Expression between projection onto vector in base B and base B'

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I would like to prove the following equality between one projection in arbitrary basis $B=(e_i)$ and the same projection in another basis $B'=(e'_i)$ :

The projection projects a vector $vecw$ onto $vecv$ direction. $w$ and $v$ are expressed in $B$ basis whereas $w'$ and $v'$ are the same vectors but expressed in $B'$ basis.

Then, My goal is to get :

$$textproj_vecv(vecw) = P,textproj_vecv'(vecw')quadquad(1)$$

with the matrix $P$ which is the transfer matrix between $B'$ and $B$ : so I have $vecv=P,vecv'$, $vecv^*=P^-1,vecv'^*$, $vecw=P,vecw'$.

To start, I take the expression of $textproj_vecv(vecw)$ :

$$textproj_vecv(vecw)=vv^*w=P,vecv',P^-1,vecv'^*, P, vecw'quadquad(2)$$

and from this, I would like to have :

$$textproj_vecv'(vecw')=vecv'vecv'^*vecw'$$

but I don't how to make disappear the matrices $P$ and $P^-1$ in equation $(2)$ since the product of matrices is not commutative.

If this was commutative, I could write from $(2)$ :

$$textproj_vecv(vecw)=vv^*w=P,vecv',P^-1,P,vecv'^*, vecw'=P,vecv',vecv'^*, vecw'=P,textproj_vecv'(vecw')quadquad(3)$$

Unfortunately, this swapping is not allowed.

How to circumvent this issue in order to get equation $(1)$ ?

EDIT 1:

Maybe I should write : $vecv^*=vecv'^*,P^-1$ instead of $vecv^*=P^-1,vecv'^*$, so it comes :

$$textproj_vecv(vecw)=vv^*w=P,vecv',vecv'^*,P^-1,P,vecw'=P,vecv',vecv'^*,vecw'=P,textproj_vecv'(vecw')quadquad(4)$$

By this way, I could find the simple expression between coordinates "$X$" of a vector expressed in "$B$" basis and its coordinates "$X'$" expressed in "$B'$" basis , like this :

$$X=PX'$$

Is equation(4) correct ?

EDIT 2: I have doubts about the following expression (see $(1))$ :

$$bigg(textproj_vecv(vecw)bigg)_B = P,bigg(textproj_vecv'(vecw')bigg)_B'$$

with $P=textMat_BB'$ the transfer matrix.

Why couldn't we remove the $P$ matrix and write simply :

$$bigg(textproj_vecv(vecw)bigg)_B = bigg(textproj_vecv'(vecw')bigg)_B'quadquad(5)$$

It seems the projected vector into one basis $B$ is the same than projected vector but expressed into another basis $B'$.

Then, if I follow $(5)$, I get :

$$vv^*w=w^ivece_i=v'v'^*w'=w'^jvece'_j=w'^jP_ijvece_i$$

So one gets :

$w^i=w'^jP_ij$

This is the relation between contravariants components.

What expression is right :

$$bigg(textproj_vecv(vecw)bigg)_B = bigg(textproj_vecv'(vecw')bigg)_B'quadquad(6)$$

OR

$$bigg(textproj_vecv(vecw)bigg)_B = P, bigg(textproj_vecv'(vecw')bigg)_B'quadquad(7)$$

???

EDIT 3:

Following the notations of @amd , my goal is not to write $$[pi_v]_mathcal B^mathcal B = P [pi_v]_mathcal B'^mathcal B'$$

but rather : $$[pi_v w]_mathcal B = P,[pi_v w]_mathcal B'$$.

Indeed, we have $$<[v]_B,[w]_B> = <[v]_B',[w]_B'>$$ and $$[v]_B=P,[v]_B'$$,

so finally we can write : $$[pi_v w]_mathcal B=P,[pi_v w]_mathcal B'$$, is it right ?

With my notations, this would take the form :

$$bigg(textproj_(vecv)_B(vecw)_Bbigg)_B = P,bigg(textproj_(vecv)_B'(vecw)_B')bigg)_B'$$

wouldn't it ?

linear-algebra matrices projection-matrices

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edited Aug 3 at 22:39

asked Jul 30 at 7:54

youpilat13

6911

• 
  
  
  

  
  

  
  
  
  
  
  

  
  How did you arrive at $v^*=P^-1v'^*$?
  – amd
  Jul 31 at 0:25
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  -@amd . I think $v^*=P^−1v'^*$ is not the right expression. Considering $v^*$ and $v'^*$ 2 linear forms expressed respectively into $B=(e^i*)$ and $B′=(e′^i*)$, with $P$ the transfer matrix, we could write maybe : $v^*=v′^* P^-1$ but I do confusions between these 2 expresions, I am not really sure, that's why I would like to solve this issue. All that I know is covariant coordinates are transformed like their basis vectors, isn't it ?
  – youpilat13
  Jul 31 at 13:21
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $(Pv')^*ne P^-1v'^*$, nor is it $v'^*P^-1$. That aside, why would you want to eliminate the various $P$’s in the expression? They are essential for expressing the inner product relative to $B'$. Remember that you must use the same inner product throughout. Just because it can be expressed as $v^*w$, there’s no reason to believe that it’s also expressible as the (matrix) product $v'^*w'$.
  – amd
  Aug 2 at 8:53
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  -@amd What relation can I get to do the link between $textproj_vecv'(vecw')$ and $textproj_vecv(vecw)$ : these 2 projections produced a vector which is actually the same, but expressed into basis $(v'_i)$ for the first one and into basis $(v_i)$ for the second : that's why I would like to associate them by the transfer Matrix $P$. Is it clearer ?
  – youpilat13
  Aug 2 at 13:03
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You misunderstand my question. You seem to think that some of the $P$’s internal to the resulting expression must somehow cancel, but they don’t in general. The change of basis also affects the formula for the inner product that’s inherent in $operatornameproj$. You should expect to see a Gram matrix somewhere in there.
  – amd
  Aug 2 at 20:28
  

  
  
  
  

 | 
show 2 more comments

up vote
0
down vote

favorite

I would like to prove the following equality between one projection in arbitrary basis $B=(e_i)$ and the same projection in another basis $B'=(e'_i)$ :

The projection projects a vector $vecw$ onto $vecv$ direction. $w$ and $v$ are expressed in $B$ basis whereas $w'$ and $v'$ are the same vectors but expressed in $B'$ basis.

Then, My goal is to get :

$$textproj_vecv(vecw) = P,textproj_vecv'(vecw')quadquad(1)$$

with the matrix $P$ which is the transfer matrix between $B'$ and $B$ : so I have $vecv=P,vecv'$, $vecv^*=P^-1,vecv'^*$, $vecw=P,vecw'$.

To start, I take the expression of $textproj_vecv(vecw)$ :

$$textproj_vecv(vecw)=vv^*w=P,vecv',P^-1,vecv'^*, P, vecw'quadquad(2)$$

and from this, I would like to have :

$$textproj_vecv'(vecw')=vecv'vecv'^*vecw'$$

but I don't how to make disappear the matrices $P$ and $P^-1$ in equation $(2)$ since the product of matrices is not commutative.

If this was commutative, I could write from $(2)$ :

$$textproj_vecv(vecw)=vv^*w=P,vecv',P^-1,P,vecv'^*, vecw'=P,vecv',vecv'^*, vecw'=P,textproj_vecv'(vecw')quadquad(3)$$

Unfortunately, this swapping is not allowed.

How to circumvent this issue in order to get equation $(1)$ ?

EDIT 1:

Maybe I should write : $vecv^*=vecv'^*,P^-1$ instead of $vecv^*=P^-1,vecv'^*$, so it comes :

$$textproj_vecv(vecw)=vv^*w=P,vecv',vecv'^*,P^-1,P,vecw'=P,vecv',vecv'^*,vecw'=P,textproj_vecv'(vecw')quadquad(4)$$

By this way, I could find the simple expression between coordinates "$X$" of a vector expressed in "$B$" basis and its coordinates "$X'$" expressed in "$B'$" basis , like this :

$$X=PX'$$

Is equation(4) correct ?

EDIT 2: I have doubts about the following expression (see $(1))$ :

$$bigg(textproj_vecv(vecw)bigg)_B = P,bigg(textproj_vecv'(vecw')bigg)_B'$$

with $P=textMat_BB'$ the transfer matrix.

Why couldn't we remove the $P$ matrix and write simply :

$$bigg(textproj_vecv(vecw)bigg)_B = bigg(textproj_vecv'(vecw')bigg)_B'quadquad(5)$$

It seems the projected vector into one basis $B$ is the same than projected vector but expressed into another basis $B'$.

