How to block diagonalize a real skew-symmetric matrix of 3*3
$begingroup$
Suppose $t = [t_1,t_2,t_3]^Tin mathbb R^3,t neq 0$. Then define
$$t^{land} = begin{bmatrix} 0 & -t_3 & t_2 \
t_3 & 0 & -t_1\
-t_2 & t_1 & 0end{bmatrix},
Z= begin{bmatrix} 0 & 1 \
-1 & 0end{bmatrix}.$$
According to exponential map, there exist a corresponding rotation matrix
$$R = exp(t^{land})$$
if $t^{land}$ can be block diagonalized, that is
$$t^{land} = Ubegin{bmatrix} aZ & 0 \
0 & 0end{bmatrix}U^T,a>0$$
then we get $R = exp(t^{land})=Ubegin{bmatrix}cos(|t|_2)&sin(|t|_2) & 0\
-sin(|t|_2)&cos(|t|_2)&0 \
0 & 0 &1end{bmatrix}U^T,a=|t|_2$.
(refer to R.Hartley & A.Zisserman Multiple View Geometry in Computer Vision page 583 to 584.)
So how to do block diagonalization in 3*3 case? And what is $U$ like?
matrices lie-groups lie-algebras numerical-linear-algebra rotations
asked Dec 7 '18 at 9:37
FinleyFinley
375213
$endgroup$
add a comment |
$begingroup$
Suppose $t = [t_1,t_2,t_3]^Tin mathbb R^3,t neq 0$. Then define
$$t^{land} = begin{bmatrix} 0 & -t_3 & t_2 \
t_3 & 0 & -t_1\
-t_2 & t_1 & 0end{bmatrix},
Z= begin{bmatrix} 0 & 1 \
-1 & 0end{bmatrix}.$$
According to exponential map, there exist a corresponding rotation matrix
$$R = exp(t^{land})$$
if $t^{land}$ can be block diagonalized, that is
$$t^{land} = Ubegin{bmatrix} aZ & 0 \
0 & 0end{bmatrix}U^T,a>0$$
then we get $R = exp(t^{land})=Ubegin{bmatrix}cos(|t|_2)&sin(|t|_2) & 0\
-sin(|t|_2)&cos(|t|_2)&0 \
0 & 0 &1end{bmatrix}U^T,a=|t|_2$.
(refer to R.Hartley & A.Zisserman Multiple View Geometry in Computer Vision page 583 to 584.)
So how to do block diagonalization in 3*3 case? And what is $U$ like?
matrices lie-groups lie-algebras numerical-linear-algebra rotations
asked Dec 7 '18 at 9:37
FinleyFinley
375213
$endgroup$
1
$begingroup$
Extend $w=t/|t|_2$ to an orthonormal basis ${u,v,w}$. Then take $U=pmatrix{u,v,w}$.
$endgroup$
– user1551
Dec 7 '18 at 9:40
$begingroup$
@user1551 hello, i apply your thread(see following post of mine) while leads to a contradiction, can you help me out?
$endgroup$
– Finley
Dec 7 '18 at 15:37
$begingroup$
I didn't noticed that you need $a>0$. If you want to enforce this constraint and you find that with your chosen $u,v$ the cross product $vtimes u$ is $-w$ rather than $w$, you may simply interchange $u$ and $v$. Alternatively, replace $v$ by $-v$.
$endgroup$
– user1551
Dec 7 '18 at 16:39
$begingroup$
@user1551 But $det(U) = -1$ once I interchange $u text{ and } v$ or negate $v$. Is it reasonable for an orthonormal matrix?
$endgroup$
– Finley
Dec 8 '18 at 2:11
$begingroup$
Why not? Your question doesn't require that $det(U)=1$. Anyway, since it sets $Z$ to $pmatrix{0&1\ -1&0}$ rather than $pmatrix{0&-1\ 1&0}$, the determinant of $U$ has to be $-1$: when $u$ is orthogonal to $t$, the effect of $ttimes u$ in the tangential direction is an anticlockwise rotation of $u$ about the directed axis $t$. However, $Z=pmatrix{0&1\ -1&0}$ represents a clockwise rotation. So, in the change-of-basis matrix $U$, you must incorporate a reflection to preserve the sign, meaning that $det(U)$ has to be $-1$.
