A question about rotation of a plane
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Suppose for any subspace $F$ of $mathbb{R}^d$(with the usual Euclidean norm), $pi_F$ denote the orthogonal projection onto $F$. Let $R$ be a rotation of $F$ and $F^prime = RF$. Prove that $forall p in mathbb{R}^d$,
$d(pi_F(p), pi_{F^prime}(p)) leq d(pi_F(p), Rpi_F(p))$.
This relation is used in this paper, equation 4.5.
linear-algebra euclidean-geometry rotations
asked Dec 2 '18 at 18:59
Sudipta RoySudipta Roy
29518
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add a comment |
$begingroup$
Suppose for any subspace $F$ of $mathbb{R}^d$(with the usual Euclidean norm), $pi_F$ denote the orthogonal projection onto $F$. Let $R$ be a rotation of $F$ and $F^prime = RF$. Prove that $forall p in mathbb{R}^d$,
$d(pi_F(p), pi_{F^prime}(p)) leq d(pi_F(p), Rpi_F(p))$.
This relation is used in this paper, equation 4.5.
linear-algebra euclidean-geometry rotations
asked Dec 2 '18 at 18:59
Sudipta RoySudipta Roy
29518
$endgroup$
$begingroup$
Are you sure that the inequality is correct? Take $P=(x,2x)$ in $mathbb{R}^2$, with $F$ the $x-$axis and $R$ the rotation of $pi/2$.
$endgroup$
– Emilio Novati
Dec 2 '18 at 20:46
add a comment |
$begingroup$
Suppose for any subspace $F$ of $mathbb{R}^d$(with the usual Euclidean norm), $pi_F$ denote the orthogonal projection onto $F$. Let $R$ be a rotation of $F$ and $F^prime = RF$. Prove that $forall p in mathbb{R}^d$,
$d(pi_F(p), pi_{F^prime}(p)) leq d(pi_F(p), Rpi_F(p))$.
This relation is used in this paper, equation 4.5.
linear-algebra euclidean-geometry rotations
asked Dec 2 '18 at 18:59
Sudipta RoySudipta Roy
29518
$endgroup$
Suppose for any subspace $F$ of $mathbb{R}^d$(with the usual Euclidean norm), $pi_F$ denote the orthogonal projection onto $F$. Let $R$ be a rotation of $F$ and $F^prime = RF$. Prove that $forall p in mathbb{R}^d$,
$d(pi_F(p), pi_{F^prime}(p)) leq d(pi_F(p), Rpi_F(p))$.
This relation is used in this paper, equation 4.5.
linear-algebra euclidean-geometry rotations
linear-algebra euclidean-geometry rotations
asked Dec 2 '18 at 18:59
Sudipta RoySudipta Roy
29518
asked Dec 2 '18 at 18:59
Sudipta RoySudipta Roy
29518
asked Dec 2 '18 at 18:59
Sudipta RoySudipta Roy
29518
asked Dec 2 '18 at 18:59
Sudipta RoySudipta Roy
29518
asked Dec 2 '18 at 18:59
Sudipta RoySudipta Roy
29518
29518
$begingroup$
Are you sure that the inequality is correct? Take $P=(x,2x)$ in $mathbb{R}^2$, with $F$ the $x-$axis and $R$ the rotation of $pi/2$.
$endgroup$
– Emilio Novati
Dec 2 '18 at 20:46
add a comment |
$begingroup$
Are you sure that the inequality is correct? Take $P=(x,2x)$ in $mathbb{R}^2$, with $F$ the $x-$axis and $R$ the rotation of $pi/2$.
$endgroup$
– Emilio Novati
Dec 2 '18 at 20:46
$begingroup$
Are you sure that the inequality is correct? Take $P=(x,2x)$ in $mathbb{R}^2$, with $F$ the $x-$axis and $R$ the rotation of $pi/2$.
$endgroup$
– Emilio Novati
Dec 2 '18 at 20:46
$begingroup$
Are you sure that the inequality is correct? Take $P=(x,2x)$ in $mathbb{R}^2$, with $F$ the $x-$axis and $R$ the rotation of $pi/2$.
$endgroup$
– Emilio Novati
Dec 2 '18 at 20:46
add a comment |
1 Answer
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$begingroup$
If I well understand the question the inequality is not true in general.
As suggested in my comment a simple counterexample can be done in $mathbb{R}^2$, as we can see in this figure.
