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Minimal rotation matrix for two vectors in $mathbbR^n$
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Suppose we have two normal vectors $v, u in mathbbR^n, |u|_2=1, |v|_2=1$. We would like to find a rotation matrix $RinmathbbR^n times n$ that satisfies
$R u = v$. This clearly does not uniquely determine matrix $R$ because there are only $n$ equations while there are $n(n-1)/2$ degrees of freedom for $R$. However we can make it unique by forcing it to only rotate vectors in the $Spanu,v$. In other words:
$$forall w in Spanu,v^perp : R w = w$$
Intuitively, it only rotates vectors in the $Spanu,v$ plane while leaving every other direction untouched. In this sense, the rotation matrix is minimal.
In the very simple case that $Spanu,v = Spane_i,e_j$, i.e., the linear space spanned by $u,v$ is the same as the one spanned by $e_i, e_j$ we can compute $R$ as follows: $R = I_n - M$, in which $Min mathbbR^ntimes n$ is only zeros except for its $i$-th and $j$-th rows and columns:
$$M_i, jtimes i, j = beginbmatrix
1-cos theta & sin theta\
-sintheta & 1-costheta \
endbmatrix$$
in which $theta = cos^-1 langle u, v rangle$
Is there a way to compute the rotation matrix for general normal vectors $u, v$?
linear-algebra rotations
asked Jul 25 at 16:15
kvphxga
1163
add a comment |Â
up vote
3
down vote
favorite
Suppose we have two normal vectors $v, u in mathbbR^n, |u|_2=1, |v|_2=1$. We would like to find a rotation matrix $RinmathbbR^n times n$ that satisfies
$R u = v$. This clearly does not uniquely determine matrix $R$ because there are only $n$ equations while there are $n(n-1)/2$ degrees of freedom for $R$. However we can make it unique by forcing it to only rotate vectors in the $Spanu,v$. In other words:
$$forall w in Spanu,v^perp : R w = w$$
Intuitively, it only rotates vectors in the $Spanu,v$ plane while leaving every other direction untouched. In this sense, the rotation matrix is minimal.
In the very simple case that $Spanu,v = Spane_i,e_j$, i.e., the linear space spanned by $u,v$ is the same as the one spanned by $e_i, e_j$ we can compute $R$ as follows: $R = I_n - M$, in which $Min mathbbR^ntimes n$ is only zeros except for its $i$-th and $j$-th rows and columns:
$$M_i, jtimes i, j = beginbmatrix
1-cos theta & sin theta\
-sintheta & 1-costheta \
endbmatrix$$
in which $theta = cos^-1 langle u, v rangle$
Is there a way to compute the rotation matrix for general normal vectors $u, v$?
linear-algebra rotations
asked Jul 25 at 16:15
kvphxga
1163
Your rotation can be computed as a product of two Householder reflections
– Omnomnomnom
Jul 25 at 16:31
@MikeEarnest that's what I thought at first too. But it seems $R R^T = I + v v^T - u u^T$, while it's supposed to be $I$, am I right?
– kvphxga
Jul 25 at 16:32
1
The first reflection is in the angle bisector of $u$ and $v$.
– amd
Jul 25 at 16:57
1
Have a look at this math.stackexchange.com/a/598782/485657
– Mauricio Cele Lopez Belon
Jul 25 at 21:55
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Suppose we have two normal vectors $v, u in mathbbR^n, |u|_2=1, |v|_2=1$. We would like to find a rotation matrix $RinmathbbR^n times n$ that satisfies
$R u = v$. This clearly does not uniquely determine matrix $R$ because there are only $n$ equations while there are $n(n-1)/2$ degrees of freedom for $R$. However we can make it unique by forcing it to only rotate vectors in the $Spanu,v$. In other words:
$$forall w in Spanu,v^perp : R w = w$$
Intuitively, it only rotates vectors in the $Spanu,v$ plane while leaving every other direction untouched. In this sense, the rotation matrix is minimal.
In the very simple case that $Spanu,v = Spane_i,e_j$, i.e., the linear space spanned by $u,v$ is the same as the one spanned by $e_i, e_j$ we can compute $R$ as follows: $R = I_n - M$, in which $Min mathbbR^ntimes n$ is only zeros except for its $i$-th and $j$-th rows and columns:
$$M_i, jtimes i, j = beginbmatrix
1-cos theta & sin theta\
-sintheta & 1-costheta \
endbmatrix$$
in which $theta = cos^-1 langle u, v rangle$
Is there a way to compute the rotation matrix for general normal vectors $u, v$?
linear-algebra rotations
asked Jul 25 at 16:15
kvphxga
1163
Suppose we have two normal vectors $v, u in mathbbR^n, |u|_2=1, |v|_2=1$. We would like to find a rotation matrix $RinmathbbR^n times n$ that satisfies
$R u = v$. This clearly does not uniquely determine matrix $R$ because there are only $n$ equations while there are $n(n-1)/2$ degrees of freedom for $R$. However we can make it unique by forcing it to only rotate vectors in the $Spanu,v$. In other words:
$$forall w in Spanu,v^perp : R w = w$$
Intuitively, it only rotates vectors in the $Spanu,v$ plane while leaving every other direction untouched. In this sense, the rotation matrix is minimal.