$endgroup$
– user1551
Dec 8 '18 at 6:07
add a comment |
$begingroup$
Suppose $t = [t_1,t_2,t_3]^Tin mathbb R^3,t neq 0$. Then define
$$t^{land} = begin{bmatrix} 0 & -t_3 & t_2 \
t_3 & 0 & -t_1\
-t_2 & t_1 & 0end{bmatrix},
Z= begin{bmatrix} 0 & 1 \
-1 & 0end{bmatrix}.$$
According to exponential map, there exist a corresponding rotation matrix
$$R = exp(t^{land})$$
if $t^{land}$ can be block diagonalized, that is
$$t^{land} = Ubegin{bmatrix} aZ & 0 \
0 & 0end{bmatrix}U^T,a>0$$
then we get $R = exp(t^{land})=Ubegin{bmatrix}cos(|t|_2)&sin(|t|_2) & 0\
-sin(|t|_2)&cos(|t|_2)&0 \
0 & 0 &1end{bmatrix}U^T,a=|t|_2$.
(refer to R.Hartley & A.Zisserman Multiple View Geometry in Computer Vision page 583 to 584.)
So how to do block diagonalization in 3*3 case? And what is $U$ like?
matrices lie-groups lie-algebras numerical-linear-algebra rotations
asked Dec 7 '18 at 9:37
FinleyFinley
375213
$endgroup$
Suppose $t = [t_1,t_2,t_3]^Tin mathbb R^3,t neq 0$. Then define
$$t^{land} = begin{bmatrix} 0 & -t_3 & t_2 \
t_3 & 0 & -t_1\
-t_2 & t_1 & 0end{bmatrix},
Z= begin{bmatrix} 0 & 1 \
-1 & 0end{bmatrix}.$$
According to exponential map, there exist a corresponding rotation matrix
$$R = exp(t^{land})$$
if $t^{land}$ can be block diagonalized, that is
$$t^{land} = Ubegin{bmatrix} aZ & 0 \
0 & 0end{bmatrix}U^T,a>0$$
then we get $R = exp(t^{land})=Ubegin{bmatrix}cos(|t|_2)&sin(|t|_2) & 0\
-sin(|t|_2)&cos(|t|_2)&0 \
0 & 0 &1end{bmatrix}U^T,a=|t|_2$.
(refer to R.Hartley & A.Zisserman Multiple View Geometry in Computer Vision page 583 to 584.)
So how to do block diagonalization in 3*3 case? And what is $U$ like?
matrices lie-groups lie-algebras numerical-linear-algebra rotations
matrices lie-groups lie-algebras numerical-linear-algebra rotations
asked Dec 7 '18 at 9:37
FinleyFinley
375213
asked Dec 7 '18 at 9:37
FinleyFinley
375213
asked Dec 7 '18 at 9:37
FinleyFinley
375213
asked Dec 7 '18 at 9:37
FinleyFinley
375213
asked Dec 7 '18 at 9:37
FinleyFinley
375213
375213
1
$begingroup$
Extend $w=t/|t|_2$ to an orthonormal basis ${u,v,w}$. Then take $U=pmatrix{u,v,w}$.
$endgroup$
– user1551
Dec 7 '18 at 9:40
$begingroup$
@user1551 hello, i apply your thread(see following post of mine) while leads to a contradiction, can you help me out?
$endgroup$
– Finley
Dec 7 '18 at 15:37
$begingroup$
I didn't noticed that you need $a>0$. If you want to enforce this constraint and you find that with your chosen $u,v$ the cross product $vtimes u$ is $-w$ rather than $w$, you may simply interchange $u$ and $v$. Alternatively, replace $v$ by $-v$.
$endgroup$
– user1551
Dec 7 '18 at 16:39
$begingroup$
@user1551 But $det(U) = -1$ once I interchange $u text{ and } v$ or negate $v$. Is it reasonable for an orthonormal matrix?
$endgroup$
– Finley
Dec 8 '18 at 2:11
$begingroup$
Why not? Your question doesn't require that $det(U)=1$. Anyway, since it sets $Z$ to $pmatrix{0&1\ -1&0}$ rather than $pmatrix{0&-1\ 1&0}$, the determinant of $U$ has to be $-1$: when $u$ is orthogonal to $t$, the effect of $ttimes u$ in the tangential direction is an anticlockwise rotation of $u$ about the directed axis $t$. However, $Z=pmatrix{0&1\ -1&0}$ represents a clockwise rotation. So, in the change-of-basis matrix $U$, you must incorporate a reflection to preserve the sign, meaning that $det(U)$ has to be $-1$.
