What is the set of all isometric matrix in $mathbb{R}^{k times d}$?
$begingroup$
An isometry from metric space $X=mathbb{R}^{d}$ to metric space $Y=mathbb{R}^{k}$ with usual norm for both spaces is the following:
$$
Phi: mathbb{R}^{d} rightarrow mathbb{R}^{k}
$$
where $Phi(x)=Px$ and $P in mathbb{R}^{k times d}$ such that $|x|_2^2=|Px|_2^2$. The set of all $P$'s can be written as
$$
mathcal{P}={P in mathbb{R}^{k times d} mid |x|_2^2=|Px|_2^2 ,,,, forall x in mathbb{R}^{d}}
$$
My questions:
Is $mathcal{P}$ the set of all $P in mathbb{R}^{k times d}$ where each column is a Euclidean basis $e_i=[0,0,cdots,0,overbrace{1}^{i},0,cdots,0]^{T}$?
If so, why this is true and we have only zero and 1 in the isometry?
real-analysis linear-algebra isometry projection-matrices
asked Jan 2 at 15:54
SaeedSaeed
1,080310
$endgroup$
|
show 2 more comments
$begingroup$
An isometry from metric space $X=mathbb{R}^{d}$ to metric space $Y=mathbb{R}^{k}$ with usual norm for both spaces is the following:
$$
Phi: mathbb{R}^{d} rightarrow mathbb{R}^{k}
$$
where $Phi(x)=Px$ and $P in mathbb{R}^{k times d}$ such that $|x|_2^2=|Px|_2^2$. The set of all $P$'s can be written as
$$
mathcal{P}={P in mathbb{R}^{k times d} mid |x|_2^2=|Px|_2^2 ,,,, forall x in mathbb{R}^{d}}
$$
My questions:
Is $mathcal{P}$ the set of all $P in mathbb{R}^{k times d}$ where each column is a Euclidean basis $e_i=[0,0,cdots,0,overbrace{1}^{i},0,cdots,0]^{T}$?
If so, why this is true and we have only zero and 1 in the isometry?
real-analysis linear-algebra isometry projection-matrices
asked Jan 2 at 15:54
SaeedSaeed
1,080310
$endgroup$
1
$begingroup$
No. $mathcal P$ is the set of all $P$ satisfying $P^TP = I_d$, where $I_d$ is the $d times d$ identity matrix.
$endgroup$
– Omnomnomnom
Jan 2 at 16:15
$begingroup$
But $P$ is $d times k$, shouldn't be $P^TP=I_k$?
$endgroup$
– Saeed
Jan 2 at 19:19
$begingroup$
If $Phi$ goes from $X$ to $Y$, then $P$ should be $ktimes d$.
$endgroup$
– Omnomnomnom
Jan 2 at 19:35
$begingroup$
@Omnomnomnom: Yes. you are right. I edited it. $P$ can be any matrix with $d$ colums including orthonormal vectors in $mathbb{R}^k$?
$endgroup$
– Saeed
Jan 2 at 19:57
$begingroup$
that's exactly right
$endgroup$
– Omnomnomnom
Jan 2 at 20:02
|
show 2 more comments
$begingroup$
An isometry from metric space $X=mathbb{R}^{d}$ to metric space $Y=mathbb{R}^{k}$ with usual norm for both spaces is the following:
$$
Phi: mathbb{R}^{d} rightarrow mathbb{R}^{k}
$$
where $Phi(x)=Px$ and $P in mathbb{R}^{k times d}$ such that $|x|_2^2=|Px|_2^2$. The set of all $P$'s can be written as
$$
mathcal{P}={P in mathbb{R}^{k times d} mid |x|_2^2=|Px|_2^2 ,,,, forall x in mathbb{R}^{d}}
$$
My questions:
Is $mathcal{P}$ the set of all $P in mathbb{R}^{k times d}$ where each column is a Euclidean basis $e_i=[0,0,cdots,0,overbrace{1}^{i},0,cdots,0]^{T}$?
If so, why this is true and we have only zero and 1 in the isometry?
real-analysis linear-algebra isometry projection-matrices
asked Jan 2 at 15:54
SaeedSaeed
1,080310
$endgroup$
An isometry from metric space $X=mathbb{R}^{d}$ to metric space $Y=mathbb{R}^{k}$ with usual norm for both spaces is the following:
$$
Phi: mathbb{R}^{d} rightarrow mathbb{R}^{k}
$$
where $Phi(x)=Px$ and $P in mathbb{R}^{k times d}$ such that $|x|_2^2=|Px|_2^2$. The set of all $P$'s can be written as
$$
mathcal{P}={P in mathbb{R}^{k times d} mid |x|_2^2=|Px|_2^2 ,,,, forall x in mathbb{R}^{d}}
$$
My questions:
Is $mathcal{P}$ the set of all $P in mathbb{R}^{k times d}$ where each column is a Euclidean basis $e_i=[0,0,cdots,0,overbrace{1}^{i},0,cdots,0]^{T}$?