In the very simple case that $Spanu,v = Spane_i,e_j$, i.e., the linear space spanned by $u,v$ is the same as the one spanned by $e_i, e_j$ we can compute $R$ as follows: $R = I_n - M$, in which $Min mathbbR^ntimes n$ is only zeros except for its $i$-th and $j$-th rows and columns:
$$M_i, jtimes i, j = beginbmatrix
1-cos theta & sin theta\
-sintheta & 1-costheta \
endbmatrix$$
in which $theta = cos^-1 langle u, v rangle$
Is there a way to compute the rotation matrix for general normal vectors $u, v$?
linear-algebra rotations
asked Jul 25 at 16:15
kvphxga
1163
edited Jul 25 at 16:34
asked Jul 25 at 16:15
kvphxga
1163
asked Jul 25 at 16:15
kvphxga
1163
asked Jul 25 at 16:15
kvphxga
1163
1163
Your rotation can be computed as a product of two Householder reflections
– Omnomnomnom
Jul 25 at 16:31
@MikeEarnest that's what I thought at first too. But it seems $R R^T = I + v v^T - u u^T$, while it's supposed to be $I$, am I right?
– kvphxga
Jul 25 at 16:32
1
The first reflection is in the angle bisector of $u$ and $v$.
– amd
Jul 25 at 16:57
1
Have a look at this math.stackexchange.com/a/598782/485657
– Mauricio Cele Lopez Belon
Jul 25 at 21:55
add a comment |Â
Your rotation can be computed as a product of two Householder reflections
– Omnomnomnom
Jul 25 at 16:31
@MikeEarnest that's what I thought at first too. But it seems $R R^T = I + v v^T - u u^T$, while it's supposed to be $I$, am I right?
– kvphxga
Jul 25 at 16:32
1
The first reflection is in the angle bisector of $u$ and $v$.
– amd
Jul 25 at 16:57
1
Have a look at this math.stackexchange.com/a/598782/485657
– Mauricio Cele Lopez Belon
Jul 25 at 21:55
Your rotation can be computed as a product of two Householder reflections
– Omnomnomnom
Jul 25 at 16:31
Your rotation can be computed as a product of two Householder reflections
– Omnomnomnom
Jul 25 at 16:31
@MikeEarnest that's what I thought at first too. But it seems $R R^T = I + v v^T - u u^T$, while it's supposed to be $I$, am I right?
– kvphxga
Jul 25 at 16:32
@MikeEarnest that's what I thought at first too. But it seems $R R^T = I + v v^T - u u^T$, while it's supposed to be $I$, am I right?
– kvphxga
Jul 25 at 16:32
1
1
The first reflection is in the angle bisector of $u$ and $v$.
– amd
Jul 25 at 16:57
The first reflection is in the angle bisector of $u$ and $v$.
– amd
Jul 25 at 16:57
1
1
Have a look at this math.stackexchange.com/a/598782/485657
– Mauricio Cele Lopez Belon
Jul 25 at 21:55
Have a look at this math.stackexchange.com/a/598782/485657
– Mauricio Cele Lopez Belon
Jul 25 at 21:55
add a comment |Â
1 Answer
1
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0
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Too long for a comment... Let $a^top$ be the projection of $a$ onto the span of $u$ and $v$ such that $a=a^top+a^perp$. Then rotate $a^top$ by $theta$ and add $a^perp$ again.
Alternatively: Extend $u,v$ to a basis. We want $u$ to be mapped to $v$ and $v$ mapped to $2langle u,vrangle v-u$. In respect to this basis the matrix of $R$ is the identity except for the upper left $2times2$ entries; they read
$$beginpmatrix0&-1\ 1&2langle u,vrangleendpmatrix.$$
Even more alternatively: call $c=langle u,vrangle$. Now compute as usual — with a little help of the Gram-matrix of $u$ and $v$ — the tangential component
$$a^top=frac11-c^2bigl((langle u,arangle-clangle v,arangle)u
+(langle v,arangle-clangle u,arangle)vbigr).$$
To obtain $Ra^top$ change $u$ to $v$ and $v$ to $2cv-u$. Now compute
$$Ra=a^perp+Ra^top=a-a^top+Ra^top
=a-frac11+cbigl(langle u+v,arangle u+langle v-(1+2c)u,arangle vbigr).$$
answered Jul 25 at 17:09
Michael Hoppe
9,55631432
doesn't $a=⟨a,u⟩u+⟨a,v⟩v+a^perp$ imply $u$ and $v$ must be orthogonal?