Then, if I follow $(5)$, I get :

$$vv^*w=w^ivece_i=v'v'^*w'=w'^jvece'_j=w'^jP_ijvece_i$$

So one gets :

$w^i=w'^jP_ij$

This is the relation between contravariants components.

What expression is right :

$$bigg(textproj_vecv(vecw)bigg)_B = bigg(textproj_vecv'(vecw')bigg)_B'quadquad(6)$$

OR

$$bigg(textproj_vecv(vecw)bigg)_B = P, bigg(textproj_vecv'(vecw')bigg)_B'quadquad(7)$$

???

EDIT 3:

Following the notations of @amd , my goal is not to write $$[pi_v]_mathcal B^mathcal B = P [pi_v]_mathcal B'^mathcal B'$$

but rather : $$[pi_v w]_mathcal B = P,[pi_v w]_mathcal B'$$.

Indeed, we have $$<[v]_B,[w]_B> = <[v]_B',[w]_B'>$$ and $$[v]_B=P,[v]_B'$$,

so finally we can write : $$[pi_v w]_mathcal B=P,[pi_v w]_mathcal B'$$, is it right ?

With my notations, this would take the form :

$$bigg(textproj_(vecv)_B(vecw)_Bbigg)_B = P,bigg(textproj_(vecv)_B'(vecw)_B')bigg)_B'$$

wouldn't it ?

linear-algebra matrices projection-matrices

share|cite|improve this question

edited Aug 3 at 22:39

asked Jul 30 at 7:54

youpilat13

6911

• 
  
  
  

  
  

  
  
  
  
  
  

  
  How did you arrive at $v^*=P^-1v'^*$?
  – amd
  Jul 31 at 0:25
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  -@amd . I think $v^*=P^−1v'^*$ is not the right expression. Considering $v^*$ and $v'^*$ 2 linear forms expressed respectively into $B=(e^i*)$ and $B′=(e′^i*)$, with $P$ the transfer matrix, we could write maybe : $v^*=v′^* P^-1$ but I do confusions between these 2 expresions, I am not really sure, that's why I would like to solve this issue. All that I know is covariant coordinates are transformed like their basis vectors, isn't it ?
  – youpilat13
  Jul 31 at 13:21
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $(Pv')^*ne P^-1v'^*$, nor is it $v'^*P^-1$. That aside, why would you want to eliminate the various $P$’s in the expression? They are essential for expressing the inner product relative to $B'$. Remember that you must use the same inner product throughout. Just because it can be expressed as $v^*w$, there’s no reason to believe that it’s also expressible as the (matrix) product $v'^*w'$.
  – amd
  Aug 2 at 8:53
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  -@amd What relation can I get to do the link between $textproj_vecv'(vecw')$ and $textproj_vecv(vecw)$ : these 2 projections produced a vector which is actually the same, but expressed into basis $(v'_i)$ for the first one and into basis $(v_i)$ for the second : that's why I would like to associate them by the transfer Matrix $P$. Is it clearer ?
  – youpilat13
  Aug 2 at 13:03
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You misunderstand my question. You seem to think that some of the $P$’s internal to the resulting expression must somehow cancel, but they don’t in general. The change of basis also affects the formula for the inner product that’s inherent in $operatornameproj$. You should expect to see a Gram matrix somewhere in there.
  – amd
  Aug 2 at 20:28
  

  
  
  
  

 | 
show 2 more comments

up vote
0
down vote

favorite

up vote
0
down vote

favorite

I would like to prove the following equality between one projection in arbitrary basis $B=(e_i)$ and the same projection in another basis $B'=(e'_i)$ :

The projection projects a vector $vecw$ onto $vecv$ direction. $w$ and $v$ are expressed in $B$ basis whereas $w'$ and $v'$ are the same vectors but expressed in $B'$ basis.

Then, My goal is to get :

$$textproj_vecv(vecw) = P,textproj_vecv'(vecw')quadquad(1)$$

with the matrix $P$ which is the transfer matrix between $B'$ and $B$ : so I have $vecv=P,vecv'$, $vecv^*=P^-1,vecv'^*$, $vecw=P,vecw'$.

To start, I take the expression of $textproj_vecv(vecw)$ :

$$textproj_vecv(vecw)=vv^*w=P,vecv',P^-1,vecv'^*, P, vecw'quadquad(2)$$

and from this, I would like to have :

$$textproj_vecv'(vecw')=vecv'vecv'^*vecw'$$

but I don't how to make disappear the matrices $P$ and $P^-1$ in equation $(2)$ since the product of matrices is not commutative.

If this was commutative, I could write from $(2)$ :

$$textproj_vecv(vecw)=vv^*w=P,vecv',P^-1,P,vecv'^*, vecw'=P,vecv',vecv'^*, vecw'=P,textproj_vecv'(vecw')quadquad(3)$$

Unfortunately, this swapping is not allowed.

How to circumvent this issue in order to get equation $(1)$ ?

EDIT 1:

Maybe I should write : $vecv^*=vecv'^*,P^-1$ instead of $vecv^*=P^-1,vecv'^*$, so it comes :

$$textproj_vecv(vecw)=vv^*w=P,vecv',vecv'^*,P^-1,P,vecw'=P,vecv',vecv'^*,vecw'=P,textproj_vecv'(vecw')quadquad(4)$$

By this way, I could find the simple expression between coordinates "$X$" of a vector expressed in "$B$" basis and its coordinates "$X'$" expressed in "$B'$" basis , like this :

$$X=PX'$$

Is equation(4) correct ?

EDIT 2: I have doubts about the following expression (see $(1))$ :

$$bigg(textproj_vecv(vecw)bigg)_B = P,bigg(textproj_vecv'(vecw')bigg)_B'$$

with $P=textMat_BB'$ the transfer matrix.

Why couldn't we remove the $P$ matrix and write simply :

$$bigg(textproj_vecv(vecw)bigg)_B = bigg(textproj_vecv'(vecw')bigg)_B'quadquad(5)$$

It seems the projected vector into one basis $B$ is the same than projected vector but expressed into another basis $B'$.

Then, if I follow $(5)$, I get :

$$vv^*w=w^ivece_i=v'v'^*w'=w'^jvece'_j=w'^jP_ijvece_i$$

So one gets :

$w^i=w'^jP_ij$

This is the relation between contravariants components.

What expression is right :

$$bigg(textproj_vecv(vecw)bigg)_B = bigg(textproj_vecv'(vecw')bigg)_B'quadquad(6)$$

OR

$$bigg(textproj_vecv(vecw)bigg)_B = P, bigg(textproj_vecv'(vecw')bigg)_B'quadquad(7)$$

???

EDIT 3:

Following the notations of @amd , my goal is not to write $$[pi_v]_mathcal B^mathcal B = P [pi_v]_mathcal B'^mathcal B'$$

but rather : $$[pi_v w]_mathcal B = P,[pi_v w]_mathcal B'$$.

Indeed, we have $$<[v]_B,[w]_B> = <[v]_B',[w]_B'>$$ and $$[v]_B=P,[v]_B'$$,

so finally we can write : $$[pi_v w]_mathcal B=P,[pi_v w]_mathcal B'$$, is it right ?

With my notations, this would take the form :

$$bigg(textproj_(vecv)_B(vecw)_Bbigg)_B = P,bigg(textproj_(vecv)_B'(vecw)_B')bigg)_B'$$

wouldn't it ?

linear-algebra matrices projection-matrices

share|cite|improve this question

edited Aug 3 at 22:39

asked Jul 30 at 7:54

youpilat13

6911

I would like to prove the following equality between one projection in arbitrary basis $B=(e_i)$ and the same projection in another basis $B'=(e'_i)$ :

The projection projects a vector $vecw$ onto $vecv$ direction. $w$ and $v$ are expressed in $B$ basis whereas $w'$ and $v'$ are the same vectors but expressed in $B'$ basis.

Then, My goal is to get :

$$textproj_vecv(vecw) = P,textproj_vecv'(vecw')quadquad(1)$$

with the matrix $P$ which is the transfer matrix between $B'$ and $B$ : so I have $vecv=P,vecv'$, $vecv^*=P^-1,vecv'^*$, $vecw=P,vecw'$.

To start, I take the expression of $textproj_vecv(vecw)$ :

$$textproj_vecv(vecw)=vv^*w=P,vecv',P^-1,vecv'^*, P, vecw'quadquad(2)$$

and from this, I would like to have :

$$textproj_vecv'(vecw')=vecv'vecv'^*vecw'$$

but I don't how to make disappear the matrices $P$ and $P^-1$ in equation $(2)$ since the product of matrices is not commutative.

