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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 5 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 5 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 5 hours ago

Dr.StoneDr.Stone

676

$endgroup$

share|cite|improve this question

edited 1 hour ago

Carmeister

2,8792924

asked 5 hours ago

Dr.StoneDr.Stone

676

share|cite|improve this question

edited 1 hour ago

Carmeister

2,8792924

asked 5 hours ago

Dr.StoneDr.Stone

676

edited 1 hour ago

Carmeister

2,8792924

edited 1 hour ago

Carmeister

2,8792924

asked 5 hours ago

Dr.StoneDr.Stone

676

asked 5 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 5 hours ago

MaxMax

16.6k11144

$endgroup$

add a comment | 

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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 5 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 5 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 5 hours ago

MaxMax

16.6k11144

$endgroup$

share|cite|improve this answer

answered 5 hours ago

MaxMax

16.6k11144

answered 5 hours ago

MaxMax

16.6k11144

answered 5 hours ago

MaxMax

16.6k11144