$endgroup$
– user1551
Dec 8 '18 at 6:07
add a comment |
1
$begingroup$
Extend $w=t/|t|_2$ to an orthonormal basis ${u,v,w}$. Then take $U=pmatrix{u,v,w}$.
$endgroup$
– user1551
Dec 7 '18 at 9:40
$begingroup$
@user1551 hello, i apply your thread(see following post of mine) while leads to a contradiction, can you help me out?
$endgroup$
– Finley
Dec 7 '18 at 15:37
$begingroup$
I didn't noticed that you need $a>0$. If you want to enforce this constraint and you find that with your chosen $u,v$ the cross product $vtimes u$ is $-w$ rather than $w$, you may simply interchange $u$ and $v$. Alternatively, replace $v$ by $-v$.
$endgroup$
– user1551
Dec 7 '18 at 16:39
$begingroup$
@user1551 But $det(U) = -1$ once I interchange $u text{ and } v$ or negate $v$. Is it reasonable for an orthonormal matrix?
$endgroup$
– Finley
Dec 8 '18 at 2:11
$begingroup$
Why not? Your question doesn't require that $det(U)=1$. Anyway, since it sets $Z$ to $pmatrix{0&1\ -1&0}$ rather than $pmatrix{0&-1\ 1&0}$, the determinant of $U$ has to be $-1$: when $u$ is orthogonal to $t$, the effect of $ttimes u$ in the tangential direction is an anticlockwise rotation of $u$ about the directed axis $t$. However, $Z=pmatrix{0&1\ -1&0}$ represents a clockwise rotation. So, in the change-of-basis matrix $U$, you must incorporate a reflection to preserve the sign, meaning that $det(U)$ has to be $-1$.
$endgroup$
– user1551
Dec 8 '18 at 6:07
1
1
$begingroup$
Extend $w=t/|t|_2$ to an orthonormal basis ${u,v,w}$. Then take $U=pmatrix{u,v,w}$.
$endgroup$
– user1551
Dec 7 '18 at 9:40
$begingroup$
Extend $w=t/|t|_2$ to an orthonormal basis ${u,v,w}$. Then take $U=pmatrix{u,v,w}$.
$endgroup$
– user1551
Dec 7 '18 at 9:40
$begingroup$
@user1551 hello, i apply your thread(see following post of mine) while leads to a contradiction, can you help me out?
$endgroup$
– Finley
Dec 7 '18 at 15:37
$begingroup$
@user1551 hello, i apply your thread(see following post of mine) while leads to a contradiction, can you help me out?
$endgroup$
– Finley
Dec 7 '18 at 15:37
$begingroup$
I didn't noticed that you need $a>0$. If you want to enforce this constraint and you find that with your chosen $u,v$ the cross product $vtimes u$ is $-w$ rather than $w$, you may simply interchange $u$ and $v$. Alternatively, replace $v$ by $-v$.
$endgroup$
– user1551
Dec 7 '18 at 16:39
$begingroup$
I didn't noticed that you need $a>0$. If you want to enforce this constraint and you find that with your chosen $u,v$ the cross product $vtimes u$ is $-w$ rather than $w$, you may simply interchange $u$ and $v$. Alternatively, replace $v$ by $-v$.
$endgroup$
– user1551
Dec 7 '18 at 16:39
$begingroup$
@user1551 But $det(U) = -1$ once I interchange $u text{ and } v$ or negate $v$. Is it reasonable for an orthonormal matrix?
$endgroup$
– Finley
Dec 8 '18 at 2:11
$begingroup$
@user1551 But $det(U) = -1$ once I interchange $u text{ and } v$ or negate $v$. Is it reasonable for an orthonormal matrix?
$endgroup$
– Finley
Dec 8 '18 at 2:11
$begingroup$
Why not? Your question doesn't require that $det(U)=1$. Anyway, since it sets $Z$ to $pmatrix{0&1\ -1&0}$ rather than $pmatrix{0&-1\ 1&0}$, the determinant of $U$ has to be $-1$: when $u$ is orthogonal to $t$, the effect of $ttimes u$ in the tangential direction is an anticlockwise rotation of $u$ about the directed axis $t$. However, $Z=pmatrix{0&1\ -1&0}$ represents a clockwise rotation. So, in the change-of-basis matrix $U$, you must incorporate a reflection to preserve the sign, meaning that $det(U)$ has to be $-1$.