If this was commutative, I could write from $(2)$ :

$$textproj_vecv(vecw)=vv^*w=P,vecv',P^-1,P,vecv'^*, vecw'=P,vecv',vecv'^*, vecw'=P,textproj_vecv'(vecw')quadquad(3)$$

Unfortunately, this swapping is not allowed.

How to circumvent this issue in order to get equation $(1)$ ?

EDIT 1:

Maybe I should write : $vecv^*=vecv'^*,P^-1$ instead of $vecv^*=P^-1,vecv'^*$, so it comes :

$$textproj_vecv(vecw)=vv^*w=P,vecv',vecv'^*,P^-1,P,vecw'=P,vecv',vecv'^*,vecw'=P,textproj_vecv'(vecw')quadquad(4)$$

By this way, I could find the simple expression between coordinates "$X$" of a vector expressed in "$B$" basis and its coordinates "$X'$" expressed in "$B'$" basis , like this :

$$X=PX'$$

Is equation(4) correct ?

EDIT 2: I have doubts about the following expression (see $(1))$ :

$$bigg(textproj_vecv(vecw)bigg)_B = P,bigg(textproj_vecv'(vecw')bigg)_B'$$

with $P=textMat_BB'$ the transfer matrix.

Why couldn't we remove the $P$ matrix and write simply :

$$bigg(textproj_vecv(vecw)bigg)_B = bigg(textproj_vecv'(vecw')bigg)_B'quadquad(5)$$

It seems the projected vector into one basis $B$ is the same than projected vector but expressed into another basis $B'$.

Then, if I follow $(5)$, I get :

$$vv^*w=w^ivece_i=v'v'^*w'=w'^jvece'_j=w'^jP_ijvece_i$$

So one gets :

$w^i=w'^jP_ij$

This is the relation between contravariants components.

What expression is right :

$$bigg(textproj_vecv(vecw)bigg)_B = bigg(textproj_vecv'(vecw')bigg)_B'quadquad(6)$$

OR

$$bigg(textproj_vecv(vecw)bigg)_B = P, bigg(textproj_vecv'(vecw')bigg)_B'quadquad(7)$$

???

EDIT 3:

Following the notations of @amd , my goal is not to write $$[pi_v]_mathcal B^mathcal B = P [pi_v]_mathcal B'^mathcal B'$$

but rather : $$[pi_v w]_mathcal B = P,[pi_v w]_mathcal B'$$.

Indeed, we have $$<[v]_B,[w]_B> = <[v]_B',[w]_B'>$$ and $$[v]_B=P,[v]_B'$$,

so finally we can write : $$[pi_v w]_mathcal B=P,[pi_v w]_mathcal B'$$, is it right ?

With my notations, this would take the form :

$$bigg(textproj_(vecv)_B(vecw)_Bbigg)_B = P,bigg(textproj_(vecv)_B'(vecw)_B')bigg)_B'$$

wouldn't it ?

linear-algebra matrices projection-matrices

share|cite|improve this question

edited Aug 3 at 22:39

asked Jul 30 at 7:54

youpilat13

6911

share|cite|improve this question

share|cite|improve this question

edited Aug 3 at 22:39

edited Aug 3 at 22:39

edited Aug 3 at 22:39

asked Jul 30 at 7:54

youpilat13

6911

asked Jul 30 at 7:54

youpilat13

6911

asked Jul 30 at 7:54

youpilat13

6911

6911

• 
  
  
  

  
  

  
  
  
  
  
  

  
  How did you arrive at $v^*=P^-1v'^*$?
  – amd
  Jul 31 at 0:25
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  -@amd . I think $v^*=P^−1v'^*$ is not the right expression. Considering $v^*$ and $v'^*$ 2 linear forms expressed respectively into $B=(e^i*)$ and $B′=(e′^i*)$, with $P$ the transfer matrix, we could write maybe : $v^*=v′^* P^-1$ but I do confusions between these 2 expresions, I am not really sure, that's why I would like to solve this issue. All that I know is covariant coordinates are transformed like their basis vectors, isn't it ?
  – youpilat13
  Jul 31 at 13:21
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $(Pv')^*ne P^-1v'^*$, nor is it $v'^*P^-1$. That aside, why would you want to eliminate the various $P$’s in the expression? They are essential for expressing the inner product relative to $B'$. Remember that you must use the same inner product throughout. Just because it can be expressed as $v^*w$, there’s no reason to believe that it’s also expressible as the (matrix) product $v'^*w'$.
  – amd
  Aug 2 at 8:53
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  -@amd What relation can I get to do the link between $textproj_vecv'(vecw')$ and $textproj_vecv(vecw)$ : these 2 projections produced a vector which is actually the same, but expressed into basis $(v'_i)$ for the first one and into basis $(v_i)$ for the second : that's why I would like to associate them by the transfer Matrix $P$. Is it clearer ?
  – youpilat13
  Aug 2 at 13:03
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You misunderstand my question. You seem to think that some of the $P$’s internal to the resulting expression must somehow cancel, but they don’t in general. The change of basis also affects the formula for the inner product that’s inherent in $operatornameproj$. You should expect to see a Gram matrix somewhere in there.
  – amd
  Aug 2 at 20:28
  

  
  
  
  

 | 
show 2 more comments

• 
  
  
  

  
  

  
  
  
  
  
  

  
  How did you arrive at $v^*=P^-1v'^*$?
  – amd
  Jul 31 at 0:25
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  -@amd . I think $v^*=P^−1v'^*$ is not the right expression. Considering $v^*$ and $v'^*$ 2 linear forms expressed respectively into $B=(e^i*)$ and $B′=(e′^i*)$, with $P$ the transfer matrix, we could write maybe : $v^*=v′^* P^-1$ but I do confusions between these 2 expresions, I am not really sure, that's why I would like to solve this issue. All that I know is covariant coordinates are transformed like their basis vectors, isn't it ?
  – youpilat13
  Jul 31 at 13:21
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $(Pv')^*ne P^-1v'^*$, nor is it $v'^*P^-1$. That aside, why would you want to eliminate the various $P$’s in the expression? They are essential for expressing the inner product relative to $B'$. Remember that you must use the same inner product throughout. Just because it can be expressed as $v^*w$, there’s no reason to believe that it’s also expressible as the (matrix) product $v'^*w'$.
  – amd
  Aug 2 at 8:53
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  -@amd What relation can I get to do the link between $textproj_vecv'(vecw')$ and $textproj_vecv(vecw)$ : these 2 projections produced a vector which is actually the same, but expressed into basis $(v'_i)$ for the first one and into basis $(v_i)$ for the second : that's why I would like to associate them by the transfer Matrix $P$. Is it clearer ?
  – youpilat13
  Aug 2 at 13:03
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You misunderstand my question. You seem to think that some of the $P$’s internal to the resulting expression must somehow cancel, but they don’t in general. The change of basis also affects the formula for the inner product that’s inherent in $operatornameproj$. You should expect to see a Gram matrix somewhere in there.
  – amd
  Aug 2 at 20:28
  

  
  
  
  

How did you arrive at $v^*=P^-1v'^*$?
– amd
Jul 31 at 0:25

How did you arrive at $v^*=P^-1v'^*$?
– amd
Jul 31 at 0:25

-@amd . I think $v^*=P^−1v'^*$ is not the right expression. Considering $v^*$ and $v'^*$ 2 linear forms expressed respectively into $B=(e^i*)$ and $B′=(e′^i*)$, with $P$ the transfer matrix, we could write maybe : $v^*=v′^* P^-1$ but I do confusions between these 2 expresions, I am not really sure, that's why I would like to solve this issue. All that I know is covariant coordinates are transformed like their basis vectors, isn't it ?
– youpilat13
Jul 31 at 13:21

-@amd . I think $v^*=P^−1v'^*$ is not the right expression. Considering $v^*$ and $v'^*$ 2 linear forms expressed respectively into $B=(e^i*)$ and $B′=(e′^i*)$, with $P$ the transfer matrix, we could write maybe : $v^*=v′^* P^-1$ but I do confusions between these 2 expresions, I am not really sure, that's why I would like to solve this issue. All that I know is covariant coordinates are transformed like their basis vectors, isn't it ?
– youpilat13
Jul 31 at 13:21

$(Pv')^*ne P^-1v'^*$, nor is it $v'^*P^-1$. That aside, why would you want to eliminate the various $P$’s in the expression? They are essential for expressing the inner product relative to $B'$. Remember that you must use the same inner product throughout. Just because it can be expressed as $v^*w$, there’s no reason to believe that it’s also expressible as the (matrix) product $v'^*w'$.
– amd
Aug 2 at 8:53

