111

$endgroup$

share|cite|improve this answer

answered 3 hours ago

littleOlittleO

30.6k649111

answered 3 hours ago

littleOlittleO

30.6k649111

answered 3 hours ago

littleOlittleO

30.6k649111

1

$begingroup$

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 4 hours ago

MaxMax

16.6k11144

$endgroup$

add a comment | 

1

$begingroup$

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 4 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 4 hours ago

MaxMax

16.6k11144

$endgroup$

share|cite|improve this answer

answered 4 hours ago

MaxMax

16.6k11144

answered 4 hours ago

MaxMax

16.6k11144

answered 4 hours ago

MaxMax

16.6k11144