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Projection independent from basis
Clash Royale CLAN TAG#URR8PPP
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I would like to get a simple demo to proove that a projection defined for example by : $vecvvecv^*vecw$ where $vecw$ is projected onto vector $vecv$.
First, if I start from the definition of an inner product between $X$ and $Y$ vectors, the definition is :
$$Phi(X,Y)=X^T M Y$$
with $M$ the tensor metric (that I assimilate to the projection matrix $vecvvecv^*$ above) :
If I introduce a transformation matrix $P$ to do the link between $(e'_i)$ and $(e'_i)$ such as for vectors $X$ and $Y$, I can write :
$$X=PX'$$
$$Y=PY'$$
Introduce this in definition of bilinear form $Phi$ :
$$Phi(X,Y) = Phi(PX',PY')= <X'^T P^T M P Y'>$$
Any help is welcome
EDIT 2 : The formula $P^T M P$ seems to be the general expression of matrix $M$ into ($vece'_i$) vectors basis : is it correct ?
bilinear-form projection-matrices
asked Jul 29 at 1:13
youpilat13
6911
add a comment |Â
up vote
0
down vote
favorite
I would like to get a simple demo to proove that a projection defined for example by : $vecvvecv^*vecw$ where $vecw$ is projected onto vector $vecv$.
First, if I start from the definition of an inner product between $X$ and $Y$ vectors, the definition is :
$$Phi(X,Y)=X^T M Y$$
with $M$ the tensor metric (that I assimilate to the projection matrix $vecvvecv^*$ above) :
If I introduce a transformation matrix $P$ to do the link between $(e'_i)$ and $(e'_i)$ such as for vectors $X$ and $Y$, I can write :
$$X=PX'$$
$$Y=PY'$$
Introduce this in definition of bilinear form $Phi$ :
$$Phi(X,Y) = Phi(PX',PY')= <X'^T P^T M P Y'>$$
Any help is welcome
EDIT 2 : The formula $P^T M P$ seems to be the general expression of matrix $M$ into ($vece'_i$) vectors basis : is it correct ?
bilinear-form projection-matrices
asked Jul 29 at 1:13
youpilat13
6911
Sorry, I'm not quite sure what you are trying to do. Is P the projection? If so projections don't preserve norms (since you are making the vector smaller by only keeping the parts parallel to v).
– Ruvi Lecamwasam
Jul 29 at 1:20
-@RuviLecamwasam. Sorry, I have not been explicit enough. The projection matrix is represented by the tensorial product $vecvvecv^*$ at the begining of post : $vecv$ is a classical vector and $vecv^*$ its dual. I want to use this matrix to proove that in any vectors basis, the projected vector has always the same values, independently of basis in which this "projected vector" is expressed. Do you understand better ?
– youpilat13
Jul 29 at 1:29
-@RuviLecamwasam Matrix P is the matrix which expresses basis vectors $vece'_i$ as a function of $vece_i$ (transfer Matrix is called ?).
– youpilat13
Jul 29 at 1:33
Ok, but are your $e_i'$ the projected $e_i$, i.e. does $e_i'=vv^*e_i$? Because in that case you will have $Phi(X,Y)le Phi(X',Y')$ rather than equality.
– Ruvi Lecamwasam
Jul 29 at 1:46
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I would like to get a simple demo to proove that a projection defined for example by : $vecvvecv^*vecw$ where $vecw$ is projected onto vector $vecv$.
First, if I start from the definition of an inner product between $X$ and $Y$ vectors, the definition is :
$$Phi(X,Y)=X^T M Y$$
with $M$ the tensor metric (that I assimilate to the projection matrix $vecvvecv^*$ above) :
If I introduce a transformation matrix $P$ to do the link between $(e'_i)$ and $(e'_i)$ such as for vectors $X$ and $Y$, I can write :
$$X=PX'$$
$$Y=PY'$$
Introduce this in definition of bilinear form $Phi$ :
$$Phi(X,Y) = Phi(PX',PY')= <X'^T P^T M P Y'>$$
Any help is welcome
EDIT 2 : The formula $P^T M P$ seems to be the general expression of matrix $M$ into ($vece'_i$) vectors basis : is it correct ?
bilinear-form projection-matrices
asked Jul 29 at 1:13
youpilat13
6911
I would like to get a simple demo to proove that a projection defined for example by : $vecvvecv^*vecw$ where $vecw$ is projected onto vector $vecv$.
First, if I start from the definition of an inner product between $X$ and $Y$ vectors, the definition is :
$$Phi(X,Y)=X^T M Y$$
with $M$ the tensor metric (that I assimilate to the projection matrix $vecvvecv^*$ above) :
If I introduce a transformation matrix $P$ to do the link between $(e'_i)$ and $(e'_i)$ such as for vectors $X$ and $Y$, I can write :
$$X=PX'$$
$$Y=PY'$$
Introduce this in definition of bilinear form $Phi$ :
$$Phi(X,Y) = Phi(PX',PY')= <X'^T P^T M P Y'>$$
Any help is welcome
EDIT 2 : The formula $P^T M P$ seems to be the general expression of matrix $M$ into ($vece'_i$) vectors basis : is it correct ?
bilinear-form projection-matrices
asked Jul 29 at 1:13
youpilat13
6911
edited Jul 29 at 19:00
asked Jul 29 at 1:13
youpilat13
6911
asked Jul 29 at 1:13
youpilat13
6911
asked Jul 29 at 1:13
youpilat13
6911
6911
Sorry, I'm not quite sure what you are trying to do. Is P the projection? If so projections don't preserve norms (since you are making the vector smaller by only keeping the parts parallel to v).