$(Pv')^*ne P^-1v'^*$, nor is it $v'^*P^-1$. That aside, why would you want to eliminate the various $P$’s in the expression? They are essential for expressing the inner product relative to $B'$. Remember that you must use the same inner product throughout. Just because it can be expressed as $v^*w$, there’s no reason to believe that it’s also expressible as the (matrix) product $v'^*w'$.
– amd
Aug 2 at 8:53

-@amd What relation can I get to do the link between $textproj_vecv'(vecw')$ and $textproj_vecv(vecw)$ : these 2 projections produced a vector which is actually the same, but expressed into basis $(v'_i)$ for the first one and into basis $(v_i)$ for the second : that's why I would like to associate them by the transfer Matrix $P$. Is it clearer ?
– youpilat13
Aug 2 at 13:03

-@amd What relation can I get to do the link between $textproj_vecv'(vecw')$ and $textproj_vecv(vecw)$ : these 2 projections produced a vector which is actually the same, but expressed into basis $(v'_i)$ for the first one and into basis $(v_i)$ for the second : that's why I would like to associate them by the transfer Matrix $P$. Is it clearer ?
– youpilat13
Aug 2 at 13:03

You misunderstand my question. You seem to think that some of the $P$’s internal to the resulting expression must somehow cancel, but they don’t in general. The change of basis also affects the formula for the inner product that’s inherent in $operatornameproj$. You should expect to see a Gram matrix somewhere in there.
– amd
Aug 2 at 20:28

You misunderstand my question. You seem to think that some of the $P$’s internal to the resulting expression must somehow cancel, but they don’t in general. The change of basis also affects the formula for the inner product that’s inherent in $operatornameproj$. You should expect to see a Gram matrix somewhere in there.
– amd
Aug 2 at 20:28

 | 
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accepted

There’s not really anything tricky here. In particular, you don’t need to try so hard to eliminate the $P$’s in the resulting expression because they actually belong there. They’re essential to the inner product that’s inherent in the projection.

I’ll use somewhat different notation from that in the question because it’s important to distinguish the vectors themselves—elements of the inner product space $V$—from their coordinate tuples—elements of various copies of $mathbb K^n$. The symbols $v$ and $w$ stand for vectors, i.e., $v,win V$. Their coordinate tuples relative to the ordered basis $mathcal B$ are denoted $[v]_mathcal B$ and $[w]_mathcal B$, respectively. These are elements of $mathbb K^n$. Similarly, if $L:Vto V$ is an automorphism of $V$, then its matrix relative to the “input” basis $mathcal B'$ and “output” basis $mathcal B$ is denoted by $[L]_mathcal B^mathcal B'$. This notation allows you to “cancel” a superscript against the subscript of the next term to its right in a multiplication. From this point of view, a change of basis doesn’t change the vector that the two coordinate tuples both represent, so a transfer matrix is a representation of the identity map $operatornameid$.

The map $pi_v:Vto V$ that orthogonally projects onto some fixed $vin V$ is given by the formula $$pi_v: w mapsto langle w, vrangle over langle v, vrangle v.tag1$$ Here, $langlecdot,cdotrangle : Vtimes Vtomathbb K$ stands for the inner product used to define orthogonality. Now, suppose that $mathcal B$ is a basis in which $langle w, vrangle = [v]_mathcal B^* [w]_mathcal B$. Then, if $P=[operatornameid]_mathcal B^mathcal B'$, we have $[v]_mathcal B=P[v]_mathcal B'$ and $[w]_mathcal B=P[w]_mathcal B'$, and so $$langle w, vrangle = left(P[v]_mathcal B'right)^* left(P[w]_mathcal B'right) = [v]_mathcal B'^* P^*P [w]_mathcal B'.tag2$$ This is nothing more than an application of the change-of-basis formula for quadratic forms: if $Q$ is the matrix of a quadratic form relative to the basis $mathcal B$ and $P = [operatornameid]_mathcal B^mathcal B'$ the transfer matrix from another basis $mathcal B'$, then the matrix of the quadratic form relative to $mathcal B'$ is $P^*QP$. This is one of the ways in which you erred: $langle w,vrangle = [v]_mathcal B^* [w]_mathcal B$ for all $v$ and $w$ iff $mathcal B$ is orthonormal. Otherwise, some positive-definite matrix other than the identity must come into play.

To reduce clutter, assume now that $|v|=1$. The derivation is similar, but messier, in the general case, for which there’ll be a $1/langle v,vrangle$ term floating around. I’m also assuming that $mathcal B$ is orthonormal. For full generality, we’d have to start with $langle w,vrangle = [v]_mathcal B^* Q [w]_mathcal B$ for some positive-definite matrix $Q$. From (1) we have $$[pi_v w]_mathcal B^mathcal B = [v]_mathcal B^*[w]_mathcal B[v]_mathcal B = [v]_mathcal B [v]_mathcal B^*[w]_mathcal B,$$ so $[pi_v]_mathcal B^mathcal B = [v]_mathcal B [v]_mathcal B^*$. Similarly, $$[pi_v]_mathcal B'^mathcal B' = [v]_mathcal B'^* P^*P [w]_mathcal B' [v]_mathcal B' = [v]_mathcal B' [v]_mathcal B'^* P^*P [w]_mathcal B'$$ and $[pi_v]_mathcal B'^mathcal B' = [v]_mathcal B' [v]_mathcal B'^* P^*P$. The two matrices are related by the usual change-of-basis formula: $$[pi_v]_mathcal B^mathcal B = [operatornameid]_mathcal B^mathcal B'[pi_v]_mathcal B'^mathcal B' [operatornameid]_mathcal B'^mathcal B = P[pi_v]_mathcal B'^mathcal B' P^-1.$$ On the other hand, $$[pi_v]_mathcal B^mathcal B = [v]_mathcal B [v]_mathcal B^* = left(P [v]_mathcal B'right) left(P [v]_mathcal B'right)^* = P [v]_mathcal B' [v]_mathcal B'^* P^* = P [pi_v]_mathcal B'^mathcal B' P^-1.$$ If $mathcal B'$ is also orthonormal, then $P^*P = I$ and this term will vanish, as you were trying to make it do, but you still won’t have $[pi_v]_mathcal B^mathcal B = P [pi_v]_mathcal B'^mathcal B'$ because of what’s effectively a type mismatch: the left-hand side expects its inputs expressed relative to $mathcal B$, but the right-hand side expects inputs relative to $mathcal B'$.

share|cite|improve this answer

edited Aug 3 at 22:38

answered Aug 3 at 8:48

amd

25.7k2943

• 
  
  
  

  
  

  
  
  
  
  
  

  
  -@amd Thanks for your detailed answer. Following your notations, my goal is not to write $[pi_v]_mathcal B^mathcal B = P [pi_v]_mathcal B'^mathcal B'$ but rather : $$[pi_v w]_mathcal B^mathcal B = P,[pi_v w]_mathcal B'^mathcal B'$$. Indeed, we have $<[v]_B,[w]_B> = <[v]_B',[w]_B'>$ and $[v]_B=P,[v]_B'$, so finally we can write : $$[pi_v w]_mathcal B^mathcal B=P,[pi_v w]_mathcal B'^mathcal B'$$, is it right ?
  – youpilat13
  Aug 3 at 21:38
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  -@amd . I think there is a little error : $$langle w, vrangle = left(P[v]_mathcal B'right)^* left(P[w]_mathcal B'right) = [v]_mathcal B'^* P^*P [w]_mathcal B'$$
  – youpilat13
  Aug 3 at 21:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @youpilat13 Not really. Let $V$ be the inner product space over the field $mathbb K$. Then $pi_vwin V$ is a vector; $[pi_vw]_mathcal Binmathbb K^n$ is the coordinate tuple of $pi_vw$ relative to the basis $mathcal B$, and the expression $[pi_vw]_mathcal B^mathcal B$ doesn’t make sense because, unlike the matrix of a linear transformation, there’s only one basis to consider. The identity $[pi_vw]_mathcal B=P[pi_vw]_mathcal B'$ is a trivial matter of applying the transfer matrix $P$ to the $mathcal B'$-coordinates of $pi_vw$.
  – amd
  Aug 3 at 22:10
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  -@amd If $B$ and $B'$ are orthonormal, then we could write :$$ langle w, vrangle = left(P[v]_mathcal B'right)^*,Q, left(P[w]_mathcal B'right) = [v]_mathcal B'^* P^*,Q,P [w]_mathcal B'$$ and so with $Q=I_d$ and $P^*=P^-1$, we get : $$langle w,vrangle = [v]_mathcal B'^* [w]_mathcal B'$$ : is this correct ? What does it imply on $P$ transfer matrix and $Q$ matrix if one has $B$ and $B'$ orthonormal ? i.e just $Q=I_d$ ?
  – youpilat13
  Aug 4 at 14:07
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  -@amd Please could you take a look at my last comment above, i.e to confirm it ?
  – youpilat13
  Aug 5 at 9:28
  

  
  
  
  

 | 
show 2 more comments

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There’s not really anything tricky here. In particular, you don’t need to try so hard to eliminate the $P$’s in the resulting expression because they actually belong there. They’re essential to the inner product that’s inherent in the projection.