$endgroup$
– user1551
Dec 8 '18 at 6:07
$begingroup$
Why not? Your question doesn't require that $det(U)=1$. Anyway, since it sets $Z$ to $pmatrix{0&1\ -1&0}$ rather than $pmatrix{0&-1\ 1&0}$, the determinant of $U$ has to be $-1$: when $u$ is orthogonal to $t$, the effect of $ttimes u$ in the tangential direction is an anticlockwise rotation of $u$ about the directed axis $t$. However, $Z=pmatrix{0&1\ -1&0}$ represents a clockwise rotation. So, in the change-of-basis matrix $U$, you must incorporate a reflection to preserve the sign, meaning that $det(U)$ has to be $-1$.
$endgroup$
– user1551
Dec 8 '18 at 6:07
add a comment |
1 Answer
1
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$begingroup$
Update
let $w=t/|t|$, then we take three orthonormal basis which form a orthonormal matrix $U = [u,v,w]$.Therefore
$$begin{align} Udiag(aZ,0)U^T&=[u,v,w]begin{bmatrix}
0 & a & 0 \
-a & 0 & 0 \
0 & 0 & 0end{bmatrix}begin{bmatrix} u^T \ v^T \ w^T end{bmatrix} \
&=a(uv^T-vu^T) \
&=abegin{bmatrix}
0 & u_1v_2-u_2v_1 & u_1v_3-u_3v_1 \
u_2v_1-u_1v_2 & 0 & u_2v_3-u_3v_2 \
u_3v_1-u_1v_3 & u_3v_2-u_2v_3 & 0end{bmatrix} \
&=a(v times u)^{land}end{align} $$
So above derivation leads to
$$t^{land} = a(v times u)^{land}$$that is
$$t = a*(-w) = |t|(-w)$$
but it contradicts with $t = |t|w$, who can help?
answered Dec 7 '18 at 15:35
FinleyFinley
375213
$endgroup$
add a comment |
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$begingroup$
Update
let $w=t/|t|$, then we take three orthonormal basis which form a orthonormal matrix $U = [u,v,w]$.Therefore
$$begin{align} Udiag(aZ,0)U^T&=[u,v,w]begin{bmatrix}
0 & a & 0 \
-a & 0 & 0 \
0 & 0 & 0end{bmatrix}begin{bmatrix} u^T \ v^T \ w^T end{bmatrix} \
&=a(uv^T-vu^T) \
&=abegin{bmatrix}
0 & u_1v_2-u_2v_1 & u_1v_3-u_3v_1 \
u_2v_1-u_1v_2 & 0 & u_2v_3-u_3v_2 \
u_3v_1-u_1v_3 & u_3v_2-u_2v_3 & 0end{bmatrix} \
&=a(v times u)^{land}end{align} $$
So above derivation leads to
$$t^{land} = a(v times u)^{land}$$that is
$$t = a*(-w) = |t|(-w)$$
but it contradicts with $t = |t|w$, who can help?
answered Dec 7 '18 at 15:35
FinleyFinley
375213
$endgroup$
add a comment |
$begingroup$
Update
let $w=t/|t|$, then we take three orthonormal basis which form a orthonormal matrix $U = [u,v,w]$.Therefore
$$begin{align} Udiag(aZ,0)U^T&=[u,v,w]begin{bmatrix}
0 & a & 0 \
-a & 0 & 0 \
0 & 0 & 0end{bmatrix}begin{bmatrix} u^T \ v^T \ w^T end{bmatrix} \
&=a(uv^T-vu^T) \
&=abegin{bmatrix}
0 & u_1v_2-u_2v_1 & u_1v_3-u_3v_1 \
u_2v_1-u_1v_2 & 0 & u_2v_3-u_3v_2 \
u_3v_1-u_1v_3 & u_3v_2-u_2v_3 & 0end{bmatrix} \
&=a(v times u)^{land}end{align} $$
So above derivation leads to
$$t^{land} = a(v times u)^{land}$$that is
$$t = a*(-w) = |t|(-w)$$
but it contradicts with $t = |t|w$, who can help?