If so, why this is true and we have only zero and 1 in the isometry?
real-analysis linear-algebra isometry projection-matrices
real-analysis linear-algebra isometry projection-matrices
asked Jan 2 at 15:54
SaeedSaeed
1,080310
asked Jan 2 at 15:54
SaeedSaeed
1,080310
edited Jan 2 at 19:50
Saeed
asked Jan 2 at 15:54
SaeedSaeed
1,080310
asked Jan 2 at 15:54
SaeedSaeed
1,080310
asked Jan 2 at 15:54
SaeedSaeed
1,080310
1,080310
1
$begingroup$
No. $mathcal P$ is the set of all $P$ satisfying $P^TP = I_d$, where $I_d$ is the $d times d$ identity matrix.
$endgroup$
– Omnomnomnom
Jan 2 at 16:15
$begingroup$
But $P$ is $d times k$, shouldn't be $P^TP=I_k$?
$endgroup$
– Saeed
Jan 2 at 19:19
$begingroup$
If $Phi$ goes from $X$ to $Y$, then $P$ should be $ktimes d$.
$endgroup$
– Omnomnomnom
Jan 2 at 19:35
$begingroup$
@Omnomnomnom: Yes. you are right. I edited it. $P$ can be any matrix with $d$ colums including orthonormal vectors in $mathbb{R}^k$?
$endgroup$
– Saeed
Jan 2 at 19:57
$begingroup$
that's exactly right
$endgroup$
– Omnomnomnom
Jan 2 at 20:02
|
show 2 more comments
1
$begingroup$
No. $mathcal P$ is the set of all $P$ satisfying $P^TP = I_d$, where $I_d$ is the $d times d$ identity matrix.
$endgroup$
– Omnomnomnom
Jan 2 at 16:15
$begingroup$
But $P$ is $d times k$, shouldn't be $P^TP=I_k$?
$endgroup$
– Saeed
Jan 2 at 19:19
$begingroup$
If $Phi$ goes from $X$ to $Y$, then $P$ should be $ktimes d$.
$endgroup$
– Omnomnomnom
Jan 2 at 19:35
$begingroup$
@Omnomnomnom: Yes. you are right. I edited it. $P$ can be any matrix with $d$ colums including orthonormal vectors in $mathbb{R}^k$?
$endgroup$
– Saeed
Jan 2 at 19:57
$begingroup$
that's exactly right
$endgroup$
– Omnomnomnom
Jan 2 at 20:02
1
1
$begingroup$
No. $mathcal P$ is the set of all $P$ satisfying $P^TP = I_d$, where $I_d$ is the $d times d$ identity matrix.
$endgroup$
– Omnomnomnom
Jan 2 at 16:15
$begingroup$
No. $mathcal P$ is the set of all $P$ satisfying $P^TP = I_d$, where $I_d$ is the $d times d$ identity matrix.
$endgroup$
– Omnomnomnom
Jan 2 at 16:15
$begingroup$
But $P$ is $d times k$, shouldn't be $P^TP=I_k$?
$endgroup$
– Saeed
Jan 2 at 19:19
$begingroup$
But $P$ is $d times k$, shouldn't be $P^TP=I_k$?
$endgroup$
– Saeed
Jan 2 at 19:19
$begingroup$
If $Phi$ goes from $X$ to $Y$, then $P$ should be $ktimes d$.
$endgroup$
– Omnomnomnom
Jan 2 at 19:35
$begingroup$
If $Phi$ goes from $X$ to $Y$, then $P$ should be $ktimes d$.
$endgroup$
– Omnomnomnom
Jan 2 at 19:35
$begingroup$
@Omnomnomnom: Yes. you are right. I edited it. $P$ can be any matrix with $d$ colums including orthonormal vectors in $mathbb{R}^k$?
$endgroup$
– Saeed
Jan 2 at 19:57
$begingroup$
@Omnomnomnom: Yes. you are right. I edited it. $P$ can be any matrix with $d$ colums including orthonormal vectors in $mathbb{R}^k$?
$endgroup$
– Saeed
Jan 2 at 19:57
$begingroup$
that's exactly right
$endgroup$
– Omnomnomnom
Jan 2 at 20:02
$begingroup$
that's exactly right
$endgroup$
– Omnomnomnom
Jan 2 at 20:02
|
show 2 more comments
0
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$begingroup$
No. $mathcal P$ is the set of all $P$ satisfying $P^TP = I_d$, where $I_d$ is the $d times d$ identity matrix.
$endgroup$
– Omnomnomnom
Jan 2 at 16:15
$begingroup$
But $P$ is $d times k$, shouldn't be $P^TP=I_k$?
$endgroup$
– Saeed
Jan 2 at 19:19
$begingroup$
If $Phi$ goes from $X$ to $Y$, then $P$ should be $ktimes d$.
$endgroup$
– Omnomnomnom
Jan 2 at 19:35
$begingroup$
@Omnomnomnom: Yes. you are right. I edited it. $P$ can be any matrix with $d$ colums including orthonormal vectors in $mathbb{R}^k$?
$endgroup$
– Saeed
Jan 2 at 19:57
$begingroup$
that's exactly right
$endgroup$
– Omnomnomnom
Jan 2 at 20:02