I’ll use somewhat different notation from that in the question because it’s important to distinguish the vectors themselves—elements of the inner product space $V$—from their coordinate tuples—elements of various copies of $mathbb K^n$. The symbols $v$ and $w$ stand for vectors, i.e., $v,win V$. Their coordinate tuples relative to the ordered basis $mathcal B$ are denoted $[v]_mathcal B$ and $[w]_mathcal B$, respectively. These are elements of $mathbb K^n$. Similarly, if $L:Vto V$ is an automorphism of $V$, then its matrix relative to the “input” basis $mathcal B'$ and “output” basis $mathcal B$ is denoted by $[L]_mathcal B^mathcal B'$. This notation allows you to “cancel” a superscript against the subscript of the next term to its right in a multiplication. From this point of view, a change of basis doesn’t change the vector that the two coordinate tuples both represent, so a transfer matrix is a representation of the identity map $operatornameid$.

The map $pi_v:Vto V$ that orthogonally projects onto some fixed $vin V$ is given by the formula $$pi_v: w mapsto langle w, vrangle over langle v, vrangle v.tag1$$ Here, $langlecdot,cdotrangle : Vtimes Vtomathbb K$ stands for the inner product used to define orthogonality. Now, suppose that $mathcal B$ is a basis in which $langle w, vrangle = [v]_mathcal B^* [w]_mathcal B$. Then, if $P=[operatornameid]_mathcal B^mathcal B'$, we have $[v]_mathcal B=P[v]_mathcal B'$ and $[w]_mathcal B=P[w]_mathcal B'$, and so $$langle w, vrangle = left(P[v]_mathcal B'right)^* left(P[w]_mathcal B'right) = [v]_mathcal B'^* P^*P [w]_mathcal B'.tag2$$ This is nothing more than an application of the change-of-basis formula for quadratic forms: if $Q$ is the matrix of a quadratic form relative to the basis $mathcal B$ and $P = [operatornameid]_mathcal B^mathcal B'$ the transfer matrix from another basis $mathcal B'$, then the matrix of the quadratic form relative to $mathcal B'$ is $P^*QP$. This is one of the ways in which you erred: $langle w,vrangle = [v]_mathcal B^* [w]_mathcal B$ for all $v$ and $w$ iff $mathcal B$ is orthonormal. Otherwise, some positive-definite matrix other than the identity must come into play.

To reduce clutter, assume now that $|v|=1$. The derivation is similar, but messier, in the general case, for which there’ll be a $1/langle v,vrangle$ term floating around. I’m also assuming that $mathcal B$ is orthonormal. For full generality, we’d have to start with $langle w,vrangle = [v]_mathcal B^* Q [w]_mathcal B$ for some positive-definite matrix $Q$. From (1) we have $$[pi_v w]_mathcal B^mathcal B = [v]_mathcal B^*[w]_mathcal B[v]_mathcal B = [v]_mathcal B [v]_mathcal B^*[w]_mathcal B,$$ so $[pi_v]_mathcal B^mathcal B = [v]_mathcal B [v]_mathcal B^*$. Similarly, $$[pi_v]_mathcal B'^mathcal B' = [v]_mathcal B'^* P^*P [w]_mathcal B' [v]_mathcal B' = [v]_mathcal B' [v]_mathcal B'^* P^*P [w]_mathcal B'$$ and $[pi_v]_mathcal B'^mathcal B' = [v]_mathcal B' [v]_mathcal B'^* P^*P$. The two matrices are related by the usual change-of-basis formula: $$[pi_v]_mathcal B^mathcal B = [operatornameid]_mathcal B^mathcal B'[pi_v]_mathcal B'^mathcal B' [operatornameid]_mathcal B'^mathcal B = P[pi_v]_mathcal B'^mathcal B' P^-1.$$ On the other hand, $$[pi_v]_mathcal B^mathcal B = [v]_mathcal B [v]_mathcal B^* = left(P [v]_mathcal B'right) left(P [v]_mathcal B'right)^* = P [v]_mathcal B' [v]_mathcal B'^* P^* = P [pi_v]_mathcal B'^mathcal B' P^-1.$$ If $mathcal B'$ is also orthonormal, then $P^*P = I$ and this term will vanish, as you were trying to make it do, but you still won’t have $[pi_v]_mathcal B^mathcal B = P [pi_v]_mathcal B'^mathcal B'$ because of what’s effectively a type mismatch: the left-hand side expects its inputs expressed relative to $mathcal B$, but the right-hand side expects inputs relative to $mathcal B'$.

share|cite|improve this answer

edited Aug 3 at 22:38

answered Aug 3 at 8:48

amd

25.7k2943

• 
  
  
  

  
  

  
  
  
  
  
  

  
  -@amd Thanks for your detailed answer. Following your notations, my goal is not to write $[pi_v]_mathcal B^mathcal B = P [pi_v]_mathcal B'^mathcal B'$ but rather : $$[pi_v w]_mathcal B^mathcal B = P,[pi_v w]_mathcal B'^mathcal B'$$. Indeed, we have $<[v]_B,[w]_B> = <[v]_B',[w]_B'>$ and $[v]_B=P,[v]_B'$, so finally we can write : $$[pi_v w]_mathcal B^mathcal B=P,[pi_v w]_mathcal B'^mathcal B'$$, is it right ?
  – youpilat13
  Aug 3 at 21:38
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  -@amd . I think there is a little error : $$langle w, vrangle = left(P[v]_mathcal B'right)^* left(P[w]_mathcal B'right) = [v]_mathcal B'^* P^*P [w]_mathcal B'$$
  – youpilat13
  Aug 3 at 21:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @youpilat13 Not really. Let $V$ be the inner product space over the field $mathbb K$. Then $pi_vwin V$ is a vector; $[pi_vw]_mathcal Binmathbb K^n$ is the coordinate tuple of $pi_vw$ relative to the basis $mathcal B$, and the expression $[pi_vw]_mathcal B^mathcal B$ doesn’t make sense because, unlike the matrix of a linear transformation, there’s only one basis to consider. The identity $[pi_vw]_mathcal B=P[pi_vw]_mathcal B'$ is a trivial matter of applying the transfer matrix $P$ to the $mathcal B'$-coordinates of $pi_vw$.
  – amd
  Aug 3 at 22:10
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  -@amd If $B$ and $B'$ are orthonormal, then we could write :$$ langle w, vrangle = left(P[v]_mathcal B'right)^*,Q, left(P[w]_mathcal B'right) = [v]_mathcal B'^* P^*,Q,P [w]_mathcal B'$$ and so with $Q=I_d$ and $P^*=P^-1$, we get : $$langle w,vrangle = [v]_mathcal B'^* [w]_mathcal B'$$ : is this correct ? What does it imply on $P$ transfer matrix and $Q$ matrix if one has $B$ and $B'$ orthonormal ? i.e just $Q=I_d$ ?
  – youpilat13
  Aug 4 at 14:07
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  -@amd Please could you take a look at my last comment above, i.e to confirm it ?
  – youpilat13
  Aug 5 at 9:28
  

  
  
  
  

 | 
show 2 more comments

up vote
0
down vote



accepted

There’s not really anything tricky here. In particular, you don’t need to try so hard to eliminate the $P$’s in the resulting expression because they actually belong there. They’re essential to the inner product that’s inherent in the projection.

I’ll use somewhat different notation from that in the question because it’s important to distinguish the vectors themselves—elements of the inner product space $V$—from their coordinate tuples—elements of various copies of $mathbb K^n$. The symbols $v$ and $w$ stand for vectors, i.e., $v,win V$. Their coordinate tuples relative to the ordered basis $mathcal B$ are denoted $[v]_mathcal B$ and $[w]_mathcal B$, respectively. These are elements of $mathbb K^n$. Similarly, if $L:Vto V$ is an automorphism of $V$, then its matrix relative to the “input” basis $mathcal B'$ and “output” basis $mathcal B$ is denoted by $[L]_mathcal B^mathcal B'$. This notation allows you to “cancel” a superscript against the subscript of the next term to its right in a multiplication. From this point of view, a change of basis doesn’t change the vector that the two coordinate tuples both represent, so a transfer matrix is a representation of the identity map $operatornameid$.

