What is the set of all isometric matrix in $mathbb{R}^{k times d}$?












1












$begingroup$


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?










share|cite|improve this question











$endgroup$








  • 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
















1












$begingroup$


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?










share|cite|improve this question











$endgroup$








  • 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














1












1








1





$begingroup$


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?










share|cite|improve this question











$endgroup$




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






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share|cite|improve this question













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share|cite|improve this question








edited Jan 2 at 19:50







Saeed

















asked Jan 2 at 15:54









SaeedSaeed

1,124310




1,124310








  • 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




    $begingroup$
    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










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