– Ruvi Lecamwasam
Jul 29 at 1:20
-@RuviLecamwasam. Sorry, I have not been explicit enough. The projection matrix is represented by the tensorial product $vecvvecv^*$ at the begining of post : $vecv$ is a classical vector and $vecv^*$ its dual. I want to use this matrix to proove that in any vectors basis, the projected vector has always the same values, independently of basis in which this "projected vector" is expressed. Do you understand better ?
– youpilat13
Jul 29 at 1:29
-@RuviLecamwasam Matrix P is the matrix which expresses basis vectors $vece'_i$ as a function of $vece_i$ (transfer Matrix is called ?).
– youpilat13
Jul 29 at 1:33
Ok, but are your $e_i'$ the projected $e_i$, i.e. does $e_i'=vv^*e_i$? Because in that case you will have $Phi(X,Y)le Phi(X',Y')$ rather than equality.
– Ruvi Lecamwasam
Jul 29 at 1:46
add a comment |Â
Sorry, I'm not quite sure what you are trying to do. Is P the projection? If so projections don't preserve norms (since you are making the vector smaller by only keeping the parts parallel to v).
– Ruvi Lecamwasam
Jul 29 at 1:20
-@RuviLecamwasam. Sorry, I have not been explicit enough. The projection matrix is represented by the tensorial product $vecvvecv^*$ at the begining of post : $vecv$ is a classical vector and $vecv^*$ its dual. I want to use this matrix to proove that in any vectors basis, the projected vector has always the same values, independently of basis in which this "projected vector" is expressed. Do you understand better ?
– youpilat13
Jul 29 at 1:29
-@RuviLecamwasam Matrix P is the matrix which expresses basis vectors $vece'_i$ as a function of $vece_i$ (transfer Matrix is called ?).
– youpilat13
Jul 29 at 1:33
Ok, but are your $e_i'$ the projected $e_i$, i.e. does $e_i'=vv^*e_i$? Because in that case you will have $Phi(X,Y)le Phi(X',Y')$ rather than equality.
– Ruvi Lecamwasam
Jul 29 at 1:46
Sorry, I'm not quite sure what you are trying to do. Is P the projection? If so projections don't preserve norms (since you are making the vector smaller by only keeping the parts parallel to v).
– Ruvi Lecamwasam
Jul 29 at 1:20
Sorry, I'm not quite sure what you are trying to do. Is P the projection? If so projections don't preserve norms (since you are making the vector smaller by only keeping the parts parallel to v).
– Ruvi Lecamwasam
Jul 29 at 1:20
-@RuviLecamwasam. Sorry, I have not been explicit enough. The projection matrix is represented by the tensorial product $vecvvecv^*$ at the begining of post : $vecv$ is a classical vector and $vecv^*$ its dual. I want to use this matrix to proove that in any vectors basis, the projected vector has always the same values, independently of basis in which this "projected vector" is expressed. Do you understand better ?
– youpilat13
Jul 29 at 1:29
-@RuviLecamwasam. Sorry, I have not been explicit enough. The projection matrix is represented by the tensorial product $vecvvecv^*$ at the begining of post : $vecv$ is a classical vector and $vecv^*$ its dual. I want to use this matrix to proove that in any vectors basis, the projected vector has always the same values, independently of basis in which this "projected vector" is expressed. Do you understand better ?
– youpilat13
Jul 29 at 1:29
-@RuviLecamwasam Matrix P is the matrix which expresses basis vectors $vece'_i$ as a function of $vece_i$ (transfer Matrix is called ?).
– youpilat13
Jul 29 at 1:33
-@RuviLecamwasam Matrix P is the matrix which expresses basis vectors $vece'_i$ as a function of $vece_i$ (transfer Matrix is called ?).
– youpilat13
Jul 29 at 1:33
Ok, but are your $e_i'$ the projected $e_i$, i.e. does $e_i'=vv^*e_i$? Because in that case you will have $Phi(X,Y)le Phi(X',Y')$ rather than equality.
– Ruvi Lecamwasam
Jul 29 at 1:46
Ok, but are your $e_i'$ the projected $e_i$, i.e. does $e_i'=vv^*e_i$? Because in that case you will have $Phi(X,Y)le Phi(X',Y')$ rather than equality.
– Ruvi Lecamwasam
Jul 29 at 1:46
add a comment |Â
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Sorry, I'm not quite sure what you are trying to do. Is P the projection? If so projections don't preserve norms (since you are making the vector smaller by only keeping the parts parallel to v).
– Ruvi Lecamwasam
Jul 29 at 1:20
-@RuviLecamwasam. Sorry, I have not been explicit enough. The projection matrix is represented by the tensorial product $vecvvecv^*$ at the begining of post : $vecv$ is a classical vector and $vecv^*$ its dual. I want to use this matrix to proove that in any vectors basis, the projected vector has always the same values, independently of basis in which this "projected vector" is expressed. Do you understand better ?
– youpilat13
Jul 29 at 1:29
-@RuviLecamwasam Matrix P is the matrix which expresses basis vectors $vece'_i$ as a function of $vece_i$ (transfer Matrix is called ?).
– youpilat13
Jul 29 at 1:33
Ok, but are your $e_i'$ the projected $e_i$, i.e. does $e_i'=vv^*e_i$? Because in that case you will have $Phi(X,Y)le Phi(X',Y')$ rather than equality.
– Ruvi Lecamwasam
Jul 29 at 1:46