answered Dec 7 '18 at 15:35
FinleyFinley
375213
$endgroup$
add a comment |
$begingroup$
Update
let $w=t/|t|$, then we take three orthonormal basis which form a orthonormal matrix $U = [u,v,w]$.Therefore
$$begin{align} Udiag(aZ,0)U^T&=[u,v,w]begin{bmatrix}
0 & a & 0 \
-a & 0 & 0 \
0 & 0 & 0end{bmatrix}begin{bmatrix} u^T \ v^T \ w^T end{bmatrix} \
&=a(uv^T-vu^T) \
&=abegin{bmatrix}
0 & u_1v_2-u_2v_1 & u_1v_3-u_3v_1 \
u_2v_1-u_1v_2 & 0 & u_2v_3-u_3v_2 \
u_3v_1-u_1v_3 & u_3v_2-u_2v_3 & 0end{bmatrix} \
&=a(v times u)^{land}end{align} $$
So above derivation leads to
$$t^{land} = a(v times u)^{land}$$that is
$$t = a*(-w) = |t|(-w)$$
but it contradicts with $t = |t|w$, who can help?
answered Dec 7 '18 at 15:35
FinleyFinley
375213
$endgroup$
Update
let $w=t/|t|$, then we take three orthonormal basis which form a orthonormal matrix $U = [u,v,w]$.Therefore
$$begin{align} Udiag(aZ,0)U^T&=[u,v,w]begin{bmatrix}
0 & a & 0 \
-a & 0 & 0 \
0 & 0 & 0end{bmatrix}begin{bmatrix} u^T \ v^T \ w^T end{bmatrix} \
&=a(uv^T-vu^T) \
&=abegin{bmatrix}
0 & u_1v_2-u_2v_1 & u_1v_3-u_3v_1 \
u_2v_1-u_1v_2 & 0 & u_2v_3-u_3v_2 \
u_3v_1-u_1v_3 & u_3v_2-u_2v_3 & 0end{bmatrix} \
&=a(v times u)^{land}end{align} $$
So above derivation leads to
$$t^{land} = a(v times u)^{land}$$that is
$$t = a*(-w) = |t|(-w)$$
but it contradicts with $t = |t|w$, who can help?
answered Dec 7 '18 at 15:35
FinleyFinley
375213
answered Dec 7 '18 at 15:35
FinleyFinley
375213
answered Dec 7 '18 at 15:35
FinleyFinley
375213
answered Dec 7 '18 at 15:35
FinleyFinley
375213
375213
add a comment |
add a comment |
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1
$begingroup$
Extend $w=t/|t|_2$ to an orthonormal basis ${u,v,w}$. Then take $U=pmatrix{u,v,w}$.
$endgroup$
– user1551
Dec 7 '18 at 9:40
$begingroup$
@user1551 hello, i apply your thread(see following post of mine) while leads to a contradiction, can you help me out?
$endgroup$
– Finley
Dec 7 '18 at 15:37
$begingroup$
I didn't noticed that you need $a>0$. If you want to enforce this constraint and you find that with your chosen $u,v$ the cross product $vtimes u$ is $-w$ rather than $w$, you may simply interchange $u$ and $v$. Alternatively, replace $v$ by $-v$.
$endgroup$
– user1551
Dec 7 '18 at 16:39
$begingroup$
@user1551 But $det(U) = -1$ once I interchange $u text{ and } v$ or negate $v$. Is it reasonable for an orthonormal matrix?
$endgroup$
– Finley
Dec 8 '18 at 2:11
$begingroup$
Why not? Your question doesn't require that $det(U)=1$. Anyway, since it sets $Z$ to $pmatrix{0&1\ -1&0}$ rather than $pmatrix{0&-1\ 1&0}$, the determinant of $U$ has to be $-1$: when $u$ is orthogonal to $t$, the effect of $ttimes u$ in the tangential direction is an anticlockwise rotation of $u$ about the directed axis $t$. However, $Z=pmatrix{0&1\ -1&0}$ represents a clockwise rotation. So, in the change-of-basis matrix $U$, you must incorporate a reflection to preserve the sign, meaning that $det(U)$ has to be $-1$.
$endgroup$
– user1551
Dec 8 '18 at 6:07