The map $pi_v:Vto V$ that orthogonally projects onto some fixed $vin V$ is given by the formula $$pi_v: w mapsto langle w, vrangle over langle v, vrangle v.tag1$$ Here, $langlecdot,cdotrangle : Vtimes Vtomathbb K$ stands for the inner product used to define orthogonality. Now, suppose that $mathcal B$ is a basis in which $langle w, vrangle = [v]_mathcal B^* [w]_mathcal B$. Then, if $P=[operatornameid]_mathcal B^mathcal B'$, we have $[v]_mathcal B=P[v]_mathcal B'$ and $[w]_mathcal B=P[w]_mathcal B'$, and so $$langle w, vrangle = left(P[v]_mathcal B'right)^* left(P[w]_mathcal B'right) = [v]_mathcal B'^* P^*P [w]_mathcal B'.tag2$$ This is nothing more than an application of the change-of-basis formula for quadratic forms: if $Q$ is the matrix of a quadratic form relative to the basis $mathcal B$ and $P = [operatornameid]_mathcal B^mathcal B'$ the transfer matrix from another basis $mathcal B'$, then the matrix of the quadratic form relative to $mathcal B'$ is $P^*QP$. This is one of the ways in which you erred: $langle w,vrangle = [v]_mathcal B^* [w]_mathcal B$ for all $v$ and $w$ iff $mathcal B$ is orthonormal. Otherwise, some positive-definite matrix other than the identity must come into play.

To reduce clutter, assume now that $|v|=1$. The derivation is similar, but messier, in the general case, for which there’ll be a $1/langle v,vrangle$ term floating around. I’m also assuming that $mathcal B$ is orthonormal. For full generality, we’d have to start with $langle w,vrangle = [v]_mathcal B^* Q [w]_mathcal B$ for some positive-definite matrix $Q$. From (1) we have $$[pi_v w]_mathcal B^mathcal B = [v]_mathcal B^*[w]_mathcal B[v]_mathcal B = [v]_mathcal B [v]_mathcal B^*[w]_mathcal B,$$ so $[pi_v]_mathcal B^mathcal B = [v]_mathcal B [v]_mathcal B^*$. Similarly, $$[pi_v]_mathcal B'^mathcal B' = [v]_mathcal B'^* P^*P [w]_mathcal B' [v]_mathcal B' = [v]_mathcal B' [v]_mathcal B'^* P^*P [w]_mathcal B'$$ and $[pi_v]_mathcal B'^mathcal B' = [v]_mathcal B' [v]_mathcal B'^* P^*P$. The two matrices are related by the usual change-of-basis formula: $$[pi_v]_mathcal B^mathcal B = [operatornameid]_mathcal B^mathcal B'[pi_v]_mathcal B'^mathcal B' [operatornameid]_mathcal B'^mathcal B = P[pi_v]_mathcal B'^mathcal B' P^-1.$$ On the other hand, $$[pi_v]_mathcal B^mathcal B = [v]_mathcal B [v]_mathcal B^* = left(P [v]_mathcal B'right) left(P [v]_mathcal B'right)^* = P [v]_mathcal B' [v]_mathcal B'^* P^* = P [pi_v]_mathcal B'^mathcal B' P^-1.$$ If $mathcal B'$ is also orthonormal, then $P^*P = I$ and this term will vanish, as you were trying to make it do, but you still won’t have $[pi_v]_mathcal B^mathcal B = P [pi_v]_mathcal B'^mathcal B'$ because of what’s effectively a type mismatch: the left-hand side expects its inputs expressed relative to $mathcal B$, but the right-hand side expects inputs relative to $mathcal B'$.

share|cite|improve this answer

edited Aug 3 at 22:38

answered Aug 3 at 8:48

amd

25.7k2943

• 
  
  
  

  
  

  
  
  
  
  
  

  
  -@amd Thanks for your detailed answer. Following your notations, my goal is not to write $[pi_v]_mathcal B^mathcal B = P [pi_v]_mathcal B'^mathcal B'$ but rather : $$[pi_v w]_mathcal B^mathcal B = P,[pi_v w]_mathcal B'^mathcal B'$$. Indeed, we have $<[v]_B,[w]_B> = <[v]_B',[w]_B'>$ and $[v]_B=P,[v]_B'$, so finally we can write : $$[pi_v w]_mathcal B^mathcal B=P,[pi_v w]_mathcal B'^mathcal B'$$, is it right ?
  – youpilat13
  Aug 3 at 21:38
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  -@amd . I think there is a little error : $$langle w, vrangle = left(P[v]_mathcal B'right)^* left(P[w]_mathcal B'right) = [v]_mathcal B'^* P^*P [w]_mathcal B'$$
  – youpilat13
  Aug 3 at 21:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @youpilat13 Not really. Let $V$ be the inner product space over the field $mathbb K$. Then $pi_vwin V$ is a vector; $[pi_vw]_mathcal Binmathbb K^n$ is the coordinate tuple of $pi_vw$ relative to the basis $mathcal B$, and the expression $[pi_vw]_mathcal B^mathcal B$ doesn’t make sense because, unlike the matrix of a linear transformation, there’s only one basis to consider. The identity $[pi_vw]_mathcal B=P[pi_vw]_mathcal B'$ is a trivial matter of applying the transfer matrix $P$ to the $mathcal B'$-coordinates of $pi_vw$.
  – amd
  Aug 3 at 22:10
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  -@amd If $B$ and $B'$ are orthonormal, then we could write :$$ langle w, vrangle = left(P[v]_mathcal B'right)^*,Q, left(P[w]_mathcal B'right) = [v]_mathcal B'^* P^*,Q,P [w]_mathcal B'$$ and so with $Q=I_d$ and $P^*=P^-1$, we get : $$langle w,vrangle = [v]_mathcal B'^* [w]_mathcal B'$$ : is this correct ? What does it imply on $P$ transfer matrix and $Q$ matrix if one has $B$ and $B'$ orthonormal ? i.e just $Q=I_d$ ?
  – youpilat13
  Aug 4 at 14:07
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  -@amd Please could you take a look at my last comment above, i.e to confirm it ?
  – youpilat13
  Aug 5 at 9:28
  

  
  
  
  

 | 
show 2 more comments

up vote
0
down vote



accepted

up vote
0
down vote



accepted

There’s not really anything tricky here. In particular, you don’t need to try so hard to eliminate the $P$’s in the resulting expression because they actually belong there. They’re essential to the inner product that’s inherent in the projection.

I’ll use somewhat different notation from that in the question because it’s important to distinguish the vectors themselves—elements of the inner product space $V$—from their coordinate tuples—elements of various copies of $mathbb K^n$. The symbols $v$ and $w$ stand for vectors, i.e., $v,win V$. Their coordinate tuples relative to the ordered basis $mathcal B$ are denoted $[v]_mathcal B$ and $[w]_mathcal B$, respectively. These are elements of $mathbb K^n$. Similarly, if $L:Vto V$ is an automorphism of $V$, then its matrix relative to the “input” basis $mathcal B'$ and “output” basis $mathcal B$ is denoted by $[L]_mathcal B^mathcal B'$. This notation allows you to “cancel” a superscript against the subscript of the next term to its right in a multiplication. From this point of view, a change of basis doesn’t change the vector that the two coordinate tuples both represent, so a transfer matrix is a representation of the identity map $operatornameid$.

The map $pi_v:Vto V$ that orthogonally projects onto some fixed $vin V$ is given by the formula $$pi_v: w mapsto langle w, vrangle over langle v, vrangle v.tag1$$ Here, $langlecdot,cdotrangle : Vtimes Vtomathbb K$ stands for the inner product used to define orthogonality. Now, suppose that $mathcal B$ is a basis in which $langle w, vrangle = [v]_mathcal B^* [w]_mathcal B$. Then, if $P=[operatornameid]_mathcal B^mathcal B'$, we have $[v]_mathcal B=P[v]_mathcal B'$ and $[w]_mathcal B=P[w]_mathcal B'$, and so $$langle w, vrangle = left(P[v]_mathcal B'right)^* left(P[w]_mathcal B'right) = [v]_mathcal B'^* P^*P [w]_mathcal B'.tag2$$ This is nothing more than an application of the change-of-basis formula for quadratic forms: if $Q$ is the matrix of a quadratic form relative to the basis $mathcal B$ and $P = [operatornameid]_mathcal B^mathcal B'$ the transfer matrix from another basis $mathcal B'$, then the matrix of the quadratic form relative to $mathcal B'$ is $P^*QP$. This is one of the ways in which you erred: $langle w,vrangle = [v]_mathcal B^* [w]_mathcal B$ for all $v$ and $w$ iff $mathcal B$ is orthonormal. Otherwise, some positive-definite matrix other than the identity must come into play.