Here the line $OC$ is the subspace $F$ and the line $OD$ is the subspace $F'=R_alpha F$. For a point $P$ we have the projections $E_1=pi_F(P)$ and $E_2=pi_{F'}(P)$, and the point $G=R_alpha(E_1)$.
We can see that $d(E_1E_2)=d(E_1Q)$ if $P$ is a point of the bisector of the angle $alpha$, otherwise we can have $d(E_1E_2)>d(E_1Q)$ or $d(E_1E_2)<d(E_1Q)$ depending on the position of $P$.
answered Dec 14 '18 at 13:46
Emilio NovatiEmilio Novati
51.6k43473
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add a comment |
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1 Answer
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$begingroup$
If I well understand the question the inequality is not true in general.
As suggested in my comment a simple counterexample can be done in $mathbb{R}^2$, as we can see in this figure.
Here the line $OC$ is the subspace $F$ and the line $OD$ is the subspace $F'=R_alpha F$. For a point $P$ we have the projections $E_1=pi_F(P)$ and $E_2=pi_{F'}(P)$, and the point $G=R_alpha(E_1)$.
We can see that $d(E_1E_2)=d(E_1Q)$ if $P$ is a point of the bisector of the angle $alpha$, otherwise we can have $d(E_1E_2)>d(E_1Q)$ or $d(E_1E_2)<d(E_1Q)$ depending on the position of $P$.
answered Dec 14 '18 at 13:46
Emilio NovatiEmilio Novati
51.6k43473
$endgroup$
add a comment |
$begingroup$
If I well understand the question the inequality is not true in general.
As suggested in my comment a simple counterexample can be done in $mathbb{R}^2$, as we can see in this figure.
Here the line $OC$ is the subspace $F$ and the line $OD$ is the subspace $F'=R_alpha F$. For a point $P$ we have the projections $E_1=pi_F(P)$ and $E_2=pi_{F'}(P)$, and the point $G=R_alpha(E_1)$.
We can see that $d(E_1E_2)=d(E_1Q)$ if $P$ is a point of the bisector of the angle $alpha$, otherwise we can have $d(E_1E_2)>d(E_1Q)$ or $d(E_1E_2)<d(E_1Q)$ depending on the position of $P$.
answered Dec 14 '18 at 13:46
Emilio NovatiEmilio Novati
51.6k43473
$endgroup$
add a comment |
$begingroup$
If I well understand the question the inequality is not true in general.
As suggested in my comment a simple counterexample can be done in $mathbb{R}^2$, as we can see in this figure.
Here the line $OC$ is the subspace $F$ and the line $OD$ is the subspace $F'=R_alpha F$. For a point $P$ we have the projections $E_1=pi_F(P)$ and $E_2=pi_{F'}(P)$, and the point $G=R_alpha(E_1)$.
We can see that $d(E_1E_2)=d(E_1Q)$ if $P$ is a point of the bisector of the angle $alpha$, otherwise we can have $d(E_1E_2)>d(E_1Q)$ or $d(E_1E_2)<d(E_1Q)$ depending on the position of $P$.
answered Dec 14 '18 at 13:46
Emilio NovatiEmilio Novati
51.6k43473
$endgroup$
If I well understand the question the inequality is not true in general.
As suggested in my comment a simple counterexample can be done in $mathbb{R}^2$, as we can see in this figure.
Here the line $OC$ is the subspace $F$ and the line $OD$ is the subspace $F'=R_alpha F$. For a point $P$ we have the projections $E_1=pi_F(P)$ and $E_2=pi_{F'}(P)$, and the point $G=R_alpha(E_1)$.
We can see that $d(E_1E_2)=d(E_1Q)$ if $P$ is a point of the bisector of the angle $alpha$, otherwise we can have $d(E_1E_2)>d(E_1Q)$ or $d(E_1E_2)<d(E_1Q)$ depending on the position of $P$.
answered Dec 14 '18 at 13:46
Emilio NovatiEmilio Novati
51.6k43473
answered Dec 14 '18 at 13:46
Emilio NovatiEmilio Novati
51.6k43473
answered Dec 14 '18 at 13:46
Emilio NovatiEmilio Novati
51.6k43473
answered Dec 14 '18 at 13:46
Emilio NovatiEmilio Novati
51.6k43473
51.6k43473
add a comment |
add a comment |
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$begingroup$
Are you sure that the inequality is correct? Take $P=(x,2x)$ in $mathbb{R}^2$, with $F$ the $x-$axis and $R$ the rotation of $pi/2$.
$endgroup$
– Emilio Novati
Dec 2 '18 at 20:46