– kvphxga
Jul 25 at 17:21
Yes, you’re right, I will edit.
– Michael Hoppe
Jul 25 at 17:40
I actually thought of this and tried to find a closed form, but it turned out to be surprisingly messy. But maybe I miscalculated something.
– kvphxga
Jul 25 at 18:30
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Too long for a comment... Let $a^top$ be the projection of $a$ onto the span of $u$ and $v$ such that $a=a^top+a^perp$. Then rotate $a^top$ by $theta$ and add $a^perp$ again.
Alternatively: Extend $u,v$ to a basis. We want $u$ to be mapped to $v$ and $v$ mapped to $2langle u,vrangle v-u$. In respect to this basis the matrix of $R$ is the identity except for the upper left $2times2$ entries; they read
$$beginpmatrix0&-1\ 1&2langle u,vrangleendpmatrix.$$
Even more alternatively: call $c=langle u,vrangle$. Now compute as usual — with a little help of the Gram-matrix of $u$ and $v$ — the tangential component
$$a^top=frac11-c^2bigl((langle u,arangle-clangle v,arangle)u
+(langle v,arangle-clangle u,arangle)vbigr).$$
To obtain $Ra^top$ change $u$ to $v$ and $v$ to $2cv-u$. Now compute
$$Ra=a^perp+Ra^top=a-a^top+Ra^top
=a-frac11+cbigl(langle u+v,arangle u+langle v-(1+2c)u,arangle vbigr).$$
answered Jul 25 at 17:09
Michael Hoppe
9,55631432
doesn't $a=⟨a,u⟩u+⟨a,v⟩v+a^perp$ imply $u$ and $v$ must be orthogonal?
– kvphxga
Jul 25 at 17:21
Yes, you’re right, I will edit.
– Michael Hoppe
Jul 25 at 17:40
I actually thought of this and tried to find a closed form, but it turned out to be surprisingly messy. But maybe I miscalculated something.
– kvphxga
Jul 25 at 18:30
add a comment |Â
up vote
0
down vote
Too long for a comment... Let $a^top$ be the projection of $a$ onto the span of $u$ and $v$ such that $a=a^top+a^perp$. Then rotate $a^top$ by $theta$ and add $a^perp$ again.
Alternatively: Extend $u,v$ to a basis. We want $u$ to be mapped to $v$ and $v$ mapped to $2langle u,vrangle v-u$. In respect to this basis the matrix of $R$ is the identity except for the upper left $2times2$ entries; they read
$$beginpmatrix0&-1\ 1&2langle u,vrangleendpmatrix.$$
Even more alternatively: call $c=langle u,vrangle$. Now compute as usual — with a little help of the Gram-matrix of $u$ and $v$ — the tangential component
$$a^top=frac11-c^2bigl((langle u,arangle-clangle v,arangle)u
+(langle v,arangle-clangle u,arangle)vbigr).$$
To obtain $Ra^top$ change $u$ to $v$ and $v$ to $2cv-u$. Now compute
$$Ra=a^perp+Ra^top=a-a^top+Ra^top
=a-frac11+cbigl(langle u+v,arangle u+langle v-(1+2c)u,arangle vbigr).$$
answered Jul 25 at 17:09
Michael Hoppe
9,55631432
doesn't $a=⟨a,u⟩u+⟨a,v⟩v+a^perp$ imply $u$ and $v$ must be orthogonal?
– kvphxga
Jul 25 at 17:21
Yes, you’re right, I will edit.
– Michael Hoppe
Jul 25 at 17:40
I actually thought of this and tried to find a closed form, but it turned out to be surprisingly messy. But maybe I miscalculated something.
– kvphxga
Jul 25 at 18:30
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Too long for a comment... Let $a^top$ be the projection of $a$ onto the span of $u$ and $v$ such that $a=a^top+a^perp$. Then rotate $a^top$ by $theta$ and add $a^perp$ again.