To reduce clutter, assume now that $|v|=1$. The derivation is similar, but messier, in the general case, for which there’ll be a $1/langle v,vrangle$ term floating around. I’m also assuming that $mathcal B$ is orthonormal. For full generality, we’d have to start with $langle w,vrangle = [v]_mathcal B^* Q [w]_mathcal B$ for some positive-definite matrix $Q$. From (1) we have $$[pi_v w]_mathcal B^mathcal B = [v]_mathcal B^*[w]_mathcal B[v]_mathcal B = [v]_mathcal B [v]_mathcal B^*[w]_mathcal B,$$ so $[pi_v]_mathcal B^mathcal B = [v]_mathcal B [v]_mathcal B^*$. Similarly, $$[pi_v]_mathcal B'^mathcal B' = [v]_mathcal B'^* P^*P [w]_mathcal B' [v]_mathcal B' = [v]_mathcal B' [v]_mathcal B'^* P^*P [w]_mathcal B'$$ and $[pi_v]_mathcal B'^mathcal B' = [v]_mathcal B' [v]_mathcal B'^* P^*P$. The two matrices are related by the usual change-of-basis formula: $$[pi_v]_mathcal B^mathcal B = [operatornameid]_mathcal B^mathcal B'[pi_v]_mathcal B'^mathcal B' [operatornameid]_mathcal B'^mathcal B = P[pi_v]_mathcal B'^mathcal B' P^-1.$$ On the other hand, $$[pi_v]_mathcal B^mathcal B = [v]_mathcal B [v]_mathcal B^* = left(P [v]_mathcal B'right) left(P [v]_mathcal B'right)^* = P [v]_mathcal B' [v]_mathcal B'^* P^* = P [pi_v]_mathcal B'^mathcal B' P^-1.$$ If $mathcal B'$ is also orthonormal, then $P^*P = I$ and this term will vanish, as you were trying to make it do, but you still won’t have $[pi_v]_mathcal B^mathcal B = P [pi_v]_mathcal B'^mathcal B'$ because of what’s effectively a type mismatch: the left-hand side expects its inputs expressed relative to $mathcal B$, but the right-hand side expects inputs relative to $mathcal B'$.

share|cite|improve this answer

edited Aug 3 at 22:38

answered Aug 3 at 8:48

amd

25.7k2943

There’s not really anything tricky here. In particular, you don’t need to try so hard to eliminate the $P$’s in the resulting expression because they actually belong there. They’re essential to the inner product that’s inherent in the projection.

I’ll use somewhat different notation from that in the question because it’s important to distinguish the vectors themselves—elements of the inner product space $V$—from their coordinate tuples—elements of various copies of $mathbb K^n$. The symbols $v$ and $w$ stand for vectors, i.e., $v,win V$. Their coordinate tuples relative to the ordered basis $mathcal B$ are denoted $[v]_mathcal B$ and $[w]_mathcal B$, respectively. These are elements of $mathbb K^n$. Similarly, if $L:Vto V$ is an automorphism of $V$, then its matrix relative to the “input” basis $mathcal B'$ and “output” basis $mathcal B$ is denoted by $[L]_mathcal B^mathcal B'$. This notation allows you to “cancel” a superscript against the subscript of the next term to its right in a multiplication. From this point of view, a change of basis doesn’t change the vector that the two coordinate tuples both represent, so a transfer matrix is a representation of the identity map $operatornameid$.

The map $pi_v:Vto V$ that orthogonally projects onto some fixed $vin V$ is given by the formula $$pi_v: w mapsto langle w, vrangle over langle v, vrangle v.tag1$$ Here, $langlecdot,cdotrangle : Vtimes Vtomathbb K$ stands for the inner product used to define orthogonality. Now, suppose that $mathcal B$ is a basis in which $langle w, vrangle = [v]_mathcal B^* [w]_mathcal B$. Then, if $P=[operatornameid]_mathcal B^mathcal B'$, we have $[v]_mathcal B=P[v]_mathcal B'$ and $[w]_mathcal B=P[w]_mathcal B'$, and so $$langle w, vrangle = left(P[v]_mathcal B'right)^* left(P[w]_mathcal B'right) = [v]_mathcal B'^* P^*P [w]_mathcal B'.tag2$$ This is nothing more than an application of the change-of-basis formula for quadratic forms: if $Q$ is the matrix of a quadratic form relative to the basis $mathcal B$ and $P = [operatornameid]_mathcal B^mathcal B'$ the transfer matrix from another basis $mathcal B'$, then the matrix of the quadratic form relative to $mathcal B'$ is $P^*QP$. This is one of the ways in which you erred: $langle w,vrangle = [v]_mathcal B^* [w]_mathcal B$ for all $v$ and $w$ iff $mathcal B$ is orthonormal. Otherwise, some positive-definite matrix other than the identity must come into play.

To reduce clutter, assume now that $|v|=1$. The derivation is similar, but messier, in the general case, for which there’ll be a $1/langle v,vrangle$ term floating around. I’m also assuming that $mathcal B$ is orthonormal. For full generality, we’d have to start with $langle w,vrangle = [v]_mathcal B^* Q [w]_mathcal B$ for some positive-definite matrix $Q$. From (1) we have $$[pi_v w]_mathcal B^mathcal B = [v]_mathcal B^*[w]_mathcal B[v]_mathcal B = [v]_mathcal B [v]_mathcal B^*[w]_mathcal B,$$ so $[pi_v]_mathcal B^mathcal B = [v]_mathcal B [v]_mathcal B^*$. Similarly, $$[pi_v]_mathcal B'^mathcal B' = [v]_mathcal B'^* P^*P [w]_mathcal B' [v]_mathcal B' = [v]_mathcal B' [v]_mathcal B'^* P^*P [w]_mathcal B'$$ and $[pi_v]_mathcal B'^mathcal B' = [v]_mathcal B' [v]_mathcal B'^* P^*P$. The two matrices are related by the usual change-of-basis formula: $$[pi_v]_mathcal B^mathcal B = [operatornameid]_mathcal B^mathcal B'[pi_v]_mathcal B'^mathcal B' [operatornameid]_mathcal B'^mathcal B = P[pi_v]_mathcal B'^mathcal B' P^-1.$$ On the other hand, $$[pi_v]_mathcal B^mathcal B = [v]_mathcal B [v]_mathcal B^* = left(P [v]_mathcal B'right) left(P [v]_mathcal B'right)^* = P [v]_mathcal B' [v]_mathcal B'^* P^* = P [pi_v]_mathcal B'^mathcal B' P^-1.$$ If $mathcal B'$ is also orthonormal, then $P^*P = I$ and this term will vanish, as you were trying to make it do, but you still won’t have $[pi_v]_mathcal B^mathcal B = P [pi_v]_mathcal B'^mathcal B'$ because of what’s effectively a type mismatch: the left-hand side expects its inputs expressed relative to $mathcal B$, but the right-hand side expects inputs relative to $mathcal B'$.

share|cite|improve this answer

edited Aug 3 at 22:38

answered Aug 3 at 8:48

amd

25.7k2943

share|cite|improve this answer

share|cite|improve this answer

edited Aug 3 at 22:38

edited Aug 3 at 22:38

edited Aug 3 at 22:38

answered Aug 3 at 8:48

amd

25.7k2943

answered Aug 3 at 8:48

amd

25.7k2943

answered Aug 3 at 8:48

amd

25.7k2943

25.7k2943

• 
  
  
  

  
  

  
  
  
  
  
  

  
  -@amd Thanks for your detailed answer. Following your notations, my goal is not to write $[pi_v]_mathcal B^mathcal B = P [pi_v]_mathcal B'^mathcal B'$ but rather : $$[pi_v w]_mathcal B^mathcal B = P,[pi_v w]_mathcal B'^mathcal B'$$. Indeed, we have $<[v]_B,[w]_B> = <[v]_B',[w]_B'>$ and $[v]_B=P,[v]_B'$, so finally we can write : $$[pi_v w]_mathcal B^mathcal B=P,[pi_v w]_mathcal B'^mathcal B'$$, is it right ?
  – youpilat13
  Aug 3 at 21:38
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  -@amd . I think there is a little error : $$langle w, vrangle = left(P[v]_mathcal B'right)^* left(P[w]_mathcal B'right) = [v]_mathcal B'^* P^*P [w]_mathcal B'$$
  – youpilat13
  Aug 3 at 21:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @youpilat13 Not really. Let $V$ be the inner product space over the field $mathbb K$. Then $pi_vwin V$ is a vector; $[pi_vw]_mathcal Binmathbb K^n$ is the coordinate tuple of $pi_vw$ relative to the basis $mathcal B$, and the expression $[pi_vw]_mathcal B^mathcal B$ doesn’t make sense because, unlike the matrix of a linear transformation, there’s only one basis to consider. The identity $[pi_vw]_mathcal B=P[pi_vw]_mathcal B'$ is a trivial matter of applying the transfer matrix $P$ to the $mathcal B'$-coordinates of $pi_vw$.
  – amd
  Aug 3 at 22:10
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  -@amd If $B$ and $B'$ are orthonormal, then we could write :$$ langle w, vrangle = left(P[v]_mathcal B'right)^*,Q, left(P[w]_mathcal B'right) = [v]_mathcal B'^* P^*,Q,P [w]_mathcal B'$$ and so with $Q=I_d$ and $P^*=P^-1$, we get : $$langle w,vrangle = [v]_mathcal B'^* [w]_mathcal B'$$ : is this correct ? What does it imply on $P$ transfer matrix and $Q$ matrix if one has $B$ and $B'$ orthonormal ? i.e just $Q=I_d$ ?
  – youpilat13
  Aug 4 at 14:07
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  -@amd Please could you take a look at my last comment above, i.e to confirm it ?
  – youpilat13
  Aug 5 at 9:28
  

  
  
  
  

 | 
show 2 more comments

• 
  
  
  

  
  

  
  
  
  
  
  

  
  -@amd Thanks for your detailed answer. Following your notations, my goal is not to write $[pi_v]_mathcal B^mathcal B = P [pi_v]_mathcal B'^mathcal B'$ but rather : $$[pi_v w]_mathcal B^mathcal B = P,[pi_v w]_mathcal B'^mathcal B'$$. Indeed, we have $<[v]_B,[w]_B> = <[v]_B',[w]_B'>$ and $[v]_B=P,[v]_B'$, so finally we can write : $$[pi_v w]_mathcal B^mathcal B=P,[pi_v w]_mathcal B'^mathcal B'$$, is it right ?
  – youpilat13
  Aug 3 at 21:38
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  -@amd . I think there is a little error : $$langle w, vrangle = left(P[v]_mathcal B'right)^* left(P[w]_mathcal B'right) = [v]_mathcal B'^* P^*P [w]_mathcal B'$$
  – youpilat13
  Aug 3 at 21:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @youpilat13 Not really. Let $V$ be the inner product space over the field $mathbb K$. Then $pi_vwin V$ is a vector; $[pi_vw]_mathcal Binmathbb K^n$ is the coordinate tuple of $pi_vw$ relative to the basis $mathcal B$, and the expression $[pi_vw]_mathcal B^mathcal B$ doesn’t make sense because, unlike the matrix of a linear transformation, there’s only one basis to consider. The identity $[pi_vw]_mathcal B=P[pi_vw]_mathcal B'$ is a trivial matter of applying the transfer matrix $P$ to the $mathcal B'$-coordinates of $pi_vw$.
  – amd
  Aug 3 at 22:10
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  -@amd If $B$ and $B'$ are orthonormal, then we could write :$$ langle w, vrangle = left(P[v]_mathcal B'right)^*,Q, left(P[w]_mathcal B'right) = [v]_mathcal B'^* P^*,Q,P [w]_mathcal B'$$ and so with $Q=I_d$ and $P^*=P^-1$, we get : $$langle w,vrangle = [v]_mathcal B'^* [w]_mathcal B'$$ : is this correct ? What does it imply on $P$ transfer matrix and $Q$ matrix if one has $B$ and $B'$ orthonormal ? i.e just $Q=I_d$ ?
  – youpilat13
  Aug 4 at 14:07
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  -@amd Please could you take a look at my last comment above, i.e to confirm it ?
  – youpilat13
  Aug 5 at 9:28
  

  
  
  
  

-@amd Thanks for your detailed answer. Following your notations, my goal is not to write $[pi_v]_mathcal B^mathcal B = P [pi_v]_mathcal B'^mathcal B'$ but rather : $$[pi_v w]_mathcal B^mathcal B = P,[pi_v w]_mathcal B'^mathcal B'$$. Indeed, we have $<[v]_B,[w]_B> = <[v]_B',[w]_B'>$ and $[v]_B=P,[v]_B'$, so finally we can write : $$[pi_v w]_mathcal B^mathcal B=P,[pi_v w]_mathcal B'^mathcal B'$$, is it right ?
– youpilat13
Aug 3 at 21:38

-@amd Thanks for your detailed answer. Following your notations, my goal is not to write $[pi_v]_mathcal B^mathcal B = P [pi_v]_mathcal B'^mathcal B'$ but rather : $$[pi_v w]_mathcal B^mathcal B = P,[pi_v w]_mathcal B'^mathcal B'$$. Indeed, we have $<[v]_B,[w]_B> = <[v]_B',[w]_B'>$ and $[v]_B=P,[v]_B'$, so finally we can write : $$[pi_v w]_mathcal B^mathcal B=P,[pi_v w]_mathcal B'^mathcal B'$$, is it right ?
– youpilat13
Aug 3 at 21:38

-@amd . I think there is a little error : $$langle w, vrangle = left(P[v]_mathcal B'right)^* left(P[w]_mathcal B'right) = [v]_mathcal B'^* P^*P [w]_mathcal B'$$
– youpilat13
Aug 3 at 21:57

-@amd . I think there is a little error : $$langle w, vrangle = left(P[v]_mathcal B'right)^* left(P[w]_mathcal B'right) = [v]_mathcal B'^* P^*P [w]_mathcal B'$$
– youpilat13
Aug 3 at 21:57

@youpilat13 Not really. Let $V$ be the inner product space over the field $mathbb K$. Then $pi_vwin V$ is a vector; $[pi_vw]_mathcal Binmathbb K^n$ is the coordinate tuple of $pi_vw$ relative to the basis $mathcal B$, and the expression $[pi_vw]_mathcal B^mathcal B$ doesn’t make sense because, unlike the matrix of a linear transformation, there’s only one basis to consider. The identity $[pi_vw]_mathcal B=P[pi_vw]_mathcal B'$ is a trivial matter of applying the transfer matrix $P$ to the $mathcal B'$-coordinates of $pi_vw$.
– amd
Aug 3 at 22:10

@youpilat13 Not really. Let $V$ be the inner product space over the field $mathbb K$. Then $pi_vwin V$ is a vector; $[pi_vw]_mathcal Binmathbb K^n$ is the coordinate tuple of $pi_vw$ relative to the basis $mathcal B$, and the expression $[pi_vw]_mathcal B^mathcal B$ doesn’t make sense because, unlike the matrix of a linear transformation, there’s only one basis to consider. The identity $[pi_vw]_mathcal B=P[pi_vw]_mathcal B'$ is a trivial matter of applying the transfer matrix $P$ to the $mathcal B'$-coordinates of $pi_vw$.
– amd
Aug 3 at 22:10

-@amd If $B$ and $B'$ are orthonormal, then we could write :$$ langle w, vrangle = left(P[v]_mathcal B'right)^*,Q, left(P[w]_mathcal B'right) = [v]_mathcal B'^* P^*,Q,P [w]_mathcal B'$$ and so with $Q=I_d$ and $P^*=P^-1$, we get : $$langle w,vrangle = [v]_mathcal B'^* [w]_mathcal B'$$ : is this correct ? What does it imply on $P$ transfer matrix and $Q$ matrix if one has $B$ and $B'$ orthonormal ? i.e just $Q=I_d$ ?
– youpilat13
Aug 4 at 14:07

-@amd If $B$ and $B'$ are orthonormal, then we could write :$$ langle w, vrangle = left(P[v]_mathcal B'right)^*,Q, left(P[w]_mathcal B'right) = [v]_mathcal B'^* P^*,Q,P [w]_mathcal B'$$ and so with $Q=I_d$ and $P^*=P^-1$, we get : $$langle w,vrangle = [v]_mathcal B'^* [w]_mathcal B'$$ : is this correct ? What does it imply on $P$ transfer matrix and $Q$ matrix if one has $B$ and $B'$ orthonormal ? i.e just $Q=I_d$ ?
– youpilat13
Aug 4 at 14:07

-@amd Please could you take a look at my last comment above, i.e to confirm it ?
– youpilat13
Aug 5 at 9:28

-@amd Please could you take a look at my last comment above, i.e to confirm it ?
– youpilat13
Aug 5 at 9:28

 | 
show 2 more comments

 

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