Alternatively: Extend $u,v$ to a basis. We want $u$ to be mapped to $v$ and $v$ mapped to $2langle u,vrangle v-u$. In respect to this basis the matrix of $R$ is the identity except for the upper left $2times2$ entries; they read
$$beginpmatrix0&-1\ 1&2langle u,vrangleendpmatrix.$$
Even more alternatively: call $c=langle u,vrangle$. Now compute as usual — with a little help of the Gram-matrix of $u$ and $v$ — the tangential component
$$a^top=frac11-c^2bigl((langle u,arangle-clangle v,arangle)u
+(langle v,arangle-clangle u,arangle)vbigr).$$
To obtain $Ra^top$ change $u$ to $v$ and $v$ to $2cv-u$. Now compute
$$Ra=a^perp+Ra^top=a-a^top+Ra^top
=a-frac11+cbigl(langle u+v,arangle u+langle v-(1+2c)u,arangle vbigr).$$
answered Jul 25 at 17:09
Michael Hoppe
9,55631432
Too long for a comment... Let $a^top$ be the projection of $a$ onto the span of $u$ and $v$ such that $a=a^top+a^perp$. Then rotate $a^top$ by $theta$ and add $a^perp$ again.
Alternatively: Extend $u,v$ to a basis. We want $u$ to be mapped to $v$ and $v$ mapped to $2langle u,vrangle v-u$. In respect to this basis the matrix of $R$ is the identity except for the upper left $2times2$ entries; they read
$$beginpmatrix0&-1\ 1&2langle u,vrangleendpmatrix.$$
Even more alternatively: call $c=langle u,vrangle$. Now compute as usual — with a little help of the Gram-matrix of $u$ and $v$ — the tangential component
$$a^top=frac11-c^2bigl((langle u,arangle-clangle v,arangle)u
+(langle v,arangle-clangle u,arangle)vbigr).$$
To obtain $Ra^top$ change $u$ to $v$ and $v$ to $2cv-u$. Now compute
$$Ra=a^perp+Ra^top=a-a^top+Ra^top
=a-frac11+cbigl(langle u+v,arangle u+langle v-(1+2c)u,arangle vbigr).$$
answered Jul 25 at 17:09
Michael Hoppe
9,55631432
edited Jul 26 at 16:54
answered Jul 25 at 17:09
Michael Hoppe
9,55631432
answered Jul 25 at 17:09
Michael Hoppe
9,55631432
answered Jul 25 at 17:09
Michael Hoppe
9,55631432
9,55631432
doesn't $a=⟨a,u⟩u+⟨a,v⟩v+a^perp$ imply $u$ and $v$ must be orthogonal?
– kvphxga
Jul 25 at 17:21
Yes, you’re right, I will edit.
– Michael Hoppe
Jul 25 at 17:40
I actually thought of this and tried to find a closed form, but it turned out to be surprisingly messy. But maybe I miscalculated something.
– kvphxga
Jul 25 at 18:30
add a comment |Â
doesn't $a=⟨a,u⟩u+⟨a,v⟩v+a^perp$ imply $u$ and $v$ must be orthogonal?
– kvphxga
Jul 25 at 17:21
Yes, you’re right, I will edit.
– Michael Hoppe
Jul 25 at 17:40
I actually thought of this and tried to find a closed form, but it turned out to be surprisingly messy. But maybe I miscalculated something.
– kvphxga
Jul 25 at 18:30
doesn't $a=⟨a,u⟩u+⟨a,v⟩v+a^perp$ imply $u$ and $v$ must be orthogonal?
– kvphxga
Jul 25 at 17:21
doesn't $a=⟨a,u⟩u+⟨a,v⟩v+a^perp$ imply $u$ and $v$ must be orthogonal?
– kvphxga
Jul 25 at 17:21
Yes, you’re right, I will edit.
– Michael Hoppe
Jul 25 at 17:40
Yes, you’re right, I will edit.
– Michael Hoppe
Jul 25 at 17:40
I actually thought of this and tried to find a closed form, but it turned out to be surprisingly messy. But maybe I miscalculated something.
– kvphxga
Jul 25 at 18:30
I actually thought of this and tried to find a closed form, but it turned out to be surprisingly messy. But maybe I miscalculated something.
– kvphxga
Jul 25 at 18:30
add a comment |Â
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Your rotation can be computed as a product of two Householder reflections
– Omnomnomnom
Jul 25 at 16:31
@MikeEarnest that's what I thought at first too. But it seems $R R^T = I + v v^T - u u^T$, while it's supposed to be $I$, am I right?
– kvphxga
Jul 25 at 16:32
1
The first reflection is in the angle bisector of $u$ and $v$.
– amd
Jul 25 at 16:57
1
Have a look at this math.stackexchange.com/a/598782/485657
– Mauricio Cele Lopez Belon
Jul 25 at 21:55